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## Kiss those Math Headaches GOODBYE!
### How to Find the GCF for Three or More Numbers
To find the GCF for three or more numbers, follow these steps:
1) Determine which of the given numbers is smallest, then find the smallest difference between any pair of numbers.
2) See what is smaller: the smallest number, or the smallest difference. Whichever one is smallest, that number is the GPGCF (Greatest Possible GCF). That means that this is the biggest number that the GCF could possibly be. Or, more formally we would say: The GCF, if it exists, must be less than or equal to the GPGCF.
3) Check if the GPGCF itself goes into all of the given numbers. If so, then it is the GCF. If not, list the factors of the GPGCF from largest to the smallest and test them until you find the largest one that does divide evenly into the given numbers. The first factor (i.e., the largest factor) that divides evenly into the given numbers is, by definition, the GCF.
EXAMPLE:
Problem: Find the GCF for 18, 30, 54.
1) Note that the smallest number is 18, and the smallest difference between the pairs is 12 [54 – 30 = 24; 54 – 18 = 36; 30 – 18 = 12] .
2) Of those four quantities (the smallest number and the three differences), 12 is the least. This means that the
GPGCF = 12.
3) Check if 12 divides evenly into the three given numbers: 18, 30 and 54. In fact, 12 doesn’t divide evenly into ANY of these numbers. Next we check the factors of 12, in order from largest to smallest. Those factors are: 6, 4, 3 and 2. The first of those that divides evenly into all three numbers is 6. [18 ÷ 6 = 3; 30 ÷ 6 = 5; 54 ÷ 6 = 9]. So the GCF = 6. And we are done.
MORE CHALLENGING PROBLEM:
Find the GCF for 24, 148, 200.
1) Note that the smallest number is 24, and that the smallest difference between the pairs is 52 [200 – 148 = 52; 200 – 24 = 176; 148 – 24 = 124] .
2) Of those four quantities (the smallest number and the three differences), 24 is the least. This means that for this problem, the GPGCF = 24.
3) Check if 24 divides evenly into the three given numbers: 24, 148 and 200. While 24 does divide evenly into 24, it does not divide evenly into 148 or 200. So next we check the factors of 24, in order from largest to smallest. Those factors are: 12, 8, 6, 4, 3 and 2. The first of those that divides evenly into the three given numbers is 4. [24 ÷ 4 = 6; 148 ÷ 4 = 37; 200 ÷ 4 = 50]. So the GCF = 4. And, once again, we are done.
The process may seem a bit long, but once you get used to it and start doing it in your mind, not on paper, you should find that it actually is quite fast. And you’ll find yourself figuring out the GCF for three or more numbers all in your mind — with no need for pencil and paper — while everyone around you will be making prime factor trees or using calculators. And surely that is a good feeling.
Josh Rappaport is the author of five books on math, including the Parents Choice-award winning Algebra Survival Guide. If you like how Josh explains these problems, you’ll certainly like the Algebra Survival Guide and companion Workbook, both of which are available on Amazon.com Just click the links in the sidebar for more information!
### From GPGCF to GCF … in two easy steps
Once you know the GPGCF, you’re two easy steps from finding the GCF.
[If you don’t know how, see my last post: “Recent Insight on the GCF (and GPGCF)”] That’s one of the benefits of finding the GPGCF — speed in getting the GCF! Here are the short, sweet steps:
1) Find all factors of the GPGCF, and list them from largest to smallest.
2) Starting with the largest factor and working your way down the list, test to find the first factor that goes into both numbers. The first (largest) to do so is the GCF. You can bet on it!
Example 1 (Easy): Find GCF for 30 and 42.
1st) GPGCF is 12. Factors of 12, greatest to least, are 12, 6, 4, 3 and 2.
2nd) Largest factor to go evenly into 30 and 42 is 6. So 6 is GCF.
Example 2 (Harder): Find GCF for 72 and 120.
1st) GPGCF is 48. Factors of 48, greatest to least, are 48, 24, 16, 12, 8, 6, 4, 3 and 2.
2nd) Largest factor to go evenly into 72 and 120 is 24. So 24 is GCF.
NOW TRY THESE —
For each pair:
1) Find GPGCF and say if it is the difference or smaller #.
2) List factors of GPGCF, greatest to least.
3) Find GCF.
a) 8 and 12
b) 16 and 40
c) 18 and 63
d) 56 and 140
a) 8 and 12
GPGCF = 4 (difference)
Factors of 4: 4 and 2
GCF = 4
b) 16 and 40
GPGCF = 16 (smaller #)
Factors of 16: 16, 8, 4 and 2
GCF = 8
c) 18 and 66
GPGCF = 18 (smaller #)
Factors of 18: 18, 9, 6, 3 and 2
GCF = 6
d) 56 and 76
GPGCF = 20 (difference)
Factors of 20: 20, 10, 5, 4 and 2
GCF = 4
### Recent insight on the GCF (and GPGCF)
A while back I wrote a post about the GCF, and mentioned that there’s a number related to it — a number that I call the GPGCF. “GPGCF” stands for the “Greatest Possible Greatest Common Factor.”
In short, the GPGCF is a number that sets an upper limit for the size of the GCF. I’ve seen many students struggle when searching for the GCF, seeking hither and yon for it. I had a sense that students were checking numbers that were too large. That’s what led me to try to figure out what must the the upper limit for the GCF.
If you check out that post (10/25/10), you’ll see that, for any two numbers, I said that the difference between those numbers has to be the GPGCF.
And I was correct, to a degree.
But I recently realized that my little theory needs modifying.
For while the difference between any two numbers can be the upper limit for the GCF, that difference is not the only quantity that can set an upper limit for the GCF. There’s another quantity that plays a role.
That other quantity, I recently realized, is the size of the smaller of the two numbers.
Take the numbers 8 and 24, for example.
The difference between these two numbers is 16, so I would have said that 16 is the upper limit for the GCF. But there’s actually another quantity that limits the size of the GCF, and that quantity is 8. For since the GCF of 8 and 24 must by definition fit into both 8 and 24, it must fit into 8. And common sense tells us that there’s no number larger than 8 that can fit into 8! So the size of this number — the smaller of the two numbers — also sets an upper limit for the size of the GCF.
So my revised theory about the GPGCF is this: when you need to find the GCF for any two numbers, look at two quantities: 1) the smaller of the two numbers, and 2) the difference between the two numbers. Both of these quantities constrains the size of the GPGCF. So therefore, whichever of these is smaller IS the GPGCF. Once you’ve found the GPGCF, that makes it easier to find the actual GCF.
I know this sounds very abstract, so let’s look at a few examples to see what I’m blabbering on about.
Example 1: What’s the GPGCF for 6 and 16?
Smaller number is 6; difference is 10.
6 and 10 both limit the size of the GCF, but
6 is less than 10, so 6 is the GPGCF.
Example 2: What’s the GPGCF for 8 and 12?
Smaller number is 8; difference is 4.
4 is less than 8, so 4 is the GPGCF.
Example 3: What’s the GPGCF for 30 and 75?
Smaller number is 30; difference is 45.
30 is less than 45, so 30 is the GPGCF.
Example 4: What’s the GPGCF for 28 and 42?
Smaller number is 28; difference is 14.
14 is less than 28, so 14 is the GPGCF.
Now, let’s go one step further. From here, how do we figure out the GCF? I’ve done a bit more thinking about this, too, and I’ll share those ideas in my next post. |
# How to Calculate the Volume of a Sphere
A sphere is a perfectly round geometrical object that is three dimensional, with every point on its surface equidistant from its center. Many commonly-used objects such as balls or globes are spheres. If you want to calculate the volume of a sphere, you just have to find its radius and plug it into a simple formula, V = ⁴⁄₃πr³.
## StepsEdit
1. 1
Write down the equation for calculating the volume of a sphere. This is the equation: V = ⁴⁄₃πr³. In this equation, "V" represents volume and "r" represents the radius of the sphere.
2. 2
Find the radius. If you're given the radius, then you can move on to the next step. If you're given the diameter, then you can just divide it by two to get the radius. Once you know what it is, write it down. Let's say the radius we're working with is 1 inch (2.5 cm).[1]
• If you're only given the surface area of the sphere, then you can find the radius by finding the square root of the surface area divided by 4π. In that case, r = root (surface area/4π)
3. 3
Cube the radius. To cube the radius, simply multiply it by itself thrice, or raise it to the third power. For example, 1 inch3 is really just 1 inch (2.5 cm) x 1 inch (2.5 cm) x 1 inch (2.5 cm). The result of 1 inch3 is really just 1, since 1 multiplied by itself any number of times will be 1. You'll reintroduce the unit of measurement, inches, when you state your final answer. After you've done this, you can plug the cubed radius into the original equation for calculating the volume of a sphere, V = ⁴⁄₃πr³. Therefore, V = ⁴⁄₃π x 1
• If the radius was 2 inches (5.1 cm), for example, then to cube it, you would find 23, which is 2 x 2 x 2, or 8.
4. 4
Multiply the cubed radius by 4/3. Now that you've plugged r3, or 1, into the equation, you can multiply this result by 4/3 to continue plugging in to the equation, V = ⁴⁄₃πr³. 4/3 x 1 = 4/3. Now, the equation will read V = ⁴⁄₃ x π x 1, or V = ⁴⁄₃π.
5. 5
Multiply the equation by π. This is the last step to finding the volume of a sphere. You can leave π as it is, stating the final answer as V = ⁴⁄₃π. Or, you can plug π into your calculator and multiply its value by 4/3. The value of π (approximately 3.14159) x 4/3 = 4.1887, which can be rounded to 4.19. Don't forget to state your units of measurement and to state the result in cubic units. The volume of a sphere with the radius of 1 is 4.19 in.3
## Community Q&A
Search
• Why is the formula not V = pi x R squared x H?
Answered by Diante Watts
• Height is not included when measuring spheres, since usually they are congruent in all directions; it wouldn't be a necessity, because the radius is included.
Thanks! 27 10
• How do I find the volume with only the circumference given?
Answered by wikiHow Contributor
• Use the circumference formula (C=2πr) and rearrange it to solve for the radius (r=C/2π). Once you have the radius (r), you can use that value in the volume of a sphere formula (V = 4/3πr³).
Thanks! 40 20
• How can I tell the volume with only the diameter given?
Answered by wikiHow Contributor
• Divide the diameter by 2, giving you the radius and continue from there.
Thanks! 45 28
• How do I find the diameter with only the volume given?
• Use the volume formula to solve for the radius, then double the radius.
Thanks! 9 3
• What is the volume of a sphere if the radius is 6?
Answered by Smile Happy
• Use the formula V= 4/3(3.14)(6^3). 6*6*6 = 216, V= 4/3*3.14*216, V=4*3.14*216= 2,712.96, 2,712.96/3 = 904.32. So the volume would be 904.32.
Thanks! 24 13
• I need to know the volume of a cylindrical water tank with ellipsoidal bases.
Answered by wikiHow Contributor
• V=pi *a*b*h where "a" and "b" are the semi-major and semi-minor axes of the ellipse, and "h" is the height of the cylinder.
Thanks! 25 15
• Between a sphere 32 mm diameter and a sphere 18 mm diameter, what is percentage difference?
• If you're asking only about the diameters, divide 18 by 32. That gives you a decimal number. Multiply it by 100 to get what percentage 18 is of 32. If you want to compare the two volumes, you'll have to cube each diameter and then perform the same division operation.
Thanks! 9 5
• How can I calculate volume from a diameter?
• Divide by 2, then take that radius and plug it into the volume formula.
Thanks! 12 8
• If one sphere is 1000 times larger in diameter than another, how many times larger is its volume?
• One billion times larger.
Thanks! 9 6
• How would I multiply 4/3 by pi or the radius?
• 4/3 πr³ means to multiply 4 by π (3.14), then multiply by the cube of the radius, then divide by 3.
Thanks! 5 3
• I need to know how to determine yards of concrete to do my side walk
200 characters left
## TipsEdit
• Don't forget to use cubed units (e.g. 31 ft³ ).
• Make sure your measurements are all in the same unit. If they aren't, you will need to convert them.
• Note that the "*" symbol is used as a multiplication sign to avoid confusion with the variable "x".
• If you only need part of a sphere, like half or a quarter, find the full volume first, then multiply by the fraction you want to find. For instance, to find the volume of half a sphere with a volume 8, you would multiply 8 by one half or divide 8 by 2 to get 4.
## Things You'll NeedEdit
• Calculator (reason: to calculate problems that would be annoying to do without it)
• Pencil and paper (not needed if you have an advanced calculator) |
NCERT Solutions for Class 10 Maths Chapter 6 Triangles are provided here with simple step-by-step explanations. These solutions for Triangles are extremely popular among Class 10 students for Maths Triangles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the NCERT Book of Class 10 Maths Chapter 6 are provided here for you for free. You will also love the ad-free experience on Meritnation’s NCERT Solutions. All NCERT Solutions for class Class 10 Maths are prepared by experts and are 100% accurate.
#### Question 1:
Choose the correct answer from the given four options:
In the Figure, ∠BAC = 90° and AD ⊥ BC. Then,
(A) BD · CD = BC2
(B) AB · AC = BC2
(C) BD · CD = AD2
(D) AB · AC = AD2
It is given that,
∠D = ∠D = 90° (∵ AD ⊥ BC)
∠DBA = ∠DAC [each angle equals to 90° $-$ ∠C ].
$⇒$$\frac{\mathrm{BD}}{\mathrm{AD}}=\frac{\mathrm{AD}}{\mathrm{CD}}$
$⇒$ BD.CD = AD2
Hence, the correct answer is option C.
#### Question 2:
Choose the correct answer from the given four options:
The lengths of the diagonals of a rhombus are 16 cm and 12 cm. Then, the length of the side of the rhombus is
(A) 9 cm
(B) 10 cm
(C) 8cm
(D) 20 cm
We know that,
A rhombus is a simple quadrilateral whose four sides are of same length and diagonals are perpendicular bisector of each other.
The figure is:
Given: AC = 16 cm and BD = 12 cm
∠AOB = 90°
∵ AC and BD bisects each other
$\mathrm{AO}=\frac{1}{2}\mathrm{AC}$ and $\mathrm{BO}=\frac{1}{2}\mathrm{BD}$
$⇒$ AO = 8 cm and BO = 6 cm
Now, in right angled ∆AOB,
As per by Pythagoras theorem,
$⇒{\mathrm{AB}}^{2}={\mathrm{AO}}^{2}+{\mathrm{OB}}^{2}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{AB}}^{2}={8}^{2}+{6}^{2}=100$
∴ AB = √100 = 10 cm
As we know that a rhombus has all four sides equal.
$⇒$Side of rhombus = 10 cm.
Hence, the correct answer is option B.
#### Question 3:
Choose the correct answer from the given four options:
If ∆ABC ~ ∆EDF and ∆ABC is not similar to ∆DEF, then which of the following is not true?
(A) BC · EF = AC · FD
(B) AB · EF = AC · DE
(C) BC · DE = AB · EF
(D) BC · DE = AB · FD
Given: ∆ABC ∼ ∆EDF
$⇒\frac{\mathrm{AB}}{\mathrm{ED}}=\frac{\mathrm{BC}}{\mathrm{DF}}=\frac{\mathrm{AC}}{\mathrm{EF}}$ (By similarity property)
Taking first two terms, we get
$\frac{\mathrm{AB}}{\mathrm{ED}}=\frac{\mathrm{BC}}{\mathrm{DF}}$
⇒ AB.DF = ED.BC
So, option (D) is true
Taking last two terms, we get
$\frac{\mathrm{BC}}{\mathrm{DF}}=\frac{\mathrm{AC}}{\mathrm{EF}}$
⇒ BC.EF = AC.DF
So, option (A) is true.
Taking first and last terms, we get
$\frac{\mathrm{AB}}{\mathrm{ED}}=\frac{\mathrm{AC}}{\mathrm{EF}}$
⇒ AB.EF = ED.AC
Hence, option (B) is true.
As, all the options are true except C.
Hence, the correct answer is option(C).
#### Question 4:
Choose the correct answer from the given four options:
If in two triangles ABC and PQR, $\frac{\mathrm{AB}}{\mathrm{QR}}=\frac{\mathrm{BC}}{\mathrm{PR}}=\frac{\mathrm{CA}}{\mathrm{PQ}},$ then
(A) ∆ PQR ~ ∆ CAB
(B) ∆ PQR ~ ∆ ABC
(C) ∆ CAB ~ ∆ PQR
(D) ∆ BCA ~ ∆ PQR
In ∆ABC and ∆PQR.
$\frac{\mathrm{AB}}{\mathrm{QR}}=\frac{\mathrm{BC}}{\mathrm{PR}}=\frac{\mathrm{CA}}{\mathrm{PQ}},$
AB/QR = BC/PR = CA/PQ
We know that if sides of one triangle are proportional to the side of the other triangle, and then their corresponding angles are also equal, so by SSS similarity, both the triangles are also similar.
So, by SSS Similarity
We have, ∆CAB $~$ ∆PQR
Hence, the correct answer is option (C).
#### Question 5:
Choose the correct answer from the given four options:
In Fig. 6.3, two line segments AC and BD intersect each other at the point P such that PA = 6 cm, PB = 3 cm, PC = 2.5 cm, PD = 5 cm, ∠APB = 50° and ∠CDP = 30°. Then, ∠PBA is equal to
(A) 50°
(B) 30°
(C) 60°
(D) 100°
In ∆APB and ∆CPD,
∠APB = ∠CPD = 50° [vertically opposite angles]
$\frac{\mathrm{AP}}{\mathrm{PD}}=\frac{6}{5}$ .....(1)
$\frac{\mathrm{BP}}{\mathrm{CP}}=\frac{3}{2.5}=\frac{6}{5}$ .....(2)
From (1) and (2)
$\frac{\mathrm{AP}}{\mathrm{PD}}=\frac{\mathrm{BP}}{\mathrm{CP}}$
∴ ∆APB ∼ ∆DPC [by SAS similarity criterion]
∴ ∠A = ∠D = 30° [corresponding angles of similar triangles]
In ∆APB,
∠A + ∠B + ∠APB = 180° [Sum of angles of a triangle = 180°]
⇒ 30° + ∠B + 50° = 180°
∴ ∠B = 180° $-$ (50° + 30°)
∠B = 180 – 80° = 100°
∠PBA = 100°
Hence, the correct answer is option D.
#### Question 6:
Choose the correct answer from the given four options:
If in two triangles DEF and PQR, ∠D = ∠Q and ∠R = ∠E, then which of the following is not true?
(A) $\frac{\mathrm{EF}}{\mathrm{PR}}=\frac{\mathrm{DF}}{\mathrm{PQ}}$
(B) $\frac{\mathrm{DE}}{\mathrm{PQ}}=\frac{\mathrm{EF}}{\mathrm{RP}}$
(C) $\frac{\mathrm{DE}}{\mathrm{QR}}=\frac{\mathrm{DF}}{\mathrm{PQ}}$
(D) $\frac{\mathrm{EF}}{\mathrm{RP}}=\frac{\mathrm{DE}}{\mathrm{QR}}$
Given: ∆DEF and ∆PQR,
∠D = ∠Q,
∠R = ∠E
∴ ∆DEF ∼ ∆QRP [ By AA similarity criteria]
$⇒\frac{\mathrm{DE}}{\mathrm{QR}}=\frac{\mathrm{DF}}{\mathrm{PQ}}=\frac{\mathrm{FE}}{\mathrm{PR}}$ [ Corresponding sides of similar triangles are in same ratio]
From, the above ratio we can see that all options are true except B.
Hence, the correct answer is option B.
#### Question 7:
Choose the correct answer from the given four options:
In triangles ABC and DEF, ∠B = ∠E, ∠F = ∠C and AB = 3 DE. Then, the two triangles are
(A) congruent but not similar
(B) similar but not congruent
(C) neither congruent nor similar
(D) congruent as well as similar
In ∆ABC and ∆DEF,
∠B = ∠E,
∠C = ∠F,
∆ABC $~$∆DEF [by AA criteria]
As, AB = 3DE
$\therefore$ AB $\ne$ DE
For, the triangles to be congruent AB has to be equal to DE.
Hence, the correct answer is option B.
#### Question 8:
Choose the correct answer from the given four options:
It is given that ∆ABC ~ ∆PQR, with $\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{1}{3}.$ Then, is equal to
(A) 9
(B) 3
(C) $\frac{1}{3}$
(D) $\frac{1}{9}$
Given that ∆ABC ~ ∆PQR, with $\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{1}{3}.$
By Similar Triangle’s Area property, the ratio of the areas of two similar triangles is equal to square of the ratio of their corresponding sides.
$⇒\frac{\mathrm{ar}\left(∆\mathrm{PQR}\right)}{\mathrm{ar}\left(∆\mathrm{ABC}\right)}=\frac{{\left(\mathrm{QR}\right)}^{2}}{{\left(\mathrm{BC}\right)}^{2}}$
$⇒\frac{\mathrm{ar}\left(∆\mathrm{PQR}\right)}{\mathrm{ar}\left(∆\mathrm{ABC}\right)}=\frac{{\left(3\right)}^{2}}{{\left(1\right)}^{2}}=\frac{9}{1}$
Hence, the correct answer is option A.
#### Question 9:
Choose the correct answer from the given four options:
It is given that ∆ABC ~ ∆DFE, ∠A = 30°, ∠C = 50°, AB = 5 cm, AC = 8 cm and DF = 7.5 cm. Then, the following is true:
(A) DE = 12 cm, ∠F = 50°
(B) DE = 12cm, ∠F = 100°
(C) EF = 12cm, ∠D = 100°
(D) EF = 12 cm, ∠D = 30°
Given: ∆ABC ∼ ∆DFE,
As ∆ABC ∼ ∆DFE so, by the property of Corresponding angles of similar triangles,
We have,
∠A = ∠D = 30° (Corresponding angles)
∠C = ∠E = 50° (Corresponding angles)
We know that sum of angles of a triangle = 180°
So,
∴ ∠B = ∠F = 180° $-$ (30° + 50°) = 100°
Also Given,
AB = 5 cm,
AC = 8 cm and
DF = 7.5 cm
$\therefore \frac{\mathrm{AB}}{\mathrm{DF}}=\frac{\mathrm{AC}}{\mathrm{DE}}$
$⇒\frac{5}{7.5}=\frac{8}{\mathrm{DE}}\phantom{\rule{0ex}{0ex}}$
$\therefore \mathrm{DE}=\frac{8×7.5}{5}=12\mathrm{cm}$
$⇒$DE = 12 cm, and ∠F = 100°
Hence, the correct answer is option B.
#### Question 10:
Choose the correct answer from the given four options:
If in triangles ABC and DEF, $\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{FD}},$ then they will be similar, when
(A) ∠B = ∠E
(B) ∠A = ∠D
(C) ∠B = ∠D
(D) ∠A = ∠F
Given,
In ∆ABC and ∆EDF, $\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{FD}}$
Now, for the triangles to be similar by SAS criteria, the angle between the sides which are proportionate must be equal.
∠B and ∠D are the angles which satisfy the above criteria.
Hence, the correct answer is option C.
#### Question 11:
Choose the correct answer from the given four options:
If ∆ABC ~ ∆QRP, AB = 18 cm and BC = 15 cm, then PR is equal to
(A) 10 cm
(B) 12 cm
(C) $\frac{20}{3}\mathrm{cm}$
(D) 8 cm
Given,
∆ABC ∼ ∆QRP
AB = 18 cm and
BC = 15 cm
By Similar triangles area property, the ratio of area of two similar triangles is equal to the ratio of square of their corresponding sides.
So, we have
$\therefore \frac{\mathrm{ar}\left(∆\mathrm{ABC}\right)}{\mathrm{ar}\left(∆\mathrm{QRP}\right)}=\frac{{\left(\mathrm{BC}\right)}^{2}}{{\left(\mathrm{RP}\right)}^{2}}$ (By similar triangles area property).
But given, $\frac{\mathrm{ar}\left(∆\mathrm{ABC}\right)}{\mathrm{ar}\left(∆\mathrm{QRP}\right)}=\frac{9}{4}$
$⇒\frac{{\left(\mathrm{BC}\right)}^{2}}{{\left(\mathrm{RP}\right)}^{2}}=\frac{9}{4}\phantom{\rule{0ex}{0ex}}⇒{\left(\mathrm{RP}\right)}^{2}=\frac{225×4}{9}=100$
∴ RP = √100 = 10 cm.
Hence, the correct answer is option A.
#### Question 12:
Choose the correct answer from the given four options:
If S is a point on side PQ of a ∆PQR such that PS = QS = RS, then
(A) PR · QR = RS2
(B) QS2 + RS2 = QR2
(C) PR2 + QR2 = PQ2
(D) PS2 + RS2 = PR2
Given,
In ∆PQR
PS = QS = RS .....(1)
In ∆PSR
PS = RS .....[from (1)]
⇒ ∠1 = ∠2 .....(2)
Similarly,
In ∆RSQ,
⇒ ∠3 = ∠4 [Angles opposite to equal sides are equal] .....(3)
[By using (2) and (3)]
Now in,
∆PQR, sum of angles = 180°
⇒ ∠P + ∠Q + ∠R = 180°
⇒ ∠2 + ∠4 + ∠1 + ∠3 = 180°
⇒ ∠1 + ∠3 + ∠1 + ∠3 = 180°
⇒2 (∠1 + ∠3) = 180°
⇒ ∠1 + ∠3 = (180°)/2 = 90°
∴ ∠R = 90°
In ∆PQR, by Pythagoras theorem,
PR2 + QR2 = PQ2
Hence, the correct answer is option B.
#### Question 1:
Is the triangle with sides 25 cm, 5 cm and 24 cm a right triangle? Give reasons for your answer.
Let us suppose, a = 25 cm, b = 5 cm and c = 24 cm
Now, if the triangle is right-angle triangle then a should be hypotenuse.
According to the Pythagoras theorem, in a right-angle triangle the sum of square of two sides must be equal to square of third side.
$⇒$b2 + c2 = (5)2 + (24)2
$⇒$b2 + c2 = 601 $\ne$ (25)2
But, given sides do not make a right triangle because it does not satisfy the property of Pythagoras theorem.
Hence, the triangle with sides 25cm, 5cm and 24cm is not a right triangle.
#### Question 2:
It is given that ∆DEF ~ ∆RPQ. Is it true to say that ∠D = ∠R and ∠F = ∠P? Why?
We know that, in similar triangles corresponding angles are also equal.
So, considering ∆DEF ~ ∆RPQ, we will have following angles equal.
∠D = ∠R,
∠E = ∠P and
∠F = ∠Q
Hence, the given statement is false.
#### Question 3:
A and B are respectively the points on the sides PQ and PR of a triangle PQR such that PQ = 12.5 cm, PA = 5 cm, BR= 6 cm and PB = 4 cm. Is AB || QR?
Given,
PQ = 12.5 cm, PA = 5 cm, BR = 6 cm and PB = 4 cm.
Then,
QA = QP – PA = 12.5 – 5 = 7.5 cm
Now, considering the ratio for the converse of basic proportionality theorem to hold true.
.....(1)
And, .....(2)
From (1) and (2),
$\frac{\mathrm{PA}}{\mathrm{AQ}}=\frac{\mathrm{PB}}{\mathrm{BR}}$
According to the converse of basic proportionality theorem, if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
Hence, AB || QR.
#### Question 4:
In Fig 6.4, BD and CE intersect each other at the point P. Is ∆PBC ~ ∆PDE? Why?
In ∆PBC and ∆PDE,
∠BPC = ∠EPD [vertically opposite angles]
Now, .....(1)
And, .....(2)
From (1) and (2), we get
$\frac{\mathrm{PB}}{\mathrm{PD}}=\frac{\mathrm{PC}}{\mathrm{CE}}$
Since, ∠BPC of ∆PBC is equal to ∠EPD of ∆PDE and the sides including these are proportional.
Hence, ∆PBC ∼ ∆PDE by SAS similarity criteria.
#### Question 5:
In triangles PQR and MST, ∠P = 55°, ∠Q = 25°, ∠M = 100° and ∠S = 25°. Is ∆QPR ~ ∆TSM? Why?
We know that,
The sum of three angles of a triangle is 180°.
In ∆PQR,
∠P + ∠Q + ∠R = 180°
⇒ 55° + 25° + ∠R = 180°
⇒ ∠R = 180° $-$ (55° + 25°) = 180° $-$ 80° = 100°
Similarly, in ∆TSM,
∠T + ∠S + ∠M = 180°
⇒ ∠T + ∠25° + 100° = 180°
⇒ ∠T = 180° $-$ (∠25° + 100°)
⇒ ∠T = 180° $-$ 125° = 55°
In ∆PQR and ∆TSM,
∠P = ∠T,
∠Q = ∠S
∠R = ∠M
So, ∆PQR ∼ ∆TSM ( by AAA criteria of similarity )
Hence, ∆QPR is not similar to ∆TSM as ∆PQR is similar to ∆TSM.
#### Question 6:
Is the following statement true? Why?
“Two quadrilaterals are similar, if their corresponding angles are equal”.
Two quadrilaterals are similar if their corresponding angles are equal and the ratio of corresponding sides are proportional.
Having corresponding angles equal isn't enough to say that the quadrilaterals would be similar.
Hence, the given statement is false.
#### Question 7:
Two sides and the perimeter of one triangle are respectively three times the corresponding sides and the perimeter of the other triangle. Are the two triangles similar? Why?
Let the sides of first triangle be x, y and z.
As per given information, sides of second triangle will be 3x, 3y and z'.
Since, the perimeter of other triangle is three times the perimeter of first triangle.
$⇒$3x + 3y + z' = 3(x + y + z)
$⇒$3x + 3y + z' = 3x + 3y + 3z
$⇒$z' = 3z
Here, the corresponding three sides of triangle are in proportion.
Hence, the two traingles are similar by SSS criteria of similarity.
#### Question 8:
If in two right triangles, one of the acute angles of one triangle is equal to an acute angle of the other triangle, can you say that the two triangles will be similar? Why?
Let two right angled triangles be, ∆LMO and ∆RST
In which,
∠L = ∠R [both 90°]
∠M = ∠S [Acute angle equal]
Hence, by AA similarity criteria, we have ∆LMO ∼ ∆RST.
Hence, we can say that the two triangles will be similar.
#### Question 9:
The ratio of the corresponding altitudes of two similar triangles is $\frac{3}{5}.$ Is it correct to say that ratio of their areas is $\frac{6}{5}$? Why?
Given: The ratio of altitudes of similar triangles is $\frac{3}{5}$
By the property of area of two similar triangles,
We have, the ratio of the area of two similar triangles equal to the square of the ratio of their altitudes.
$\left(\frac{{\mathrm{Area}}_{1}}{{\mathrm{Area}}_{2}}\right)={\left(\frac{{\mathrm{Altitude}}_{1}}{{\mathrm{Altitude}}_{2}}\right)}^{2}$
$⇒\frac{9}{25}\ne \frac{6}{5}\phantom{\rule{0ex}{0ex}}$
So, given statement is not correct because the ratio of their areas is equal to $\frac{9}{25}$.
#### Question 10:
D is a point on side QR of ∆PQR such that PD ⊥ QR. Will it be correct to say that ∆PQD ~ ∆RPD? Why?
In ∆PQD and ∆RPD,
PD = PD [common side]
∠PDQ = ∠PDR [both 90$°$
Here, neither other sides are proportional nor angles are equal, so we can say that ∆PQD is not similar to ∆RPD.
#### Question 11:
In Fig. 6.5, if ∠D = ∠C, then is it true that ∆ADE ~ ∆ACB? Why?
∠D = ∠C [given]
∠A = ∠A [common angle]
As we know sum of all the angles of a triangle is equals to 180$°$.
So, by angle sum property of triangle the third angle of both triangles must be equal.
∠E = ∠B
∴ ∆ADE ∼ ∆ACB [by AAA similarity criterion]
Hence, the given statement is true.
#### Question 12:
Is it true to say that if in two triangles, an angle of one triangle is equal to an angle of another triangle and two sides of one triangle are proportional to the two sides of the other triangle, then the triangles are similar? Give reasons for your answer.
We know that by SAS similarity criteria, if one angle of a triangle is equal to one angle of the other triangle and the sides including these are proportional, then the two triangles are similar.
Here, one angle of a triangle is equal to one angle of the other triangle but the sides which are proportional are not the ones necessarily including equal angle.
Hence, the given statement is not correct.
#### Question 1:
In a ∆PQR, PR2 – PQ2 = QR2 and M is a point on side PR such that QM ⊥ PR. Prove that
QM2 = PM × MR.
Given,
In a ∆PQR, PR2 – PQ2 = QR2
⇒ PR= PQ2 + QR2
So, ∆PQR is right angled triangle at Q. (By converse of Pythgoras Theorem)
In ∆QMR and ∆PMQ,
∠M = ∠M (Both angles are equal to 90°)
∠MQR = ∠QPM (each equal to 90°$-$∠R)
∴ ∆QMR ∼ ∆PMQ (by AA similarity criteria)
Now, using property of area of similar triangles, we get,
.....(1)
.....(2)
From (1) and (2)
Hence proved.
#### Question 2:
Find the value of x for which DE || AB in Fig. 6.8.
Given, DE || AB
$⇒\frac{x+3}{3x+19}=\frac{x}{3x+4}$
⇒ (x + 3) (3x + 4) = x (3x + 19)
⇒ 3x2 + 4x + 9x + 12 = 3x2 + 19x
⇒ 19x $-$ 13x = 12
⇒ 6x = 12
x = 2
Hence, the required value of x is 2.
#### Question 3:
In Fig. 6.9, if ∠1 = ∠2 and ∆NSQ ≅ ∆MTR, then prove that ∆PTS ~ ∆PRQ.
Given: ∆ NSQ ≅ ∆MTR
∠1 = ∠2
Since,
∆NSQ = ∆MTR
Now,
SQ = TR (By c.p.c.t. as ∆ NSQ ≅ ∆MTR) .....(1)
PT = PS (∠1 = ∠2 ,since sides opposite to equal angles are also equal) .....(2)
From (1) and (2)
$\frac{\mathrm{PS}}{\mathrm{SQ}}=\frac{\mathrm{PT}}{\mathrm{TR}}$
We get, ST || QR [ By converse of basic proportionality theorem ]
∴ ∠1 = PQR
∠2 = ∠PRQ
In ∆PTS and ∆PRQ.
∠P = ∠P (Common angles)
∠1 = ∠PQR (proved)
∠2 = ∠PRQ (proved)
∴ ∆PTS ~ ∆PRQ (By AAA similarity criteria)
Hence proved.
#### Question 4:
Diagonals of a trapezium PQRS intersect each other at the point O, PQ || RS and PQ = 3 RS. Find the ratio of the areas of triangles POQ and ROS.
Given,
PQRS is a trapezium in which PQ‖RS and PQ = 3RS
In ∆POQ and ∆ROS,
∠SOR = ∠QOP (vertically opposite angles)
∠SRP = ∠RPQ (alternate angles)
∴ ∆POQ ∼ ∆ROS (by AA similarity criterion)
By property of area of similar triangle, the ratio of areas of triangle is equal to the square of the ratio of corresponding sides.
$\frac{\mathrm{ar}\left(∆\mathrm{POQ}\right)}{\mathrm{ar}\left(∆\mathrm{SOR}\right)}=\frac{{\left(\mathrm{PQ}\right)}^{2}}{{\left(\mathrm{RS}\right)}^{2}}=\frac{{3}^{2}}{{1}^{2}}=\frac{9}{1}$
Hence, the required ratio is 9 : 1.
#### Question 5:
In Fig. 6.10, if AB || DC and AC and PQ intersect each other at the point O, prove that OA. CQ = OC. AP.
Given,
AC and PQ intersect each other at the point O and AB‖DC.
In ∆AOP and ∆COQ,
∠AOP = ∠COQ (vertically opposite angles)
∠APO = ∠CQO (since, AB‖DC and PQ is transversal, so alternate angles)
∴ ∆AOP ∼ ∆COQ (by AA similarity criterion)
Then,$\frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{AP}}{\mathrm{CQ}}$ (since, corresponding sides are proportional of similar triangle)
⇒ OA × CQ = OC × AP
Hence Proved.
#### Question 6:
Find the altitude of an equilateral triangle of side 8 cm.
Let LMN be an equilateral triangle of side 8 cm.
LM = MN = NL = 8 cm (all sides of an equilateral triangle is equal)
Draw a line LD which is perpendicular to MN.
Since, in equilateral triangle altitude and median are same,
$\therefore$ D is the mid-point of MN.
∴ MD = ND = $\frac{1}{2}$MN
$⇒$MD = $\frac{8}{2}$ = 4 cm
Now, by Pythagoras theorem
LM2 = LD2 + MD2
⇒ (8)2 = LD+ (4)2
⇒ 64 = LD2 + 16
⇒ LD = √48 = 4√3 cm.
Hence, altitude of an equilateral triangle is 4√3 cm.
#### Question 7:
If ∆ABC ~ ∆DEF, AB = 4 cm, DE = 6cm, EF = 9cm and FD = 12 cm, find the perimeter of ∆ABC.
Given,
AB = 4 cm,
DE = 6 cm
EF = 9 cm and
FD = 12 cm
Also,
∆ABC ∼ ∆DEF
The sides of the two similar triangles are in same proportion,
$\therefore \frac{\mathrm{AB}}{\mathrm{ED}}=\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{\mathrm{AC}}{\mathrm{DF}}$
$⇒\frac{4}{6}=\frac{\mathrm{BC}}{9}=\frac{\mathrm{AC}}{12}$
By taking first two terms, we have
$\frac{4}{6}=\frac{\mathrm{BC}}{9}$
And by taking last two terms, we have
$\frac{\mathrm{BC}}{9}=\frac{\mathrm{AC}}{12}$
Now,
Perimeter of ∆ABC = AB + BC + AC
= 4 + 6 + 8 = 18 cm
Thus, the perimeter of ∆ABC is 18 cm.
#### Question 8:
In Fig. 6.11, if DE || BC, find the ratio of ar (ADE) and ar (DECB).
Given,
DE || BC,
DE = 6 cm and BC = 12 cm
∠ACB = ∠AED [corresponding angle]
∠A = ∠A [common ]
∴ ∆ABC ∼ ∆AED [by AAA similarity criterion]
Then,
$\frac{\mathrm{ar}\left(∆\mathrm{ADE}\right)}{\mathrm{ar}\left(∆\mathrm{ABC}\right)}=\frac{{\left(1\right)}^{2}}{{\left(2\right)}^{2}}=\frac{1}{4}$ [By property of area of similar triangle, ratio of area of two triangles is equal to the square of the ratio of their sides.]
Then area (∆ABC) = 4k
Now,
Area (DECB) = area (ABC) – area (ADE) = 4kk = 3k
∴ Required ratio = area (ADE) : area (DECB) = k : 3k = 1 : 3
Therefore, area (ADE) : area (DECB) = 1 : 3
#### Question 9:
ABCD is a trapezium in which AB || DC and P and Q are points on AD and BC, respectively such that PQ || DC. If PD = 18 cm, BQ = 35 cm and QC = 15 cm, find AD.
Given,
A trapezium, ABCD in which AB ‖ DC, P and Q are Points on AD and BC respectively,
Such that PQ || DC.
Thus,
AB || PQ || DC.
In ∆ABD,
PO || AB (∵ PQ || AB)
(By basic proportionality theorem) .....(1)
In ∆BDC,
OQ || DC (∵ PQ || DC )
(By basic proportionality theorem)
$⇒\frac{\mathrm{QC}}{\mathrm{BQ}}=\frac{\mathrm{DO}}{\mathrm{OB}}$ .....(2)
From (1) and (2)
$\frac{\mathrm{DP}}{\mathrm{AP}}=\frac{\mathrm{QC}}{\mathrm{BQ}}$
$⇒\frac{18}{\mathrm{AP}}=\frac{15}{35}$
$⇒\mathrm{AP}=\frac{\left(18×35\right)}{15}=42$
∴ AD = AP + DP
AD = 42 + 18 = 60
#### Question 10:
Corresponding sides of two similar triangles are in the ratio of 2 : 3. If the area of the smaller triangle is 48 cm2, find the area of the larger triangle.
Given.
Ratio of corresponding sides of two similar triangles = 2 : 3
Area of smaller triangle = 48 cm2
By the property of area of two similar triangle,
Ratio of area of two triangles = (Ratio of their corresponding sides)2
$\frac{\mathrm{ar}\left(\mathrm{smaller}∆\right)}{\mathrm{ar}\left(\mathrm{larger}∆\right)}={\left(\frac{2}{3}\right)}^{2}$
Hence, the area of the larger triangle is 108 cm.
#### Question 11:
In a triangle PQR, N is a point on PR such that QN ⊥ PR. If PN. NR = QN2, prove that ∠PQR = 90°.
In ∆PQR, N is a point on PR, such that QN ⊥ PR and PN.NR = QN2
⇒ PN.NR = QN.QN
.....(1)
$⇒\frac{\mathrm{PN}}{\mathrm{QN}}=\frac{\mathrm{QN}}{\mathrm{NR}}$ [from (1)]
∠PNQ = ∠RNQ (both equal 90$°$)
∴ ∆QNP ∼ ∆RNQ (by SAS criteria of similarity)
Then, corresponding angles ∆QNP and ∆RNQ are equal.
∠PQN = ∠QRN .....(2)
∠RQN = ∠QPN .....(3)
Adding (2) and (3), we get
∠PQN + ∠RQN = ∠QRN + ∠QPN
⇒ ∠PQR = ∠QRN + ∠QPN .....(4)
We know that, sum of all angles of a triangle = 180°
In ∆PQR,
∠PQR + ∠QPR + ∠QRP = 180°
⇒∠PQR + ∠QPN + ∠QRN = 180° (∵ ∠QPR = ∠QPN and ∠QRP = ∠QRN)
⇒∠PQR + ∠PQR = 180° [using (4) ]
⇒ 2∠PQR = 180°
∴ ∠PQR = 90°
Hence Proved.
#### Question 12:
Areas of two similar triangles are 36 cm2 and 100 cm2. If the length of a side of the larger triangle is 20 cm, find the length of the corresponding side of the smaller triangle.
Area of smaller triangle = 36 cm2
Area of larger triangle = 100 cm2
Also, length of a side of the larger triangle = 20 cm
Let length of the corresponding side of the smaller triangle = x cm
By property of area of similar triangle,
$⇒{x}^{2}=\frac{{\left(20\right)}^{2}×36}{100}$
∴ x = √144 = 12 cm
Hence, the length of corresponding side of the smaller triangle is 12 cm.
#### Question 13:
In Fig. 6.12, if ∠ACB = ∠CDA, AC = 8 cm and AD = 3 cm, find BD.
Given, AC = 8 cm, AD = 3 cm and ∠ACB = ∠CDA
From figure,
∠CDA = 90°
∴ ∠ACB = ∠CDA = 90°
In right angled ∆ADC, by Pythagoras theorem
⇒ (8)2 = (3)2 + (CD)2
$⇒$CD2 = 64 – 9 = 55
⇒ CD = √55 cm
∠DBC = ∠DCA (each equal to 90° $-$ ∠A)
∴ ∠CDB ∼ ∆ADC (by AA criteria)
Then, corresponding sides of similar triangle are in same ratio
⇒ CD2 = AD × BD
$\therefore \mathrm{BD}=\frac{{\left(\mathrm{CD}\right)}^{2}}{\mathrm{AD}}=\frac{{\left(\sqrt{55}\right)}^{2}}{3}$
Hence, the value of BD is $\frac{55}{3}$ cm.
#### Question 14:
A 15 metres high tower casts a shadow 24 metres long at a certain time and at the same time, a telephone pole casts a shadow 16 metres long. Find the height of the telephone pole.
QR = 15 m (height of tower)
PQ = 24 m (shadow of tower)
At that time ∠RPQ = θ
Again, let YZ = h be a telephone pole and its shadow XY = 16 m.
At the same time ∠YXZ = θ
Here, ∆PQR and ∆XYZ both are right angles triangles.
In ∆PQR and ∆XYZ,
∠RPQ = ∠YXZ (each θ)
∠Q = ∠Y (each 90°)
∴ ∆PQR ∼ ∆XYZ (by AA similarity criterion)
Then, corresponding sides of the two triangles are proportionate.
$\frac{\mathrm{PQ}}{\mathrm{XY}}=\frac{\mathrm{QR}}{\mathrm{YZ}}$
$⇒\frac{24}{16}=\frac{15}{h}$
Hence, the height of the telephone pole is 10 m.
#### Question 15:
Foot of a 10 m long ladder leaning against a vertical wall is 6 m away from the base of the wall. Find the height of the point on the wall where the top of the ladder reaches.
Let PQ be a vertical wall and PR = 10 m is a ladder.
The top of the ladder reaches to P and distance of ladder from the base of the wall QR is 6 m.
In right angled ∆PQR,
PR2 = PQ2 + QR2 (by Pythagoras theorem)
⇒ (10)2 = PQ2 + (6)2
⇒ 100 = PQ2 + 36
⇒ PQ2 = 100 – 36 = 64
∴ PQ = √64 = 8 m
Hence, the height of the point on the wall where the top of the ladder reaches is 8 m.
#### Question 1:
In Fig. 6.16, if ∠A = ∠C, AB = 6 cm, BP = 15 cm, AP = 12 cm and CP = 4 cm, then find the lengths of PD and CD.
Given, ∠A = ∠C,
AB = 6 cm, BP = 15 cm,
AP = 12 cm
CP = 4 cm
In ∆APB and ∆CPD,
∠A = ∠C (given)
∠APB = ∠CPD (vertically opposite angles)
∆APD ∼ ∆CPD (∴ By AA similarity criteria)
The ratio of their corresponding sides will be same
$⇒\frac{12}{4}=\frac{15}{\mathrm{PD}}=\frac{6}{\mathrm{CD}}$
By taking first two terms,
$\frac{12}{4}=\frac{15}{\mathrm{PD}}$
By taking first and last terms,
$⇒\mathrm{CD}=\frac{\left(6×4\right)}{12}=2\mathrm{cm}$
Hence, length of PD = 5 cm and length of CD = 2 cm.
#### Question 2:
It is given that ∆ABC ~ ∆EDF such that AB = 5 cm, AC = 7 cm, DF = 15 cm and DE = 12 cm. Find the lengths of the remaining sides of the triangles.
Given, ∆ABC ∼ ∆EDF,
So the corresponding sides of ∆ABC and ∆EDF are in the same ratio. (Property of similar triangle)
$\frac{\mathrm{AB}}{\mathrm{ED}}=\frac{\mathrm{AC}}{\mathrm{EF}}=\frac{\mathrm{BC}}{\mathrm{DF}}$ .....(1)
Also, AB = 5 cm, AC = 7 cm, DF = 15 cm and DE = 12 cm
Substituting these values in (1), we get
$\frac{5}{12}=\frac{7}{\mathrm{EF}}=\frac{\mathrm{BC}}{15}$
On taking first and second terms, we get
$⇒\frac{5}{12}=\frac{7}{\mathrm{EF}}$
On taking first and third terms, we get
$⇒\frac{5}{12}=\frac{\mathrm{BC}}{15}$
Hence, lengths of the remaining sides of the triangles are EF = 16.8 cm and BC = 6.25 cm.
#### Question 3:
Prove that if a line is drawn parallel to one side of a triangle to intersect the other two sides, then the two sides are divided in the same ratio.
Let a ∆ABC in which a line DE parallel to BC intersects AB at D and AC at E.
To prove : DE divides the two sides in the same ratio.
$\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}$
Construction join BE, CD and draw EF ⊥ AB and DG ⊥ AC.
$\frac{\mathrm{ar}\left(∆\mathrm{ADE}\right)}{\mathrm{ar}\left(∆\mathrm{BDE}\right)}=\frac{\frac{1}{2}×\mathrm{AD}×\mathrm{EF}}{\frac{1}{2}×\mathrm{DB}×\mathrm{EF}}$ ( area of triangle = $\frac{1}{2}$× base × height)
.....(1)
Similarly,
$\frac{\mathrm{ar}\left(∆\mathrm{ADE}\right)}{\mathrm{ar}\left(∆\mathrm{DEC}\right)}=\frac{\frac{1}{2}×\mathrm{AE}×\mathrm{GD}}{\frac{1}{2}×\mathrm{EC}×\mathrm{GD}}$ ( area of triangle = $\frac{1}{2}$× base × height)
.....(2)
Since, ∆BDE and ∆DEC lie between the same parallel DE and BC and on the same base DE.
So, area(∆BDE) = area(∆DEC) .....(3)
From (1), (2) and (3),
$\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}$
Hence proved.
#### Question 4:
In Fig 6.17, if PQRS is a parallelogram and AB || PS, then prove that OC || SR.
Given, PQRS is a parallelogram,
So, PQ || SR and PS || QR.
Also, AB || PS.
To prove: OC || SR
Proof:
In ∆OPS and ∆OAB,
PS$\parallel$AB
∠POS = ∠AOB (common angle)
∠OSP = ∠OBA (corresponding angles)
∴ ∆OPS ∼ ∆OAB (by AA similarity criteria)
Then,
$\frac{\mathrm{PS}}{\mathrm{AB}}=\frac{\mathrm{OS}}{\mathrm{OB}}$ (corresponding sides of similar triangle have same ratio) .....(1)
In ∆CQR and ∆CAB,
QR || PS || AB
∠QCR = ∠ACB (common angle)
∠CRQ = ∠CBA (corresponding angles)
∴ ∆CQR ∼ ∆CAB (by AA similarity criteria)
$⇒\frac{\mathrm{QR}}{\mathrm{AB}}=\frac{\mathrm{CR}}{\mathrm{CB}}\phantom{\rule{0ex}{0ex}}$
(PS = QR Since, PQRS is a parallelogram)
From (1) and (2),
$\frac{\mathrm{OS}}{\mathrm{OB}}=\frac{\mathrm{CR}}{\mathrm{CB}}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{OB}}{\mathrm{OS}}=\frac{\mathrm{CB}}{\mathrm{CR}}$
Subtracting 1 from both sides, we get
$\frac{\mathrm{OB}}{\mathrm{OS}}-1=\frac{\mathrm{CB}}{\mathrm{CR}}-1\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{OB}-\mathrm{OS}}{\mathrm{OS}}=\frac{\mathrm{CB}-\mathrm{CR}}{\mathrm{CR}}$
$⇒\frac{\mathrm{BS}}{\mathrm{OS}}=\frac{\mathrm{BR}}{\mathrm{CR}}$
By converse of basic proportionality theorem in $∆$OBC, SR || OC
Hence proved.
#### Question 5:
A 5 m long ladder is placed leaning towards a vertical wall such that it reaches the wall at a point 4 m high. If the foot of the ladder is moved 1.6 m towards the wall, then find the distance by which the top of the ladder would slide upwards on the wall.
Let AC be the ladder of length 5 m, BC = 4 m be the height of the wall, against which ladder is placed.
In right angled ∆ABC,
EC2 = AB2 + BC2 (by Pythagoras theorem)
⇒ (5)= (AB)2 + (4)2
⇒ AB = 3
⇒ BD = AB $-$ AD
⇒ BD = 3 $-$ 1.6 = 1.4
In right angled ∆EBD,
ED2 = EB2 + BD2 (by Pythagoras theorem)
⇒ (5)= (EB)2 + (1.4)2 ( BD = 1.4)
⇒ 25 = (EB)2 + 1.96
⇒ (EB)2 = 25 – 1.96 = 23.04
⇒ EB = √23.04 = 4.8
Now, EC = EB – BC = 4.8 – 4 = 0.8
Hence, the top of the ladder would slide upwards on the wall by distance 0.8 m.
#### Question 6:
For going to a city B from city A, there is a route via city C such that AC⊥CB, AC =2x km and CB = 2 (x + 7) km. It is proposed to construct a 26 km highway which directly connects the two cities A and B. Find how much distance will be saved in reaching city B from city A after the construction of the highway.
Given,
AC⊥CB, AC = 2x km, CB = 2 (x + 7) km and AB = 26 km
On drawing the figure, we get the right angled ∆ ACB right angled at C.
Now, in ∆ACB,
AB2 = AC2 + BC2 (by Pythagoras theorem)
⇒ (26)2 = (2x)2 + {2(x + 7)}2
⇒ 676 = 4x2 + 4(x2 + 49 + 14x)
⇒ 676 = 4x2 + 4x2 + 196 + 56x
⇒ 676 = 8x2 + 56x + 196
⇒ 8x2 + 56x – 480 = 0
On dividing by 8, we get
x2 + 7x – 60 = 0
⇒ x2 + 12x $-$ 5x $-$ 60 = 0
x(x + 12) $-$ 5(x + 12) = 0
⇒ (x + 12)(x $-$ 5) = 0
x$-$12, x = 5
As the distance can’t be negative.
x = 5 (∵ x$-$12)
Now, AC = 2x = 10 km
And BC = 2(x + 7) = 2(5 + 7) = 24 km
The distance covered to each city B from city A via city C
= AC + BC
= 10 + 24
= 34 km
Distance covered to reach city B from city A after the construction of the highway = BA = 26 km
Hence, the required saved distance is 34 – 26 = 8 km.
#### Question 7:
A flag pole 18 m high casts a shadow 9.6 m long. Find the distance of the top of the pole from the far end of the shadow.
Let MN = 18 m be the flag pole and its shadow be LM = 9.6 m.
The distance of the top of the pole N from the far end L of the shadow is LN.
In right angled ∆LMN,
LN2 = LM2 + MN2 (by Pythagoras theorem)
⇒ LN2 = (9.6)+ (18)2
⇒ LN2 = 92.16 + 324
⇒ LN2 = 416.16
∴ LN = √416.16 = 20.4 m
Hence, the required distance is 20.4 m.
#### Question 8:
A street light bulb is fixed on a pole 6 m above the level of the street. If a woman of height 1.5 m casts a shadow of 3m, find how far she is away from the base of the pole.
Let P be the position of the street bulb fixed on a pole PQ = 6m.
CD = 1.5 m be the height of a woman
ED = 3 m be the shadow of woman
Let distance between pole and woman be x meter.
Here, woman and pole both are standing vertically.
So, CD || PQ
In ∆CDE and ∆PQE,
∠E = ∠E (common angle)
∠PQE = ∠CDE (each equal to 90°)
∴ ∆CDE ∼ ∆PQE (by AA similarity criterion)
Then,
(corresponding sides of similar triangles are in same ratio)
Hence, she is at the distance of 9 m from the base of the pole.
#### Question 9:
In Fig. 6.18, ABC is a triangle right angled at B and BD⊥AC. If AD = 4 cm, and CD = 5 cm, find BD and AB.
Given, ∆ABC in which ∠B = 90° , BD ⊥ AC, AD = 4 cm and CD = 5 cm
∠ADB = ∠CDB (each equal 90°)
∠BAD = ∠DBC (each equal to 90° $-$ ∠C)
∴ ∆DBA ∼ ∆DCB (by AA similarity criteria)
Then,
$\frac{\mathrm{DB}}{\mathrm{DA}}=\frac{\mathrm{DC}}{\mathrm{DB}}$ (corresponding sides of similar triangle are in proportion)
In right angled ∆BDC,
BC2 = BD2 + CD2 (by Pythagoras theorem)
$⇒$BC2 = (2√5)2 + (5)2
$⇒$BC2 = 20 + 25 = 45
$⇒$ BC = √45 = 3√5 cm
Again,
Hence, BD = 2√5 cm and AB = 6 cm.
#### Question 10:
In Fig. 6.19, PQR is a right triangle right angled at Q and QS⊥PR. If PQ = 6 cm and PS = 4 cm, find QS, RS and QR.
Given, ∆PQR in which ∠Q = 90°, QS ⊥ PR, PQ = 6cm, PS = 4cm
In ∆SQP and ∆SRQ,
∠PSQ = ∠RSQ (each equal to 90°)
∠SPQ = ∠SQR (each equal to 90° $-$ ∠R)
∴ ∆SQP ∼ ∆SRQ
Then,
$\frac{\mathrm{SQ}}{\mathrm{PS}}=\frac{\mathrm{SR}}{\mathrm{SQ}}$
$⇒{\mathrm{SQ}}^{2}=\mathrm{PS}×\mathrm{SR}$ .....(1)
In right angled ∆PSQ,
PQ2 = PS2 + QS2 (by Pythagoras theorem)
$⇒$ (6)2 = (4)2 + QS2
$⇒$ 36 = 16 + QS2
$⇒$ QS2 = 36 $-$ 16 = 20
∴ QS = √20 = 2√5 cm
On putting the value of QS in (1), we get
In right angled ∆QSR,
QR2 = QS2 + SR2
$⇒$QR2 = (2√5)2 + (5)2
$⇒$ QR2 = 20 + 25
∴ QR = √45 = 3√5cm
Hence, QS = 2√5 cm, RS = 5 cm and QR = 3√5 cm.
#### Question 11:
In ∆PQR, PD⊥QR such that D lies on QR. If PQ = a, PR = b, QD = c and DR = d, prove that (a + b) (ab) = (c + d) (cd).
Given, In ∆PQR, PD⊥QR, PQ = a, PR = b, QD = c and DR = d
In right angled ∆PDQ,
PQ2 = PD2 + QD2 (by Pythagoras theorem)
$⇒$ a2 = PD2 + c2
$⇒$ PD2 = a– c2 .....(1)
In right angled ∆PDR,
PR2 = PD2 + DR2 (by Pythagoras theorem)
$⇒$b2 = PD2 + d2
$⇒$PD2 = b2 $-$ d2 .....(2)
From (1) and (2),
a2 $-$ c2 = b2 $-$ d2
a2 $-$ b2 = c2 $-$ d2
$⇒$(ab) (a + b) = (cd) (c + d)
Hence proved.
#### Question 12:
In a quadrilateral ABCD, ∠A + ∠D = 90°. Prove that AC2 + BD2 = AD2 + BC2
Given, Quadrilateral ABCD, in which ∠A + ∠D = 90°.
Construct AB and CD to meet at E.
Also, join AC and BD.
In ∆AED,
∠A + ∠D = 90° [given]
∴∠E = 180° $-$ (∠A + ∠D) = 90° [∵ sum of angles of a triangle = 180°]
Then, By Pythagoras theorem,
In ∆BEC, by Pythagoras theorem,
BC2 = BE2 + EC2
On adding both equations, we get
AD2 + BC2 = AE2 + DE2 + BE2 + CE2 .....(1)
In ∆AEC, by Pythagoras theorem,
AC2 = AE2 + CE2
And in ∆BED, by Pythagoras theorem,
BD2 = BE2 + DE2
On adding both equations, we get
AC2 + BD2 = AE2 + CE2 + BE2 + DE2 .....(2)
From (1) and (2), we get
AC2 + BD2 = AD2 + BC2
Hence proved.
#### Question 13:
In fig. 6.20, l || m and line segments AB, CD and EF are concurrent at point P.
Prove that $\frac{\mathrm{AE}}{\mathrm{BF}}=\frac{\mathrm{AC}}{\mathrm{BD}}=\frac{\mathrm{CE}}{\mathrm{FD}}.$
Given l || m and line segments AB, CD and EF are concurrent at point P.
To Prove : $\frac{\mathrm{AE}}{\mathrm{BF}}=\frac{\mathrm{AC}}{\mathrm{BD}}=\frac{\mathrm{CE}}{\mathrm{FD}}$
Proof :
In ∆APC and ∆BPD,
∠APC = ∠BPD [vertically opposite angles]
∠PAC = ∠PBD [alternate angles]
∴ ∆APS $-$ ∆BPD [by AA similarity criterion]
Then,
$\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{\mathrm{AC}}{\mathrm{BD}}=\frac{\mathrm{PC}}{\mathrm{PD}}$ .....(1)
In ∆APE and ∆BPF,
∠APE = ∠BPF [vertically opposite angles]
∠PAE = ∠PBF [alternate angles]
∴ ∆APE ∼ ∆BPF [by AA similarity criterion]
Then,
.....(2)
In ∆PEC and ∆PED,
∠EPC = ∠FPD [vertically opposite angles]
∠PCE = ∠PDF [alternate angles]
∴ ∆PEC ∼ ∆PFD [by AA similarity criterion]
Then,
.....(3)
From (1), (2) and (3)
$\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{\mathrm{AC}}{\mathrm{BD}}=\frac{\mathrm{AE}}{\mathrm{BF}}=\frac{\mathrm{PE}}{\mathrm{PF}}=\frac{\mathrm{EC}}{\mathrm{FD}}$
$⇒\frac{\mathrm{AE}}{\mathrm{BF}}=\frac{\mathrm{AC}}{\mathrm{BD}}=\frac{\mathrm{CE}}{\mathrm{FD}}$
Hence Proved.
#### Question 14:
In Fig. 6.21, PA, QB, RC and SD are all perpendiculars to a line l, AB = 6 cm, BC = 9 cm, CD = 12 cm and SP = 36 cm. Find PQ, QR and RS.
Given, AB = 6 cm, BC = 9 cm, CD = 12 cm and SP = 36 cm
Also, PA, QB, RC and SD are all perpendiculars to line l
∴ $\mathrm{PA}\parallel \mathrm{QB}\parallel \mathrm{RC}\parallel \mathrm{SD}$
By basic proportionality theorem,
PA : QB : RS = AB : BC : CD
= 6 : 9 : 12
Let PQ = 6x, QR = 9x and RS = 12x
Since, length of PS = 36 km
∴ PQ + QR + RS = 36
⇒ 6x + 9x + 12x = 36
⇒ 27x = 36
$⇒x=\frac{36}{27}=\frac{4}{3}$
Now,
#### Question 15:
O is the point of intersection of the diagonals AC and BD of a trapezium ABCD with AB || DC. Through O, a line segment PQ is drawn parallel to AB meeting AD in P and BC in Q. Prove that PO = QO.
Given: ABCD is a trapezium, diagonals AC and BD intersect at O.
PQ || AB || DC.
To prove: PO = QO
Proof:
In ∆ABD and ∆POD,
PO || AB [∵ PQ || AB]
∠D = ∠D [common angle]
∠ABD = ∠POD [corresponding angles]
∴ ∆ABD ∼ ∆POD [by AA similarity criterion]
Then,
$\frac{\mathrm{OP}}{\mathrm{AB}}=\frac{\mathrm{PD}}{\mathrm{AD}}$ .....(1)
In ∆ABC and ∆OQC,
OQ || AB [∵ OQ || AB]
∠C = ∠C [common angle]
∠BAC = ∠QOC [corresponding angle]
∴ ∆ABC ∼ ∆OQC [by AA similarity criterion]
Then,
$\frac{\mathrm{OQ}}{\mathrm{AB}}=\frac{\mathrm{QC}}{\mathrm{BC}}$ .....(2)
Now, in ∆ADC, OP || DC
$\therefore \frac{\mathrm{AP}}{\mathrm{PD}}=\frac{\mathrm{OA}}{\mathrm{OC}}$ [by basic proportionality theorem] ....(3)
In ∆ABC, OQ || AB
$\therefore \frac{\mathrm{BQ}}{\mathrm{QC}}=\frac{\mathrm{OA}}{\mathrm{AC}}$ [by basic proportionality theorem] ....(4)
From (3) and (4),
$\frac{\mathrm{AP}}{\mathrm{PD}}=\frac{\mathrm{BQ}}{\mathrm{QC}}$
Adding 1 on both sides, we get,
$\frac{\mathrm{AP}}{\mathrm{PD}}+1=\frac{\mathrm{BQ}}{\mathrm{QC}}+1$
Hence Proved.
#### Question 16:
In Fig. 6.22, line segment DF intersect the side AC of a triangle ABC at the point E such that E is the mid-point of CA and ∠AEF = ∠AFE. Prove that $\frac{\mathrm{BD}}{\mathrm{CD}}=\frac{\mathrm{BF}}{\mathrm{CE}}.$
Given : In ∆ABC, E is the mid-point of CA and ∠AEF = ∠AFE.
To Prove : $\frac{\mathrm{BD}}{\mathrm{CD}}=\frac{\mathrm{BF}}{\mathrm{CE}}.$
Proof :
Construction: Take a point G on AB such that CG ‖ EF.
CE = AE [E is the mid-point of CA] .....(1)
In ∆ACG,
CG ‖ EF and E is mid-point of CA.
So, AF = GF [by mid-point theorem] .....(2)
Also , AE = AF [Sides opposite to equal angles are equal] .....(3)
From (1), (2) and (3)
CE = GF .....(4)
Now, in ∆BDF
CG || DF
$\therefore \frac{\mathrm{BC}}{\mathrm{CD}}=\frac{\mathrm{BG}}{\mathrm{GF}}$ [by Baisc Proportionality Theorem]
$⇒\frac{\mathrm{BC}}{\mathrm{CD}}+1=\frac{\mathrm{BG}}{\mathrm{GF}}+1\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{BC}+\mathrm{CD}}{\mathrm{CD}}=\frac{\mathrm{BG}+\mathrm{GF}}{\mathrm{GF}}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{BD}}{\mathrm{CD}}=\frac{\mathrm{BF}}{\mathrm{GF}}$
$⇒\frac{\mathrm{BD}}{\mathrm{CD}}=\frac{\mathrm{BF}}{\mathrm{CE}}$ [ from (4) ]
Hence Proved.
#### Question 17:
Prove that the area of the semicircle drawn on the hypotenuse of a right angled triangle is equal to the sum of the areas of the semicircles drawn on the other two sides of the triangle.
Let RST be a right triangle at S and RS = y, ST = x.
Three semi-circles are drawn on the sides RS, ST and RT, respectively A1, A2 and A3.
To prove : A3 = A1 + A2
In ∆RST,
RT2 = RS2 + ST2 [by Pythagoras theorem]
$⇒$RT2 = y2 + x2
$⇒\mathrm{RT}=\sqrt{\left({y}^{2}+{x}^{2}\right)}$
We know that,
Area of a semi-circle with radius r = $\frac{\mathrm{\pi }{r}^{2}}{2}$
∴ Area of semi-circle drawn on RT,
${A}_{3}=\left(\frac{\mathrm{\pi }}{2}\right)×{\left(\frac{\mathrm{RT}}{2}\right)}^{2}$
$\left(\frac{\mathrm{\pi }}{2}\right)×{\left(\frac{\sqrt{\left({x}^{2}+{y}^{2}\right)}}{2}\right)}^{2}$
$⇒{A}_{3}=\frac{\mathrm{\pi }\left({\mathrm{x}}^{2}+{\mathrm{y}}^{2}\right)}{8}$ .....(1)
Now, the area of semi-circle drawn on RS,
${A}_{1}=\left(\frac{\mathrm{\pi }}{2}\right)×{\left(\frac{\mathrm{RS}}{2}\right)}^{2}$
$⇒{A}_{1}=\frac{\mathrm{\pi }\left({\mathrm{y}}^{2}\right)}{8}$ .....(2)
And area of semi-circle drawn on ST,
${A}_{2}=\left(\frac{\mathrm{\pi }}{2}\right)×{\left(\frac{\mathrm{ST}}{2}\right)}^{2}$
$⇒{A}_{2}=\frac{\mathrm{\pi }\left({x}^{2}\right)}{8}$ .....(3)
${A}_{{}^{2}}+{A}_{1}=\frac{\mathrm{\pi }\left({x}^{2}+{y}^{2}\right)}{8}={A}_{3}$ [ from(1)]
Hence Proved.
#### Question 18:
Prove that the area of the equilateral triangle drawn on the hypotenuse of a right angled triangle is equal to the sum of the areas of the equilateral triangles drawn on the other two sides of the triangle.
Let a right triangle QPR in which $\angle \mathrm{RPQ}$ is right angle and PR = y, PQ = x.
Three equilateral triangles ∆PER, ∆PFQ and ∆RQD are drawn on the three sides of ∆PQR.
Again, let area of triangles made on PR, PQ and RQ be A1, A2 and A3, respectively.
To prove : A3 = A1 + A2
In ∆RPQ,
QR2 = PR2 + PQ2 [By Pythagoras theorem]
⇒ QR2 = y2 + x2
We know that,
Area of an equilateral triangle = $\frac{\sqrt{3}}{4}×{\left(\mathrm{side}\right)}^{2}$
∴ Area of equilateral ∆PER,
${A}_{1}=\frac{\sqrt{3}}{4}×{y}^{2}$ .....(1)
And area of equilateral ∆PFQ,
${A}_{2}=\frac{\sqrt{3}}{4}×{x}^{2}$ .....(2)
Now, area of equilateral ∆RDQ,
Hence Proved.
View NCERT Solutions for all chapters of Class 10 |
Home » Blog » Prove some formulas using mathematical induction
# Prove some formulas using mathematical induction
Use mathematical induction to prove the following:
1. Proof. The statement is true for since on the left we have and on the right, .
Assume then that the statement is true for . Then,
Hence, if the statement is true for , then it is true for . Thus, since we have shown it is true for , we have the statement is true for all
2. Proof. The statement is true for since on the left we have and on the right, .
Assume then that the statement is true for . Then,
Hence, if the statement is true for , then it is true for . Thus, since we have shown it is true for , we have the statement is true for all
3. Proof. The statement is true for since on the left we have and on the right, .
Assume then that the statement is true for . Then,
Hence, if the statement is true for , then it is true for . Thus, since we have shown it is true for , we have the statement is true for all
4. Proof. The statement is true for since the term on the left is , the term in the middle is , and the term on the right is . Since , the statement is indeed true for .
Assume then that the statement is true for some . Then, the induction hypothesis gives us,
Taking the left inequality first, we have
Thus, if the inequality on the left is true for then it is true for . Since we have shown it is true for , we have that it is true for all .
Now, for the inequality on the right
Thus, if the inequality on the right is true for , it is also true for . Hence, since we have established that it is true for , we have that it is true for all
### 7 comments
1. Oliver Golde says:
Sorry for my lack of understanding, but I don’t see how the answer for (d) makes sense. Part of the way through for the first case, you ‘add positive terms to the right’ without changing the left hand side of the inequality. In the next case you ‘make the left smaller’ without changing the other side of the inequality again. I see that you add the 6k^2 + 4k + 1 in the numerator in the first case and subtract 6k^2 +8k + 3 from the numerator in the second case. What am I missing? Any feedback would be greatly appreciated. And sorry again for not getting it.
• Oliver Golde says:
lmao nevermind
• Scattergories says:
Could u please explain? I find myself pondering on the same questions u mentioned
• Navadeep says:
Adding the terms (6k^2 + 4k + 1)/4 in first case doesn’t affect our inequality.
If 1^3+2^3+…+k^3 < (k^4+4k^3)/4 .
Then 1^3+2^3+…+k^3 will definitely be lesser than a much bigger RHS, i.e., (k^4+4k^3)/4+(6k^2 + 4k + 1)/4.
Inequality holds good even after introducing new terms to RHS of inequality.
Adding the new terms doesn't affect our inequality and after simplification proves 1^3+2^3+…+k^3 < (k+1)^4 / 4.
Similarly for second case even after subtracting (6k^2 +8k + 3)/4 from LHS of our inequality , the inequality still holds since if
(k^4+4k^3+12k^2 + 12k + 4)/4 < 1^3+2^3+…+(k+1)^3
then making LHS even much smaller by subtracting (6k^2 +8k + 3)/4 .
the new result (k^4+4k^3+6k^2 + 4k + 1)/4 will definitely be lesser than 1^3+2^3+…+(k+1)^3 .
Subtracting the new terms doesn't affect our inequality and after simplification proves (k+1)^4 / 4 < 1^3+2^3+…+(k+1)^3
• Oliver Golde says:
lmao nvm
2. Sebastian says:
Dude shouldnt the second inequality from the bottom have a 4k instead of a 6k, the one that has the (making it smaller) sentence. Cause u cant really transform it if thats not the case.
Idk maybe im wrong cause im just learning how to prove things with induction XD
Btw awesome site, i love it :D
• RoRi says:
Yeah, typo, it should be so in the numerator we have . Fixed now. |
### Density CalculationsProblems 11-25
Twenty Examples Probs #1-10 Probs #26-50 All the examples & problems, no solutions Significant Figures Menu
Problem #11: A cylindrical glass tube of length 27.75 cm and the radius 2.00 cm is filled with argon gas. The empty tube weighs 188.250 g. and the tube filled with argon weighs 188.870 g. Use the data to calculate the density of argon gas.
Solution:
1) Volume of a cylinder:
V = πr2h
V = (3.14159) (2.00 cm)2 (27.75 cm)
V = 348.7165 cm3
2) Mass:
188.870 g - 188.250 g = 0.620 g
3) Density:
0.620 g / 0.3487165 L = 1.78 g/L
Note that 348.7165 cm3 became 0.3487165 liters. Gas density is typically measured in g/L as opposed to g/cm3 or g/mL. Reminders: 1 cm3 = 1 mL and 1000 mL = 1 L.
Problem #12: The density of silver is 10.50 g/cm3 and the density of benzene is 0.8786 g/cm3. What mass of silver will have the same volume as 15.55 grams of benzene?
Solution:
1) Determine the volume of benzene:
mass / density = volume
15.55 g / 0.8786 g/cm3 = 17.6986 cm3
2) Determine the mass of silver:
density times volume = mass
(10.50 g/cm3) (17.6986 cm3) = 185.8 g
Problem #13: Calculate the mass of copper in grams (density = 8.96 g/cm3) with the same volume as 100.0 grams of gold (density = 19.31 g/cm3)
Solution:
1) Volume of gold:
100.0 g ÷ 19.31 g/cm3 = 5.17866 cm3
2) Mass of copper:
(8.96 g/cm3) (5.17866 cm3) = 46.4 g
3) Setting up the problem in dimensional analysis style:
1 cm3 8.96 g 100.0 g x ––––––– x ––––––– = 46.4 g 19.31 g 1 cm3
Problem #14: Calculate the mass of zinc in grams (density = 7.14 g/cm3) with the same volume as 100.0 grams of aluminum (density = 2.70 g/cm3)
Solution:
1) Volume of aluminum:
100.0 g ÷ (2.70 g/cm3) = 37.037 cm3
2) Mass of zinc:
(7.14 g/cm3) (37.037 cm3) = 264 g
3) Dimensional analysis:
1 cm3 7.14 g 100.0 g x ––––––– x ––––––– = 264 g 2.70 g 1 cm3
Problem #15: A spherical ball bearing has a radius of 8.50 mm and a mass of 2.315 g. Determine the density of the ball bearing in g/cm3.
Solution:
1) Convert mm to cm:
1 cm 8.50 mm x ––––––– = 0.850 cm 10 mm
2) Determine volume of sphere:
V = (4/3)πr3
(4/3) (3.14159) (0.850 cm)3 = 2.57 cm3
3) Calculate density:
2.315 g / 2.57 cm3 = 0.900 g/cm3
Problem #16: 57.0 kg of copper is drawn into a wire with a diameter of 9.50 mm. What is the length of wire in meters? Cu density = 8.96 g/cm3.
Solution:
1) Convert kg to grams:
1000 g 57.0 kg x ––––––– = 5.70 x 104 g 1 kg
2) Determine volume of the copper wire:
(5.70 x 104 g) ÷ (8.96 g/cm3) = 6361.607 cm3
3) Convert mm to cm:
1 cm 9.50 mm x ––––––– = 0.950 cm 10 mm
3) Determine length of wire:
V = πr2h
6361.607 cm3 = (3.14159) (0.475 cm)2 h
h = 8974.91 cm
To three sig figs and in meters, 89.7 m
Problem #17: In the United States, 'copper' pennies made since 1983 actually contain very little copper. If a penny contains 93.975% of its total volume zinc and 6.025% of its total volume copper, what is its apparent density? (density of Cu = 8.96 g/cm3; density of Zn = 7.14 g/cm3.)
Solution:
1) Assume the penny occupies 1.00 cm3. This means:
copper occupies 0.06025 cm3 and zinc occupies 0.93975 cm3.
2) Calculate mass of copper:
(0.06025 cm3) (8.96 g/cm3) = 0.53984 g
3) Calculate mass of zinc:
(0.93975 cm3) (7.14 g/cm3) = 6.709815 g
4) Determine apparent density:
0.53984 g + 6.709815 g = 7.249655 g
since this mass is in 1.00 cm3, the answer is 7.25 g/cm3
Problem #18: Antarctica has an ice sheet covering 1.42 x 1018 cm2 and averaging 1.61 x 105 cm deep. Calculate the total mass if ice has a density of 0.92 g/cm3.
Solution:
1) Calculate volume of ice:
(1.42 x 1018 cm2) (1.61 x 105 cm) = 2.2862 x 1023 cm3
2) Calculate mass of ice:
(2.2862 x 1023 cm3) (0.92 g/cm3) = 2.1 x 1023 g
Problem #19: Object A is less dense than object B. If both objects are the same mass, what can be said about the volume of A as compared to the volume of B?
Solution:
Object A has a larger volume than Object B.
Problem #20: An ice cube with a volume of 45.0 mL and a density of 0.900 g/cm3 floats in a liquid with a density of 1.36 g/mL. What volume of the cube is submerged in the liquid?
Solution:
The solution to this problem involves the concept of buoyancy.
1) Determine the mass of the cube:
(45.0 mL) (0.900 g/cm3) = 40.5 g
2) The cube will float when 40.5 g of liquid is displaced. We need to know what volume of the liquid weighs 40.5 g.
40.5 g) ÷ (1.36 g/mL) = 29.8 mL
This means that 29.8 mL of the cube is submerged (this is the answer to the question), displacing 40.5 g of the liquid. The rest of the cube (45.0 − 29.8) is above the surface of the liquid.
Problem #21: Copper can be drawn into thin wires. How many meters of 34-gauge wire (diameter = 6.304 x 10¯3 inches) can be produced from the copper that is in 5.88 pounds of covellite, an ore of copper that is 66% copper by mass? (Hint: treat the wire as a cylinder. The density of copper is 8.96 g cm¯3; one kg weighs 2.2046 lb; 1 inch is 2.54 cm and the volume of a cylinder is πr2h)
Solution:
a) Determine pounds of pure copper in 5.88 lbs of covellite:
5.88 lbs x 0.66 = 3.8808 lbs
b) Convert pounds to kilograms:
3.8808 lbs ÷ 2.6046 lbs kg¯1 = 1.489979 kg (I'm keeping a few guard digits)
1.489979 kg = 1489.979 g
c) Determine volume this amount of copper occupies:
8.96 g cm¯3 = 1489.979 g / x
x = 166.292 cm3
Note: this is the volume of the cylinder.
d) Convert the diameter in inches to a radius in centimeters:
dia = 6.304 x 10¯3 in; radius = 3.152 x 10¯3 in
(3.152 x 10¯3 inch) (2.54 cm / inch) = 8.00 x 10¯3 cm
e) Determine h in the volume of a cylinder:
166.292 cm3 = (3.14159) (8.00 x 10¯3 cm)2 h
h = 8.27 x 105 cm = 8.27 x 103 m
Problem #22: A copper ingot has a mass of 2.15 kg. If the copper is drawn into wire whose diameter is 2.27 mm, how many inches of copper wire can be obtained from the ingot?
Solution:
a) Determine volume of copper:
8.96 g cm¯3 = 2150 g / x
x = 239.955 cm3
Note: this is the volume of the wire.
b) Determine h in the volume of a cylinder (i.e., the wire):
239.955 cm3 = (3.14159) (0.1135 cm)2 h
h = 5929.097 cm
Note: 0.1135 cm is the radius
c) Convert cm to inch:
5929.097 cm divided by 2.54 cm/in = 2334.29 in
To three sig figs, the answer is 2330 in
Problem #23: If the copper is drawn into wire whose diameter is 8.06 mm, how many feet of copper can be obtained from a 200.0 pound ingot?
Solution:
a) Convert pounds to grams:
200.0 lb x (453.59 g/lb) = 9.0718 x 104 g
b) Determine what volume is occupied by this many grams of copper:
8.96 g/cm3 = 9.0718 x 104 g divided by x
x = 1.01248 x 104 cm3
c) Determine the height of the cylinder (Volume of a cylinder = πr2h):
1.01248 x 104 cm3 = (3.14159) (0.0403 cm)2 h
h = 1.9844 x 106 cm
d) Convert cm to inches, then to feet:
1.9822 x 106 cm x (1 in/2.54 cm) = 7.8126 x 105 in
7.8126 x 105 in x (1 ft / 12 in) = 6.51 x 104 ft (to three sig figs)
Problem #24: A cube of copper was found to have a mass of 0.630 kg. What are the dimensions of the cube? (The density of copper is 8.96 g/cm3.)
Solution:
a) Determine the volume of the cube (note that kg have been converted to g):
8.96 g/cm3 = 630. g / volume
volume = 70.3125 cm3
b) Each side of a cube is equal in length, so take cube root of the volume for length of cube side:
$\sqrt[3]{\mathrm{70.30125 cm3}}$ = 4.13 cm (to three sig figs)
Problem #25: Calculate the volume (in m3) of a 5,020 tonne iceberg. (1 tonne = 1,000 kg, the density of ice = 0.92 g/cm3)
Solution:
a) Convert tonnes to grams:
5,020 tonne x (1,000 kg / tonne) = 5.02 x 106 kg
5.02 x 106 kg x (1000 g / kg) = 5.02 x 109 g
b) Determine volume in cubic centimeters:
0.92 g/cm3 = 5.02 x 109 g / volume
volume = 5.4565 x 109 cm3
c) Convert cubic centimeters to cubic meters:
5.4565 x 109 cm3 x (1 m3 / 106 cm3) = 5.46 x 103 m3 (rounded to 3 significant figures)
Bonus Problem: A room contains 11.5 kg of argon. The air in the room consists of 0.225% argon. The density of air in the room is 1.70 g/L. What is the volume of the room in m3?
Solution #1:
1) 11.5 kg is 0.225% of the total mass in the room. Determine total mass (in grams) of air in room:
11.5 kg / 0.00225 = 5111 kg
5111 kg x (1000 g/kg) = 5.111 x 106 g
2) Using density, determine volume of room in liters:
5.111 x 106 g ÷ 1.70 g/L = 3.00 x 106 L
3) Convert liters to cubic meters:
3.00 x 106 L x (1 m3 / 1000 L) = 3.00 x 103 m3
Solution #2:
1) Determine the liters that 11.5 kg of argon occupies:
11.5 kg x (1000 g/kg) = 11,500 g
11,500 g / 1.70 g/L = 6.7647 x 103 L
2) The volume the Ar occupies represents 0.225% of the entire room. Determine the total volume of the room:
6.7647 x 103 L / 0.00225 = 3.00 x 106 L
3) Convert liters to cubic meters:
See step (3) in solution #1.
Twenty Examples Probs #1-10 Probs #26-50 All the examples & problems, no solutions Significant Figures Menu |
# Vertex form
Considering the shifting and stretching, the following general equation, also called the vertex form, arises:
$f(x)=a(x-\color{blue}{d})^2+\color{green}{c}$
From this equation one can see the vertex and how the standard parabola is shifted / stretched. The vertex is always: $S(\color{blue}{d}|\color{green}{c})$
!
### Remember
The vertex is always the highest or lowest point of a parabola:
• When the parabola is opened upwards, the vertex is the minimum point.
• When the parabola is opened downwards, the vertex is the maximum point.
Frequently, however, a quadratic equation exists in the general form $f(x)=ax^2+bx+c$. To convert these to the vertex shape, use completing the square.
i
### Method
1. Exclude coefficients before $x^2$
2. Apply completing the square: $f(x)=x^2+px+(\frac{p}{2})^2-(\frac{p}{2})^2$
3. Apply binomial formula
### Example
Change the function $f(x)=2x^2+8x+6$ to the vertex shape.
1. #### Exclude coefficients before $x^2$
$f(x)=2(x^2+4x+3)$
2. #### Apply completing the square
$f(x)=2(x^2+4x\color{red}{+(\frac{4}{2})^2-(\frac{4}{2})^2}+3)$
$f(x)=2(x^2+4x+4-4+3)$
3. #### Apply binomial formula backwards
$f(x)=2((x+2)^2-1)$
$f(x)=2(x+2)^2-2$
4. #### Specify vertex
Note that the sign changes at $d$.
$S(-2|-2)$ |
Let a curve be given by x = 2 t^3 - 3 t^2 + 1, y = 2 t^3 - 9 t^2 + 12 t - 2. a) Find the slope...
Question:
Let a curve be given by {eq}x = 2 t^3 - 3 t^2 + 1, \\ y = 2 t^3 - 9 t^2 + 12 t - 2. {/eq}
a) Find the slope of the tangent line to this curve at the point {eq}(0, 3). {/eq}
b) Find the values of the t where the tangent line is vertical, and then horizontal. Make use of L'Hopital's Rule.
Tangents and Asymptotes of a Parametrically Defined Function
First we are given a curve in parametric form. Using the formula for the derivative of a parametrically defined function we find the slope of the tangent line to the curve at a given point in part (a). Then in part (b), using the derivative function again, we find the value of the parameter t for which we have a vertical and a horizontal tangent line or asymptote. The concepts used are first derivatives and point slope form of the equation of a line from Calculus and Algebra.
(a) First we note that the value of {eq}t=1 {/eq} gives us the point of tangent {eq}(x(1),y(1))=(2-3+1,2-9+12-2)=(0,3). {/eq}
The slope of the tangent at the above point can be found by finding the value of {eq}\displaystyle \frac {dy}{dx} \; when \; t = 1. {/eq}
To that end, we calculate
{eq}\displaystyle \frac {dy}{dx} = \frac {dy/dt}{dx/dt} = \frac {6t^2-18t+12}{6t^2-6t} = \frac {6(t-1)(t-2)}{6t(t-1)} \qquad (1) {/eq}
From (1) the slope of the tangent line at (0, 3) is undefined when t = 1.
(b) The tangent line is horizontal when {eq}\displaystyle \frac {dy}{dx}=0 \implies \frac {dy}{dt}=0. {/eq}
From part (a) this happens when t = 1 and t = 2.
Similarly the tangent line is vertical when {eq}\displaystyle \frac {dy}{dx}=undefined \implies \frac {dx}{dt}=0. {/eq}
From part (a) this happens when t = 0 and t = 1. |
Jill is 6 years younger than Bill. Twelve years ago, Bill was 2 times older than Jill. How old are both of them now?
Matthew Fonda | Certified Educator
To solve this problem, we'll translate the words into equations, then solve the equations.
Let j represent Jill's current age, and let b represent Bill's current age.
1.) "Jill is 6 years less than Bill"
`\implies j = b - 6`
2.) "Twelve years ago, Bill was 2 times older than Jill"
`\implies b - 12 = 2(j - 12)`
We will plug equation (1) into (2):
`\implies b - 12 = 2((b-6) - 12)`
We will now solve for b:
`\implies b - 12 = 2(b - 6) - 24`
`\implies b - 12 = 2b - 12 - 24`
`\implies b = 2b - 24`
`\implies b - 2b = -24`
`\implies b = 24`
Now we know that Bill is 24 years old. We can take this result, and plug it back into (1):
`\implies j = 24 - 6 = 18`
We now have our answer: Bill is 24 years old and Jill is 18 years old.
William Delaney | Certified Educator
I arrived at the same answer without algebra. If Bill was 7 when Jill was 1, then when Jill was 6 Bill would have been 12, which is the only time he could have been twice Jill's age. Add 12 to each figure and you get 18 for Jill and 24 for Bill.
najm1947 | Student
Jill is six years younger than Bill
Let Jill be x years old
Bills age = x+6
12 years ago, Bill was 2 times older than bill
Age of Bill 12 years ago = x+6-12 = x-6
Age of Jill 12 years ago = x-12
As bills age was 2 times that of Jill therefore
x-6 = 2*(x-12)
x-6 = 2x-24
x-2x = -24+6
-x = -18
x = 18
Age of Jill now = 18 years
Age of Bill now = 18+6 = 24 years
The age of Jill and Bill now is 18 and 24 years respectively |
Mechanics 1.3. mc-web-mech1-3-2009 Some problems in mechanics may involve units that are not in the International System of Units (SI). In such cases, it may be necessary to convert from one system to another. In other cases the units being used may be within the same system but have a different prefix. This leaflet illustrates examples of both types of conversions.
Converting from one system of units to another In the U.S. Customary system of units (FPS) is measured in feet (ft), in pounds (lb), and in (s). The unit of , referred to as a , is therefore equal to the amount of −2 matter accelerated at 1 ft s , when acted up on by a force of 1 lb (derived from ’s law of F = ma — see mechanics sheet 2.2) this gives that 1 slug = 1 lb ft−1 . Table 1 provides some direct conversions factors between FPS and SI units.
Quantity FPS Unit of SI Force lb = 4.4482 N Mass slug = 14.5938 kg Length ft = 0.3048 m
Table 1: Conversion factors Converting from one metric unit to another metric unit If you are converting large units to smaller units, i.e. converting to , you will need to MULTIPLY by a conversion factor. For example, to find the of centimetres in 523 m, you will need to MULTIPLY 523 by 100. Thus, 523 m = 52 300 cm. If you are converting small units to larger units, i.e. converting to centimetres, you will need to DIVIDE by a conversion factor. For example, to find the number of centimetres in 846 mm, you will need to DIVIDE by 10. Thus, 846 mm = 84.6 cm. These examples are based on the system of length where 1 m is the SI unit, with there being 100 cm in 1m and 1000 mm in 1m; this implies that there are 10 mm in 1 cm. Worked example Convert the of (3.00 × 108 m s−1) to km −1. Solution Solving this type of problem involves following a succession of conversions from one metric unit to another. The first step is to convert from m s−1 to km s−1 i.e. to change from m to km. This is changing from a smaller unit to a larger one (a is 1000 m), so we need to divide by 1000. www.mathcentre.ac.uk 1 c mathcentre 2009 3.00 × 108 3.00 × 108 m s−1 = km s−1 =3.00 × 105 km s−1. 1000 The next step is to convert s−1 into year−1. A year is a longer (bigger) unit than a second so that s−1 is a bigger unit than year−1. This that we will be multiplying by a conversion factor. There are 3600 seconds in an , 24 in a and 365 days in a year (approx.) so that
1 year = 3600 × 24 × 365 s =3.1536 × 107 s.
Rearranging this will give us that 1 year =1s. 3.1536 × 107
Now taking reciprocals of both sides we get, 3.1536 × 107 year−1 =1 s−1.
So now we can convert to km year−1 by multiplying the result from the previous step with the conversion from s−1 to year−1;
3.00 × 108 m s−1 = (3.00 × 105) × (3.1536 × 107) km year−1 =9.4608 × 1012 km year−1.
So, to three significant figures, the is 9.46 × 1012 km year−1.
Commonly used non-SI units and their conversions 1) The symbol for the is l. The SI unit for is m3 (cubic ). The conversion from m3 to l is 1l =0.001 m3. 2) The cubic , symbol = cm3, where one cm3 equals one millilitre (ml). This unit is often used for measuring the volume of solids.
Exercises
1. Convert the folllowing values to SI units using Table 1 where required: a) 1297 lb, b) 1.6 slug, c) 3.281 feet 2. Convert the following into FPS units using Table 1 where required: a) 3048 m, b) 72 kg, c) 0.8 N 3. Convert 600 ft −1 into m s−1 using the step by step method.
Solutions
1. a) 5769 N. b) 23.35 kg. c) 1 m. 2. a) 10 000 feet. b) 4.9336 slug c) 0.179848 lb. 3. 3.048 m s−1. www.mathcentre.ac.uk 2 c mathcentre 2009 |
# Quick Square
A mathematical trick
Did you know that there is a quick way of squaring a two digit number which ends in 5?
Just multiply the first digit by that number plus one.... stick a 25 after your product and there's your answer .... simple ?
Example:
Q. What is 35 squared?
A. 3x4=12
.....now stick on the 25
Barry's Response - It even seems to work with three digit numbers (using the first two digit to make a two digit number) and plain old 5, using 0 as the first digit.
Oh the handy things we can do with math, games,conversions and tricks.
## Here are some other math shortcuts and tricks:
You can use these shortcuts for mental math and quick calculations.
- A two-digit number can be multiplied by 11 by adding the two digits and putting the sum in the middle. 11 x 24 is 264, for example.
- You can multiply a number by 9 by multiplying it by 10 and subtracting the original number. The answer to 9 x 7 is 70 - 7, which equals 63.
- If you wanna multiply by 5, divide a number by 2 and then add a 0 to the end. For example, 5 x 36 is 36 ÷ 2 = 18, then add a 0 to get 180.
- If you want to square a two-digit number, find the nearest multiple of 10, then calculate the difference between the number and the multiple, square this difference, and add it to the multiple squared. To find 37 squared, find the nearest multiple of 10, which is 40. Calculate 37 - 40 = -3, square -3 to get 9, and add that to 40 squared, which is 1600. Therefore, 37 squared is 1600 + 9 = 1609.
- To multiply by 25, multiply it by 100 (by adding two zeros) and then divide it by 4. To find 25 x 48, calculate 48 x 100 = 4800, then divide it by 4, which equals 1200.
- To multiply a number by 50, multiply it by 100 (by adding two zeros) and then divide the result by 2.
- A number is divisible by 3 if the sum of its digits are divisible by 3. Check with 246.
Calculate 2 + 4 + 6 = 12, which is divisible by 3, so yeah, divisible by 3.
- Same check can be used for divisible by 9.
Average Rating
Rating I really wonder of this Techniques by: Rajesh I thought it would be good and it is really good and when i am going through all the tabs so then i am keep on getting all the tips from this site. I understood that it is very useful can understand anyone easily so there several formulas if you follow but if you follow these you will find answers with in short time. This is very interesting site and i really like to use many times and i got solved all the problems here.
Rating Awesome by: Anonymous Haha, awesome! I didn't know about that trick. I wish we got taught little tricks like that in all my math classes in college.
Rating Helpful hints by: cangel This is great. After all my years of teaching I was not aware of this little help. It just shows there is always something new to learn. And with sites like yours I can keep learning. I looked at the site about metric conversion and I am going to use it in my tutoring. Please keep up the good work. I have added your site to my favorites for browsing in the future. Thank you.
That is what I do on a full-time basis. Find out if it is necessary for your project. |
2.1 Squares, Square Roots, Cube and Cube Roots
2.1 Squares, Square Roots, Cube and Cube Roots
(A) Squares
The square of a number is the answer you get when you multiply a number by itself.
Example:
(a) 13= 13 × 13 = 169
(b) (–10)= (–10) × (–10) = 100
(c) (0.4)2 = 0.4 × 0.4 = 0.16
(d) (–0.06)= (–0.06) × (–0.06) = 0.0036
$\begin{array}{l}\text{(e)}{\left(3\frac{1}{2}\right)}^{2}={\left(\frac{7}{2}\right)}^{2}=\frac{7}{2}×\frac{7}{2}=\frac{49}{4}\\ \left(\text{f}\right)\text{}{\left(-1\frac{2}{7}\right)}^{2}={\left(-\frac{9}{7}\right)}^{2}=\left(-\frac{9}{7}\right)×\left(-\frac{9}{7}\right)=\frac{81}{49}\end{array}$
(B) Perfect Squares
1. Perfect squares are the squares of whole numbers.
2. Perfect squares are formed by multiplying a whole number by itself.
Example:
4 = 2 × 2 9 = 3 × 3 16 = 4 × 4
3. The first twelve perfect squares are:
= 12, 22, 32, 42, 52, 62, 72, 82, 92, 102, 112, 122
= 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144
(C) Square Roots
1. The square root of a positive number is a number multiplied by itself whose product is equal to the given number.
Example:
$\begin{array}{l}\text{(a)}\sqrt{169}=\sqrt{13×13}=13\\ \text{(b)}\sqrt{\frac{25}{64}}=\sqrt{\frac{5×5}{8×8}}=\frac{5}{8}\\ \text{(c)}\sqrt{\frac{72}{98}}=\sqrt{\frac{\overline{)72}36}{\overline{)98}49}}=\sqrt{\frac{6×6}{7×7}}=\frac{6}{7}\\ \text{(d)}\sqrt{3\frac{1}{16}}=\sqrt{\frac{49}{16}}=\frac{7}{4}=1\frac{3}{4}\\ \text{(e)}\sqrt{1.44}=\sqrt{1\frac{\overline{)44}11}{\overline{)100}25}}=\sqrt{\frac{36}{25}}=\frac{6}{5}=1\frac{1}{5}\end{array}$
(D) Cubes
1. The cube of a number is obtained when that number is multiplied by itself twice.
Example:
The cube of 3 is written as
33 = 3 × 3 × 3
= 27
2.
The cube of a negative number is negative.
Example:
(–2)3 = (–2) × (–2) × (–2)
= –8
3. The cube of zero is zero. The cube of one is one, 13 = 1.
(E) Cube Roots
1. The cube root of a number is a number which, when multiplied by itself twice, produces the particular number. $"\sqrt[3]{}"$ is the symbol for cube root.
Example:
$\begin{array}{l}\sqrt[3]{64}=\sqrt[3]{4×4×4}\\ \text{}=4\end{array}$
$\sqrt[3]{64}$ is read as ‘cube root of sixty-four’.
2.
The cube root of a positive number is positive.
Example:
$\begin{array}{l}\sqrt[3]{125}=\sqrt[3]{5×5×5}\\ \text{}=5\end{array}$
3.
The cube root of a negative number is negative.
Example:
$\begin{array}{l}\sqrt[3]{-125}=\sqrt[3]{\left(-5\right)×\left(-5\right)×\left(-5\right)}\\ \text{}=-5\end{array}$
4.
To determine the cube roots of fractions, the fractions should be simplified to numerators and denominators that are cubes of integers.
Example:
$\begin{array}{l}\sqrt[3]{\frac{16}{250}}=\sqrt[3]{\frac{\overline{)16}8}{\overline{)250}125}}\\ \text{}=\sqrt[3]{\frac{8}{125}}\\ \text{}=\frac{2}{5}\end{array}$ |
# How do you solve 2x+3=3x-4?
Jul 19, 2016
With practice you can do these quicker as you learn the shortcut.
$x = 7$
#### Explanation:
Objective is to have a single $x$ on one side of the = and everything else on the other side.
Given:$\text{ } \textcolor{g r e e n}{2 x + 3 = 3 x - 4}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Step 1 ")color(brown)("Having all the "x" terms on one side of =}}$
Subtract $\textcolor{red}{2 x}$ from both sides
$\textcolor{g r e e n}{2 x \textcolor{red}{- 2 x} + 3 = 3 x \textcolor{red}{- 2 x} - 4}$
$0 + 3 = x - 4$
Swap round
$x - 4 = 3$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Step 2 ")color(brown)("Have just "x" on the left of =}}$
Add $\textcolor{red}{4}$ to both sides
$\textcolor{g r e e n}{x - 4 \textcolor{red}{+ 4} = 3 \textcolor{red}{+ 4}}$
$x = 7$ |
# More coloring games.
This is a continuation of yesterday’s blog Coloring Game! If you haven’t done so yet, please take the time to read it.
In the last post we discussed one of the games that is posed in graph theory, that is the four color theorem. For this problem the goal was to color a picture or a map in such a way that parts that shared a border didn’t share a color. In addition to maps, we can also look at coloring of graphs. A graph is collection of lines and points called edges and vertices, respectively. These are drawn in such a was that each edge has a point on each end, so we end up with something that looks like the example below.
In this example, we see that we have 16 edges and 8 vertices. Furthermore, we say that the degree of a vertex is the number of edges touching that vertex. Therefore, the degree of vertex 1 is 3.
Now, we have two games we can play. In the first game, our goal is to color the vertices each with a color so that no edge has the same color for both of its vertices. That is, if we move over a single edge, we should always arrive at a new color. Unlike the in the case of maps, we don’t have a nice theorem that states what the minimum number of colors is going to be. This means that even if we provide a coloring, we won’t know that we’ve used the minimum number of colors unless we look at special properties of the particular graph. As an example, we provide a vertex coloring with 4 colors below. We note that this must be the minimum number of colors used in this case because there are four vertices that all connect to each other.
If you’d like to create vertex coloring, you can draw any arbitrary graph and see what you come up with. Have fun drawing whatever shape you like. Also, in the Games page, I have a link to the game Electric Joint. While it’s set up as a trying to connect positive to negative charges, you are essentially creating a graph that has a 2 coloring.
The second coloring game we can play with graphs is to provide an edge coloring. In this situation, instead of coloring vertices, we color edges so that if any two edges meet at the same vertex, they cannot have the same color. As before, there is no nice way to know the minimum color needed for a general graph, so we have to look at these in a case by case basis. As an example, for the graph above we can have the following edge coloring with 5 colors. Note also that this is the minimum needed since there are vertices with degree 5.
An interesting case arises when we have a graph where every vertex has the same degree, say n, and the graph admits an n coloring. In this case, if we define having the same color as a relation on the edges, it will split the edges into sets of all the same color. Furthermore, each vertex touches exactly one edge of each color. Because of this, we can look at the color as a parallelism relation and treat the graph as a geometry. It is then further possible to take the set of points and the set of lines and define a multiplication on them in such way that we end up with an algebraic structure. We are then able to talk about colorings in a geometric or algebraic way giving us options when trying to find properties. Dr. Francis Pastijn and I actually explore this topic in a paper which is currently under review, so if you’d like more details on this, contact me.
I hope you have some fun with coloring.
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4
Q:
# Ten years ago, Kumar was thrice as old as Sailesh was but 10 years hence, he will be only twice as old. Find Kumar’s present age ?
A) 50 years B) 70 years C) 60 years D) 40 years
Explanation:
Let Kumar’s present age be "x" years and Sailesh’s present age be "y" years.
Then, according to the first condition,
x - 10 = 3(y - 10) => x - 3y = - 20 ........(1)
Now, Kumar’s age after 10 years = (x + 10) years.
Sailesh’s age after 10 years = (y + 10) years.
(x + 10) = 2 (y + 10) => x - 2y = 10 ......(2)
Solving (1) and (2),
we get x = 70 and y = 30
Kumar’s present age = 70 years and Sailesh’s present age = 30 years.
Q:
The age of X is 2/3 rd that of Y. After 6 years X will be 46 years old. the present age of Y is?
A) 40 B) 56 C) 60 D) 100
Explanation:
X= 2/3Y
=> Y= 3/2X
Aage of X after 6 years = 46 years
So present age of X = 46-6 = 40 years
Now, present age of Y = 3/2 * 40 = 60 years.
0 58
Q:
Ishikha got married 9 years ago. Today her age is 4/3 times her age at the time of marriage. At present her daughter’s age is one-sixth of her age. What was her daughter’s age three years ago?
A) 4 years B) 3 years C) 2 years D) 1 year
Explanation:
Let Ishika and her daughter present ages be p and q
From the given data,
p = 4/3(p - 9)
=> p = 36 years
Hence, Ishika's present age = 36 years and
her daughter's present age = 1/6 x 36 = 6 years
Now, required her daughter’s age three years ago = 6 - 3 = 3 years.
0 105
Q:
Manogna's father was 38 years of age when she was born while her mother was 36 years old when her brother, 4 years younger than her, was born. What is the difference between the ages of her parents?
A) 2 years B) 4 years C) 6 years D) 8 years
Explanation:
From the given data,
Manogna’s age be 0 when she was born.
Her father’s age = 38 years
Her mother’s age when Manogna’s brother who is 4 years younger to her was born = 36 years
=> Manogna’s mother age when Manogna was born = 36 - 4 = 32 years
Hence, the difference between the ages of Manogna’s parents = 38 - 32 = 6 years.
0 48
Q:
The difference between the present ages of Avanthi and Kapil is 9 years. After 7 years, Kapil’s age is twice of Avanthi’s age. What will be Kapil’s age after 4 years?
A) 15 B) 16 C) 18 D) 20
Explanation:
Let the present ages of Kapil and Avanthi be K & A
Given difference in their ages = K - A = 9 years ...........(1)
After 7 years, K + 7 = 2(A + 7) ......(2)
K + 7 = 2A + 14
K - 2A = 7 ............(3)
K - A = 9 ....(1)
From (1) & (3),
-A = -2
=> A = 2 years
=> K = 9 + 2 = 11 years
Required, age of Kapil after 4 years = 11 + 4 = 15 years.
0 229
Q:
Karthik is now 32 years old and his son is 7 years old. In how many years will Karthik be twice as old as the son?
A) 18 B) 25 C) 7 D) 15
Explanation:
Given Karthik’s age = 32 years
Karthik’s son age = 7 years
Difference between their ages = K - S = 32 - 7 = 25 years ….(1)
Now, required K = 2S ……(2)
Put (2) in (1)
2S - S = 25
=> S = 25 years
Hence, at the age of 25 years of Karthik’s son, Karthik will be twice as his son’s age.
25 - 7 = 18 years
Therefore, after 18 years the father be twice as old as the son.
1 331
Q:
The ages of Sunitha and Pranitha are 40 years and 60 years, respectively. How many years before the ratio of their ages was 3 : 5?
A) 10 years B) 15 years C) 18 years D) 24 years
Explanation:
Given the ages of Sunitha and Pranitha are 40 years and 60 years.
Let before 'm' years, their ages be in the ratio of 3 : 5
ATQ,
=> 200 - 5m = 180 - 3m
=>2m = 20
=> m = 10 years.
3 488
Q:
5 years ago, the mother's age was 3 times of her son's age. 5 years hence, the ratio between the ages of mother and the son becomes 11:5. Find the mother's present age?
Let the present age's of mother be 'M' and that of her son be 'S'.
According to the given data in the question,
M - 5 = 3(S - 5)
F - 3S = -10 ...............(1)
Then given ,
M + 5 : S + 5 = 11 : 5
=> 5M - 11S = 30 ..............(2)
From (1) & (2), we get
S = 20
Then by substituting S = 20 in (1) or (2), we get
M = 50.
Hence, the present age of mother is M = 50 years.
489
Q:
The age of Sohail, Afzal and Bilal are 17,16 and 12 respectively. If the age of Aslam also included the average of the ages is increased by 5. What is the age of Aslam?
A) 32 B) 33 C) 34 D) 35
Explanation:
Given S = 17 yrs
A = 16 yrs
B = 12 yrs
Average = 17 + 16 + 12/3 = 15 yrs
Given if Aslam added to them, their average increases by 5
=> Average of 4 members = 20
Let the age of Aslam be 'x'
Then,
17 + 16 + 12 + x /4 = 20
=> x = 80 - 45
=> x = 35.
Hence, the age of Aslam is 35 yrs. |
# 2.05 - Expected Value, Variance, and Standard Deviation
Analogous to the discrete case, we can define the expected value, variance, and standard deviation of a continuous random variable. These quantities have the same interpretation as in the discrete setting. The expectation of a random variable is a measure of the center of the distribution, its mean value. The variance and standard deviation are measures of the horizontal spread or dispersion of the random variable.
Expected Value, Variance, and Standard Deviation of a Continuous Random Variable
The expected value (also called the expectation or mean) of a continuous random variable X, with probability density function f(x), is the number given by
${\displaystyle \mathbb {E} (X)=\int _{-\infty }^{\infty }xf(x)dx}$.
The variance of X is:
${\displaystyle {\text{Var}}(X)=\int _{-\infty }^{\infty }{\big (}x-\mathbb {E} (X){\big )}^{2}f(x)dx}$.
As in the discrete case, the standard deviation, σ, is the positive square root of the variance:
${\displaystyle \sigma (X)={\sqrt {{\text{Var}}(X)}}}$.
## Simple Example
A random variable X is given by the following PDF. Check that this is a valid PDF and calculate the standard deviation of X.
${\displaystyle f(x)={\begin{cases}2(1-x)&{\text{if }}0\leq x\leq 1,\\0&{\text{otherwise}}\end{cases}}}$
### Solution
#### Part 1
To verify that f(x) is a valid PDF, we must check that it is everywhere nonnegative and that it integrates to 1.
We see that 2(1-x) = 2 - 2x ≥ 0 precisely when x ≤ 1; thus f(x) is everywhere nonnegative.
To check that f(x) has unit area under its graph, we calculate
{\displaystyle {\begin{aligned}\int _{-\infty }^{\infty }f(x)dx=2\int _{0}^{1}(1-x)dx=2{\Big (}x-{\frac {x^{2}}{2}}{\Big )}{\Big |}_{0}^{1}=1\end{aligned}}}
So f(x) is indeed a valid PDF.
#### Part 2
To calculate the standard deviation of X, we must first find its variance. Calculating the variance of X requires its expected value:
{\displaystyle {\begin{aligned}\mathbb {E} (X)&=\int _{-\infty }^{\infty }xf(x)dx\\&=\int _{0}^{1}x{\Big [}2(1-x){\Big ]}dx\\&=2\int _{0}^{1}{\Big (}x-x^{2}{\Big )}dx\\&=2{\Big (}{\frac {x^{2}}{2}}-{\frac {x^{3}}{3}}{\Big )}{\Big |}_{0}^{1}\\&=1/3\end{aligned}}}
Using this value, we compute the variance of X as follows
{\displaystyle {\begin{aligned}{\text{Var}}(X)&=\int _{-\infty }^{\infty }{\big (}x-\mathbb {E} (X){\big )}^{2}f(x)dx\\&=\int _{0}^{1}{\big (}x-1/3{\big )}^{2}\cdot 2(1-x)dx\\&=2\int _{0}^{1}{\big (}x^{2}-{\frac {2}{3}}x+{\frac {1}{9}}{\big )}(1-x)dx\\&=2\int _{0}^{1}{\big (}-x^{3}+{\frac {5}{3}}x^{2}-{\frac {7}{9}}x+{\frac {1}{9}}{\big )}dx\\&=2{\big (}-{\frac {1}{4}}x^{4}+{\frac {5}{9}}x^{3}-{\frac {7}{18}}x^{2}+{\frac {1}{9}}x{\big )}{\Big |}_{0}^{1}\\&=2{\big (}-{\frac {1}{4}}+{\frac {5}{9}}-{\frac {7}{18}}+{\frac {1}{9}}{\big )}\\&={\frac {1}{18}}\end{aligned}}}
Therefore, the standard deviation of X is
{\displaystyle {\begin{aligned}\sigma &={\sqrt {{\text{Var}}(X)}}\\&={\frac {1}{3{\sqrt {2}}}}\end{aligned}}}
### An Alternative Formula for Variance
There is an alternative formula for the variance of a random variable that is less tedious than the above definition.
Alternate Formula for the Variance of a Continuous Random Variable
The variance of a continuous random variable X with PDF f(x) is the number given by
${\displaystyle {\text{Var}}(X)=\mathbb {E} (X^{2})-[\mathbb {E} (X)]^{2}}$.
The derivation of this formula is a simple manipulation and has been relegated to the exercises. We should note that a completely analogous formula holds for the variance of a discrete random variable, with the integral signs replaced by sums.
## Simple Example Revisited
We can use this alternate formula for variance to find the standard deviation of the random variable X defined above.
Remembering that the expectation of X was found to be 1/3, we compute the variance of X as follows:
{\displaystyle {\begin{aligned}{\text{Var}}(X)&=\mathbb {E} (X^{2})-[\mathbb {E} (X)]^{2}\\&=\int _{-\infty }^{\infty }x^{2}f(x)dx-\left({\frac {1}{3}}\right)^{2}\\&=2\int _{0}^{1}x^{2}(1-x)dx-{\frac {1}{9}}\\&=2\int _{0}^{1}{\big (}x^{2}-x^{3}{\big )}dx-{\frac {1}{9}}\\&=2{\big (}{\frac {1}{3}}x^{3}-{\frac {1}{4}}x^{4}{\big )}{\big |}_{0}^{1}-{\frac {1}{9}}\\&=2{\big (}{\frac {1}{3}}-{\frac {1}{4}}{\big )}-{\frac {1}{9}}\\&={\frac {1}{18}}\end{aligned}}}
In the exercises, you will compute the expectations, variances and standard deviations of many of the random variables we have introduced in this chapter, as well as those of many new ones. |
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# Algebraic Expressions
Welcome to Class !!
In today’s Mathematics class, We will be discussing Algebraic Expressions. We hope you enjoy the class!
Content
• Definition with examples
• Expansion of algebraic expression
• Factorization of simple algebraic expressions
Definition with examples
In algebra, letters stand for numbers. The numbers can be whole or fractional, positive or negative.
Example
Simplify the following
• -5 x 2y
• -3a x -6b
• -14a/7
• -1/3 of 36x2
Solution
1) -5 x 2y = -5 x (+2) x y
= -(5 x 2) x y = -10y
2) -3a x -6b = (-3) x a (-6) x b
= (-3) x (-6) x a x b = 18ab
3)
4)
-1/3 of 36x2 = (+36) x x2 = – (36/3) x x2
(-3)
= -12x2
Self Evaluation
Simplify the following
1. -16x/8
2. (-1/10) of 100z
3. (-2x) x (-9y)
Removing brackets
Example
Remove brackets from the following
a. 8 (2c + 3d) (b) 4y (3x-5) (c) (7a-2b) 3a
Solution
(a) 8(2c+3d) = 8 x 2c + 8 x 3d
= 16c + 24d
(b) 4y(3x-5) = 4y x 3x – 4y x 5
= 12xy – 20y
(c) (7a-2b)3a = 7a x 3a – 2b x 3a
=21a2 – 6ab
Self Evaluation
Remove brackets from the following
1. -5x(11x – 2y)
2. -p(p – 5q)
3. (2c + 8d)(-2)
##### Expanding algebraic expressions
The expression (a+2) (b-5) means (a+2) x (b-5)
The terms in the first bracket, (a+2), multiply each term in the second bracket, b-5.
Example
Expand the following
1. (a+b) (c+d)
2. (6-x) (3+y)
3. (2p-3q) (5p-4)
Solution
(1) (a+b)(c+d) = c(a+b) + d(a+b)
= ac + bc + ad + bd
(2) (6-x)(3+y) = 3(6-x) + y (6-x)
= 18 -3x +6y – xy
(3) (2p-3q)(5p-4) = 5p(2p – 3q)-4(2p-3q)
= 10p2 – 15pq – 8p + 12q
Self Evaluation
Expand the following
(a) (3+d)(2+d)
(b) (3x+4)(x-2)
(c) (2h-k)(3h+2k)
(d) (7m-5n)(5m+3n)
### Factorization of algebraic expression
Example:
Factorize the following
(a) 12y + 8z
(b) 4n2 – 2n
(c) 24pq – 16p2
Solution
(a) 12y +8z
The HCF of 12y and 8z is 4
12y +8z =
= 4(3y + 2z)
(b) 4n2 – 2n
The HCF of 4n2 and 2n is 2n
4n2 – 2n =
= 2n (2n-1)
(c) 24pq – 16p2
The HCF of 24pq and 16p2 is 8p
24pq – 16p2
= 8p(3q – 2p)
Self Evaluation
Factorize the following:
1. 2abx + 7acx (b) 3d2e + 5d2
2. 12ax + 8bx
New General Mathematics, UBE Edition, Chapter 1, pages 20-21
Essential Mathematics by A J S Oluwasanmi, Chapter 1, pages 1-4
WEEKEND ASSIGNMENT
1. Simplify (-6x) x (-x) =_____ (A) 6x (b) 6x2 (C) -6x (D) -6x2
2. Remove brackets from -3(12a – 5) (A) 15-36a (B) 15a-36 (C) 15a + 36 (C) 36a – 15
3. Expand (a+3)(a+4) (A) a2+7a+12 (B) a2+12a+7 (C) a2+12a-7 (D) a2+7a-12
4. Factorize abc + and (A) ab(c+d) (B) ac (b+d) (C) ad(b+c) (D) abc(c+d)
5. Factorize 5a2 + 2ax (A) a(5a+2x) (B) 5(2a2+2x) (C) a(5x+2ax) (D) a2(5+2x)
THEORY
1. Expand the following:
1. (p+2q) (p+3q)
2. (5r+2s) (3r+4s)
1. Factorize the following
• -18fg – 12g
• -5xy + 10y
We have come to the end of this class. We do hope you enjoyed the class?
Should you have any further question, feel free to ask in the comment section below and trust us to respond as soon as possible.
In our next class, we will be talking about Algebraic Fractions. We are very much eager to meet you there.
Get more class notes, videos, homework help, exam practice on Android [DOWNLOAD]
Get more class notes, videos, homework help, exam practice on iPhone [DOWNLOAD]
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# More, Less or Equal? Comparing Numbers 0 to 5
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Print Lesson
## Objective
Students will be able to compare groups and numerals of five or fewer.
#### Big Idea
In this lesson, students are using their new knowledge of the numbers 0 to 5 to make comparisons to determine more, less, or the same.
## Problem of the Day
5 minutes
I start the lesson with a problem of the day to help students review skills and concepts from prior lessons and develop their ability to problem solve. I call the students up to the carpet. The students find their spots while saying this chant with me.
Criss cross, applesauce, hands in your lap, eyes on the teacher, you've got to show me that.
I project the Problem of the Day on the SMARTBoard and say to students, "This is our Problem of the Day for today. Look at the apples. This says 'Count the butterflies. Draw a picture with more than 3 butterflies.'" I say, "This problem has two parts. What is the first thing it asks us to do?" (Count the butterflies.) I have a student come up with a pointer and count the butterflies. "Listen to the direction again. 'Count the butterflies. Draw a picture with more than 3 butterflies.' What do we need to do next?" (Draw a picture with more than 3 butterflies.) I have a student come up and draw a picture of four or more butterflies.
I remind students to keep their pictures simple. The important part is that there are more than 3 butterflies. We do not need to add a lot of details to the pictures. I say this because some students focus on adding details and making their pictures perfect and then they make mistakes with their math. If you don't have a SMARTBoard, you can use the pdf of the slides in a variety of ways to reproduce this activity.
I tell students, "Today we will be comparing the numbers 0 to 5."
## Presentation of Lesson
25 minutes
To start this lesson, I call up five girls to stand in the front of the room. I then call up two boys to stand in a second group. I have the students count the girls with me. I say, "We have 5 girls." I write the number 5 on a small white board and hand it to one of the girls. I have the students count the boys with me. I say, "We have 2 boys." I write the number 2 on a small white board and hand it to one of the boys. I say, "We need to figure out which group has more. Are there more boys or more girls?" (Girls) "We know that there are more girls because 5 is a larger number than 2. When we are looking for more, we are looking for he group that has a larger number of objects." I have the students sit back down, and I pull up Comparing Numbers 1 to 5 on the SMARTBoard. I say, "We are going to count these pictures and compare how many is in each group." I point to the lions and have the students count with me. "There are 5 lions, so I am going to write the number 5 on the line." I point to the frogs and have the students count with me."There are 4 frogs, so I am going to write the number 4 on the line. Which group has more? There are more lions. Since we know there are more lions, we can say that there are fewer frogs." I continue with the next two slides. I have students help count and write the numbers.
I tell students that we will be practicing the comparing numbers on a Comparing Numbers 1 to 5 Worksheet. I show students the paper and say, "We are going to be working on the front of this paper together. You need to get out your pencil and put your name on your paper. When your name is on your paper hold your pencil in the air, that will let me know that you are ready to start." I like to have students hold up their pencils or put their hands on their heads when they are finished with a task. It makes it easy for me to see who is ready and also keeps the students from writing all over their papers while they wait for other students to finish.
I hand each student a paper for them to take back to their seats and while the students are writing their names, I turn on the projector and document camera and display the worksheet on the SMARTBoard. When all students have their pencils up, I say, "The directions on this paper say ‘Draw lines to match the objects in the two groups. Circle the group that has more. Put an X on the group that has less (fewer).'" I model how to do number 1. I count the moons and write the number on the line. I then count the suns and write the number on the line. I draw a line from each moon to a sun. I remind students that we did this in our last unit. I say, "I see that the group of suns has two extra. There are two suns that do not have a partner. That means that there are more suns." I model how to circle the group of suns and put an X on the group of moons. I continue with number two. I have students come up to help count, write the numbers, draw the lines, circle the group with more and X the group with less. I repeat this with number three, and when we get to comparing, I still ask, "Which group has more?" I let the students come up with the answer that the groups are equal. If the students disagree, I allow them to explain their thinking and respond to each others' responses.
When the students are finished, they put their papers into the paper tray in the front of the classroom and get their center.
## Practice
20 minutes
Since the students finish their papers at different times, I circulate through the room to make sure that student are completing their papers, putting it in the tray and getting their centers. This week's centers are:
Play Dough Numbers (K-5mathteachingresources.com)
Pattern Block, Lego, and Bear Count (K-5mathteachingresources.com)
Roll and Count (Makinglearningfun.com)
Number Tracing (I purchased mine, free ones are also available from WorkSheetFun
SMART Board- Online Game Scrambled Egg City (Macmillan/McGraw-Hill)
I quickly circulate to make sure students are engaged and do not have any questions about how to complete the centers. I pull three groups during centers. I pull the first group for 10 minutes and the other two groups for 5 minutes each. The first group is comprised of the students who were having trouble identifying numbers 0-5 and matching the numbers to objects. I pull the students back to my small group table to do a reteach activity using flash cards and manipulatives (for this lesson I used bug counters). I show the flash cards and have students practice identifying the numbers. I then give each student a pile of manipulative (0-5) and have them pick the number card that matches their group. I also preview tomorrow's activity with this group. Tomorrow I plan to have the students play a comparing numbers game. I let this group play it with me for a few minutes to help them prepare for the lesson tomorrow. See example here. Pre-teaching a group of struggling students can help build their confidence. It also gives them a chance to feel successful when working with their peers. The next two groups do a follow up activity that reviews comparing groups of objects. I show them two groups of objects. I have the students count how many is in each group and place a number card under it. I then have them match up the objects and tell which group has more. Prior to clean up, I check in with each table to see how the centers are going. I turn on Tidy Up by Dr. Jean
There are many wonderful transition songs to be found, for free, online if you'd like to use music for transitions too. |
# Decimals and Fractions
## Presentation on theme: "Decimals and Fractions"— Presentation transcript:
Decimals and Fractions
Applied Mathematics Jefferson College ATS
Definitions Decimal - Any number shown with a decimal point; a number based upon tenths or hundredths. ( 0.2, 0.375, 86.4 ) Fraction - One or more of the equal parts of a whole; a number usually expressed in the form a/b. ( 1/3, 2 5/8, 7/4 )
Converting Fractions to Decimals
To convert a fraction to a decimal, divide the numerator by the denominator. Do this until you get a decimal that terminates or repeats. If it repeats, place a bar (--- ) over the first number that repeats. Example 1: Convert 3/4 to a decimal 3/4 = 3÷4= 0.75 ( terminating decimal) Example 2: Convert 2/3 to a decimal 2/3= 2÷3= ( repeating decimal )
Express fractions as percents by long division
You try: Express fractions as percents by long division 87.5 % 0.875 = 125 % 1.25 = 37.5 % .375 =
1) 7/40 2) 7/50 3) 19/10 4)1/4 5)11/10 6)37/25 7)7/20 8)32/25 9)26/20
Converting Decimals to Fractions
In order to convert a decimal to a fraction, place the number that follows the decimal point over the place that it is in ( tenths, hundredths, thousandths, etc.). Then simplify as necessary. Example 1: Convert 0.7 to a fraction 0.7= 7/10 ( 7 is in the tenths place ) Example 2: Convert 0.25 to a fraction 0.25= 25/100= 1/4 ( 25 is in the hundredths place. 1/4 is the simplified form of 25/100) Example 3: Convert to a fraction 1.625= 1 625/1000= 1 5/8 ( 625 is in the thousandths place. 5/8 is the simplified form of 625/1000)
Convert decimal to base 10 fraction; simplify
You try Convert decimal to base 10 fraction; simplify 8 4 0.8 = = 10 5 4 2 1 0.04 = = = 100 25 50 36 18 9 3.36 3 3 3 = = = 100 50 25
1)0.866 = 2)0.428 = 3)0.3 = 4)0.41 = 5)0.992 =
6) = 7) = 8)0.62 = 9) 1.22 = 10)0.19 = |
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Main content
## Integrated math 1
### Course: Integrated math 1>Unit 18
Lesson 1: Distance and midpoints
# Getting ready for analytic geometry
Analytic geometry relates geometric figures to the coordinate plane and algebraic representations. Let's review the coordinate plane, distance and displacement, slope, and a few helpful arithmetic skills to get ready.
Let’s refresh some concepts that will come in handy as you start the analytic geometry unit of the high school geometry course. You’ll see a summary of each concept, along with a sample item, links for more practice, and some info about why you will need the concept for the unit ahead.
There are a lot of sections in this article because analytic geometry pulls together a lot of ideas!
This article only includes concepts from earlier courses. There are also concepts within this high school geometry course that are important to understanding analytic geometry. If you have not yet mastered the Pythagorean theorem lesson, it may be helpful for you to review that before going farther into the unit ahead.
## Points on a coordinate plane
### What is this, and why do we need it?
We use a coordinate plane to show relative position in 2D space. We describe every point on the plane with an ordered pair in the form $\left(x,y\right)$, where $x$ represents the horizontal position, and $y$ represents the vertical position. Points to the left of the
have negative $x$-coordinates, and points to the right have positive $x$-coordinates. Likewise, points below the origin have negative $y$-coordinates, and points above the origin have positive $y$-coordinates.
### Practice
Problem 1
Use the following coordinate plane to write the ordered pair for each point.
PointOrdered pair
$A$$\left($
,
$\right)$
$B$$\left($
,
$\right)$
$C$$\left($
,
$\right)$
For more practice, go to Points on the coordinate plane.
### Where will we use this?
We will use points on a coordinate plane in almost every exercise in the analytic geometry unit! Here are a few of the exercises where reviewing the coordinate plane might be helpful:
## Adding, subtracting, and squaring negative numbers
### What is this, and why do we need it?
Negative numbers let us include direction information in a number. For example, a positive vertical change means we've gone up, but a negative vertical change means we've gone down. We'll be looking for distances and slopes between points on the coordinate plane. Points with negative coordinates are to the left of or below the
.
### Practice
Problem 2.1
Add.
$-7+4=$
### Where will we use this?
Here are a few of the exercises where reviewing negative numbers might be helpful:
## Distance and displacement between points
### What is this, and why do we need it?
Distance is how far apart two points are and is always non-negative. Displacement is the amount of change to go from one point to the other, including both distance and the direction of the change.
We often break distance and displacement into their horizontal and vertical parts. When we are only working with one direction of change (only horizontal or only vertical), then the distance is the absolute value of the displacement.
We use displacement to calculate slope, and we use the horizontal and vertical distances between points to find their total distance (with a little help from the Pythagorean theorem).
### Practice
Problem 3.1
Complete the table of distances and displacements from point $A$ to point $B$.
DisplacementDistance
Horizontal
Vertical
### Where will we use this?
Here are a few of the exercises where reviewing distances and displacements might be helpful.
## Simplifying square root expressions
### What is this, and why do we need it?
For geometry, the square root function takes the area of a square as the input and give the length of a side of the square as an output. We'll use square root expressions when we use the Pythagorean theorem to find a distance. We'll use those distances to find area and perimeter of figures on the coordinate plane and to determine whether a point is part of a circle.
### Practice
Problem 4.1
Simplify.
Remove all perfect squares from inside the square root.
$\sqrt{180}=$
### Where will we use this?
Here are a couple of the exercises where reviewing square root expressions might be helpful.
## Scaling proportional relationships
### What is this, and why do we need it?
Proportional relationships are two quantities where the ratio between the two quantities always stays the same.
Slope is a kind of proportional relationship that relates the vertical displacement (or change) to the horizontal displacement. We can scale the displacements between two points to find a third point between them that divides a line segment into lengths with a given ratio.
### Practice
Problem 5
The double number line shows that to make $4$ apple pies takes $7$ kilograms $\left(\text{kg}\right)$ of apples.
Select the double number line that correctly labels the number of kilograms of apples that are needed to make $1,2,$ and $3$ pies.
Choose 1 answer:
For more practice, go to Create double number lines.
### Where will we use this?
Here is an exercise where reviewing scaling proportional relationships might be helpful:
## Slope
### What is this, and why do we need it?
Slope is one way of measuring how steep a line is. We measure slope as $\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}$, which is the ratio of the vertical displacement to the horizontal displacement.
We can use the slope of a pair of lines to prove that they are parallel (or that they're not!). Then we can tell whether we can apply all of those relationships among the angles of figures with parallel lines. If use the slope to prove that two sides of a triangle are perpendicular, we can use trigonometric ratios to relate their angle measures and side lengths.
### Practice
Problem 6.1
What is the slope of the line through $\left(-4,2\right)$ and $\left(3,-3\right)$?
Choose 1 answer:
### Where will we use this?
Here are a few of the exercises where reviewing slope might be helpful.
## Want to join the conversation?
• Any one else just utterly confused on all of this or just me
(19 votes)
• You're not alone.
(11 votes)
• why is my algebra teacher making me do this
(19 votes)
• ik man
(4 votes)
• What is the midpoint of the line segment connecting (6,4) and (3,-8)?
(4 votes)
• The midpoint of two points is the point that is halfway between them on the x-axis, and halfway between the two points on the y-axis as well. We can represent the midpoint of points (x1, y1) and (x2, y2) like this, where you take the average of each coordinate:
Mp = ((x1 + x2) / 2, (y1 + y2) / 2)
If you try it with the points (6, 4) and (3, -8), you should end up with (9/2, -2).
(13 votes)
• wth is displacement
(7 votes)
• the distance it takes to get from A to B (could be a negative)
(0 votes)
• What grade would you typically learn this in?
(4 votes)
• 8, 9, or 10 grade, depending on your math level. Maybe even less.
(1 vote)
• what is in a burber
(2 votes)
• toma's toes
(2 votes)
• thank god i finally understand analytical geometry the day before my test
(2 votes)
• Are we really gonna use all this in our day-to-day life?
(1 vote)
• this comment is fresh fresh
(1 vote)
• I don't understand problem 4.3 - how did you get that result? I keep getting 70sqrt6z^5. Please anyone, explain!
And plus, I don't understand problem 5 - where did the 14 come from?? I was able to get the result correct beacuse I got decimals and not whole numbers. I got 1:4/7 ~~ 1:0.57 ; 2:1.14 ; 3:1.71
Because you know that 4/7 is approximately 0.57, and you get that result and multiply by 1,2,and 3 right?
(0 votes)
• sqrt z^5 can be further simplified into z^2sqrt z
(3 votes)
• What is the slope of (-8,4) and (4,-6)
(0 votes)
• y=-2x-12
(3 votes) |
1. ## Percent Word Problems
We are having trouble understanding the procedures for these problems. Todays problem is:
Night came and the mangroves and palmettos closed in on their victim. If 27 percent were mangroves and 511 were palmettos, how many total shrubs were on the attack?
This is the second time we have run across this type of problem. Can you please show us the formula and procedure to complete these?
2. Let $\displaystyle x$ be the total number of shrubs. Since $\displaystyle 27\%$ of the total are mangroves, than there are $\displaystyle .27x$ total mangroves. This number added to $\displaystyle 511$, which is the number of palmettos, is equal to the total $\displaystyle x$.
$\displaystyle .27x + 511 = x$
$\displaystyle 511 = .73x$
$\displaystyle \frac{{511}}{{.73}} = x$
$\displaystyle 700 = x$
On a side note, the total number of mangroves can then be seen to be:
$\displaystyle .27 \times 700 = 189$
or
$\displaystyle 700 - 511 = 189$
3. Originally Posted by wrath27
We are having trouble understanding the procedures for these problems. Todays problem is:
Night came and the mangroves and palmettos closed in on their victim. If 27 percent were mangroves and 511 were palmettos, how many total shrubs were on the attack?
This is the second time we have run across this type of problem. Can you please show us the formula and procedure to complete these?
Hey, Wrath
You're given all the information you need in order to work the question out. I've highlighted the key bits.
Hints: (Try to work it out using this)
If 27% of shrubs are mangroves, what percentage are palmettos?
Now if 511 scrubs are just made up by the percentage of palmettos, 100% if scrubs are made up by?
Solution
27% are mangroves means that 73% are palmettos. We know that there are 511 palmettos, so if you call the total number of shrubs on attack $\displaystyle x$ then
$\displaystyle 0.73 x = 511$
so x = 700.
4. Thanks to boh of you for posting, this helped a ton!
5. ## Extension of this question
John spends 2/5 of his pocket money on a Football and 1/4 of his pocket money on sweets. If he has $1.75 left, how much pocket money did he have in the beginnning. I got to 25% and 40% respectively of the sum has been spent, leaving a balance of$1.75. However, I couldn't break it down...can anyone help please ?
Nick
Originally Posted by WWTL@WHL
Hey, Wrath
You're given all the information you need in order to work the question out. I've highlighted the key bits.
Hints: (Try to work it out using this)
If 27% of shrubs are mangroves, what percentage are palmettos?
Now if 511 scrubs are just made up by the percentage of palmettos, 100% if scrubs are made up by?
Solution
27% are mangroves means that 73% are palmettos. We know that there are 511 palmettos, so if you call the total number of shrubs on attack $\displaystyle x$ then
$\displaystyle 0.73 x = 511$
so x = 700.
6. Originally Posted by wrath27
We are having trouble understanding the procedures for these problems. Todays problem is:
Night came and the mangroves and palmettos closed in on their victim. If 27 percent were mangroves and 511 were palmettos, how many total shrubs were on the attack?
This is the second time we have run across this type of problem. Can you please show us the formula and procedure to complete these?
7. Can someone help me on where to find how to
do a percentage graph????? Getting fustrated!!!
omg..... need help!!!!
8. Originally Posted by nick2009
John spends 2/5 of his pocket money on a Football and 1/4 of his pocket money on sweets. If he has $1.75 left, how much pocket money did he have in the beginnning. I got to 25% and 40% respectively of the sum has been spent, leaving a balance of$1.75. However, I couldn't break it down...can anyone help please ?
Let $\displaystyle x$ be the amount of money you started out with. Then the amount John spent on the football is $\displaystyle \frac{2}{5}x$ and the amount he spent on the sweets is $\displaystyle \frac{1}{4}x$. He is left with $\displaystyle \$1.75$, so your equation is:$\displaystyle x-\left( \frac{2}{5}x+\frac{1}{4}x \right)=\$1.75$. This is because you start with $\displaystyle x$ and you spend $\displaystyle \left( \frac{2}{5}x+\frac{1}{4}x \right)$. Factor out the $\displaystyle x$ to get $\displaystyle x\left[ {1 - \left( {\frac{2} {5} + \frac{1} {4}} \right)} \right] = \$ 1.75 \Rightarrow \frac{7}
{{20}}x = \$1.75 \Rightarrow x = \$ 1.75\left( {\frac{{20}}
{7}} \right) = \$5.00$.
John started with $\displaystyle \$5.00\$. |
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# Find the value of integration $\int\limits_0^{\frac{\pi }{2}} {\sin x\cos xdx} .$
Last updated date: 09th Aug 2024
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Hint: Use formula $\sin 2x = 2\sin x\cos x.$
Let the value of integration be $I.$
$\Rightarrow I = \int\limits_0^{\frac{\pi }{2}} {\sin x\cos xdx} ,$
Dividing and multiplying by $2$ on right hand side, we’ll get:
$\Rightarrow I = \frac{1}{2}\int\limits_0^{\frac{\pi }{2}} {2\sin x\cos xdx} ,$
We know that $2\sin x\cos x = \sin 2x$, using this we’ll get:
$\Rightarrow I = \frac{1}{2}\int\limits_0^{\frac{\pi }{2}} {\sin 2xdx} ,$
And we also know that, $\int {\sin 2xdx = } - \frac{{\cos 2x}}{2} + C$, using this we will get:
$\Rightarrow I = \frac{1}{2}\left[ { - \frac{{\cos 2x}}{2}} \right]_0^{\frac{\pi }{2}}, \\ \Rightarrow I = - \frac{1}{4}\left[ {\cos 2x} \right]_0^{\frac{\pi }{2}} \\$
Putting limit of integration, we’ll get:
$\Rightarrow I = - \frac{1}{4}\left[ {\cos \pi - \cos 0} \right], \\ \Rightarrow I = - \frac{1}{4}( - 1 - 1) = - \frac{1}{4} \times ( - 2), \\ \Rightarrow I = \frac{1}{2}. \\$
Thus, the value of integration is $\frac{1}{2}.$
Note: We will ignore the constant of integration while putting the limit because it is a definite integration. |
# 4th Grade Math Practice 2
In 4th grade math practice 2 children can examine or increase their own knowledge before the test or exam on different topics by practicing this sheet. Questions are mainly related to simplification, finding sum, difference and products of the numbers, writing numbers in words and figures, estimating products and also questions on geometrical shapes.
Test Yourself
1. Write the smallest four digit number:
(a) Using different digits.
(b) Using different digits with a zero in the hundreds place.
2. Simplify and write the answer in words:
839 + 3049 - 487
3. Simplify:
92294 + 1828 - 6375 - 2190
4. Write the numbers in figures and find the sum:
Twenty four thousand four hundred forty five;
thirty two thousand two hundred twenty;
four thousand sixty eight.
5. The sum of two numbers is 8976. If one number is 6898, what is the other number?
6. There are 289 students in class III; 235 students in class IV and 295 students in class V. Estimate the total number of students in the three classes.
7. Find the product using the column method:
4218 × 204
8. Find the product using the expanded rotation method:
214 × 84
9. Estimate the product by rounding each number to the nearest hundred:
426 × 258
10. Fill in the blanks:
(i) Simple closed curves formed only three line segments are called ______________.
(ii) A simple closed curve does not ______________ itself.
11. How many triangles are there in this figure?
Answers for 4th Grade Math Practice 2 are given below to check the exact answers of the above questions.
1. 1023
2. Three thousand four hundred one
3. 87747
4. 60733
5. 2078
6. 800
7. 860472
8. 17976
9. 120000
10. (i) triangles
(ii) cross/intersect
11. 11
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# Simple Interest
Ankur Kulhari
Interest: It is the extra money paid by the borrower to the owner (lender) as a form of compensation for the use of the money borrowed.
Simple Interest (SI):
SI = PRT/100
P=(100*SI)/RT
R=(100*SI)/PT
T=(100*SI)/PR
Where,
P= Principle or sum (It is the money borrowed or lent out for a time period T)
R=Rate at which interest is charged on P per annum (Thus, 6 p.c. means that Rs.6 is the interest on Rs.100 in one year.
T=time period
A = Amount (The Addition of Interest and Principle i.e. A = (I + P)
Note:
Simple interest is only based on the principal amount of a loan, while compound interest is based on the principal amount and the accumulated interest.
Example 1. Find the simple interest on Rs.400 for 5 years at 6 per cent.
Solution:
SI = 400*6*5/100=120
Interest for a number of days
When the time is given in days or in years and days, 365 days are reckoned to a year.
But when the time is given in months and days, 12 months are reckoned to a year
and 30 days to the month.
The day on which the money is paid back should be include be but not the day on which it is borrowed, ie, in counting, the first day is omitted.
Example 2. Find the simple interest on Rs.306.25 from March 3rd to July 27th at 15/4% per annum.
Solution:
Interest = (306.25*(15/4)*146)/365.
Days = (28+30+31+30+27) {As 3rd March to be omitted and 27th July to be considered. }
= 4.59
Example 3. A sum of Rs.468.75 was lent out at simple interest and at the end of 1 year 8 months the total amount was Rs 500. Find the rate of interest per annum.
Solution:
SI=A-P
=> 500-468.75= 31.25
R=SI. 100/(P*T)
= (31.25*100)/(468.75*5/3)
= 781.25
Practice Questions:
1. In what time will Rs.8500 amount to Rs.15767.50 at 9/2 per cent per annum?
2. The simple interest on a sum of money is 1/9th of the principal, and the number of years is equal to the rate per cent per annum. Find the rate per cent.
3. The rate of interest for the first 2 yrs is 3% per annum, for the next 3 years is 8% per annum and for the period beyond 5 years 10% per annum and earns an interest of RS. 1520. What money did he deposit?
4. A sum of money doubles itself in 10 years at simple interest. What is the rate of interest?
5. A certain amount was lent at a rate R for 2 yrs. If it been put at 3% higher rate, it would have fetched Rs 300 more. Find the amount?
6. The simple interest on a certain sum of money at 4% per annum for 4 yrs is Rs 80 more than the interest on the same sum for 3 yrs at 5% per annum. Find the sum.
SOLUTIONS
1. SI = 15767.50 – 8500 = 7267.50
T= SI*100/(P*R)
= 7267.50*100/(8500*9/2)
= 19 years
2. T = R years, SI= P/9
R=(SI*100)/(P*T)
= ((P/9)*100)/(P*R)
R2 = 100/9
R=10/3
3. Let money deposited = P
1520 = [(P*3*2)/100] + [P*8*3/100] + [P*10*1/100] 1520 = P*40/100
P= 1520*100/40
P= 3800
4. SI = P {As amount (SI + P) = 2P)
P = P*R*10/100
R= 10%
5. Say at rate R Interest = SI, and the amount = P
SI = P*R*2/100 …………………..(1)
Now, R = R+3 and SI = SI+300
SI+300 = P*(R+3)*2/100 ………..(2)
From eq. (1) and (2)
(P*R*2/100) + 300 = P*(R+3)*2/100
30000 = 6*P
P=5000
6. 80 = [P*4*4/100] – [P*3*5/100] P=8000
Shortcut: P = Difference in SI*100/(R1*T1 + R2*T2)
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# How do you solve x - 2y = 5 and 2x - 3y = 10 using matrices?
Apr 5, 2016
Given
$\textcolor{w h i t e}{\text{XXX}} x - 2 y = 5$
$\textcolor{w h i t e}{\text{XXX}} 2 x - 3 y = 10$
Re-writing as augmented matrix:
$\textcolor{w h i t e}{\text{XXX") [(1,-2,5),(2,-3,10)]color(white)("XXX}} \left.\begin{matrix}\left[1\right] \\ \left[2\right]\end{matrix}\right.$
Subtracting $2$ times the first row from the second row:
$\textcolor{w h i t e}{\text{XXX") [(1,-2,5),(0,1,0)]color(white)("XXX}} \left.\begin{matrix}\left[1\right] \\ \left[3\right] = \left[2\right] - 2 \times \left[1\right]\end{matrix}\right.$
Adding $2$ times row [3] to the first row:
$\textcolor{w h i t e}{\text{XXX") [(1,0,5),(0,1,0)]color(white)("XXX}} \left.\begin{matrix}\left[4\right] \\ \left[3\right] = \left[1\right] + 2 \times \left[3\right]\end{matrix}\right.$
Re-writing with variables:
$\textcolor{w h i t e}{\text{XXX}} x = 5$
$\textcolor{w h i t e}{\text{XXX}} y = 0$ |
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# How to Square Fractions
Squaring fractions is one of the simplest operations you can perform on fractions. It is very similar to squaring whole numbers in that you simply multiply both the numerator and the denominator by itself.[1] There are also some instances in which simplifying the fraction before squaring makes the process easier. If you haven't yet learned this skill, this article provides an easy overview that will improve your understanding quickly.
### Part 1 Squaring Fractions
1. 1
Understand how to square whole numbers. When you see an exponent of two, you know that you need to square the number. To square a whole number, you multiply it by itself.[2] For example:
• 52 = 5 × 5 = 25
2. 2
Realize that squaring fractions works the same way. To square a fraction, you multiply the fraction by itself. Another way to think about it is to multiply the numerator by itself and then the denominator by itself.[3] For example:
• (5/2)2 = 5/2 × 5/2 or (52/22).
• Squaring each number yields (25/4).
3. 3
Multiply the numerator by itself and the denominator by itself. The actual order that you multiply these numbers by themselves doesn’t matter as long as you have squared both numbers. To keep things simple, start with the numerator: simply multiply it by itself. Then, multiply the denominator by itself.
• The numerator will stay on top of the fraction and the denominator will stay at the bottom of the fraction.
• For example: (5/2)2 = (5 x 5/2 x 2) = (25/4).
4. 4
Simplify the fraction to finish. When working with fractions, the last step is always to reduce the fraction to its most simple form or turn the improper fraction into a mixed number.[4] For our example, 25/4 is an improper fraction because the numerator is larger than the denominator.
• To convert to a mixed number, divide 4 into 25. It goes in 6 times (6 x 4 = 24) with 1 leftover. Therefore, the mixed number is 6 1/4.
### Part 2 Squaring Fractions with Negative Numbers
1. 1
Recognize the negative sign in front of the fraction. If you are working with a negative fraction, it will have a minus sign in front of it. It is good practice to always put parentheses around a negative number so you know that the “–“ sign is referring to the number and not telling you to subtract two numbers.[5]
• For example: (–2/4)
2. 2
Multiply the fraction by itself. Square the fraction as you would normally by multiplying the numerator by itself and then multiplying the denominator by itself. Alternatively, you can simply multiply the fraction by itself.
• For example: (–2/4)2 = (–2/4) x (–2/4)
3. 3
Understand that two negative numbers multiply to make a positive number. When a minus sign is present, the entire fraction is negative. When you square the fraction, you are multiplying two negative numbers together. Whenever two negative numbers are multiplied together, they make a positive number.[6]
• For example: (-2) x (-8) = (+16)
4. 4
Remove the negative sign after squaring. After you have squared the fraction, you will have multiplied two negative numbers together. This means that the squared fraction will be positive. Be sure to write your final answer without the negative sign.[7]
• Continuing the example, the resulting fraction will be a positive number.
• (–2/4) x (–2/4) = (+4/16)
• Generally, the convention is to drop the “+” sign for positive numbers.[8]
5. 5
Reduce the fraction to its simplest form. The final step when doing any calculations with a fraction is to reduce it. Improper fractions must first be simplified into mixed numbers and then reduced.
• For example: (4/16) has a common factor of four.
• Divide the fraction through by 4: 4/4 = 1, 16/4= 4
• Rewrite simplified fraction: (1/4)
### Part 3 Using Simplifications and Shortcuts
1. 1
Check to see if you can simplify the fraction before you square it. It is usually easier to reduce fractions before squaring them. Remember, to reduce a fraction means to divide it by a common factor until the number one is the only number that can be evenly divided into both the numerator and denominator.[9] Reducing the fraction first means you don’t have to reduce it at the end when the numbers will be larger.
• For example: (12/16)2
• 12 and 16 can both be divided by 4. 12/4 = 3 and 16/4 = 4; therefore, 12/16 reduces to 3/4.
• Now, you will square the fraction 3/4.
• (3/4)2 = 9/16, which cannot be reduced.
• To prove this, let’s square the original fraction without reducing:
• (12/16)2 = (12 x 12/16 x 16) = (144/256)
• (144/256) has a common factor of 16. Dividing both the numerator and denominator by 16 reduces the fraction to (9/16), the same fraction we got from reducing first.
2. 2
Learn to recognize when you should wait to reduce a fraction. When working with more complex equations, you may be able to simply cancel one of the factors. In this case, it is actually easier to wait before you reduce the fraction. Adding an additional factor to the above example illustrates this.
• For example: 16 × (12/16)2
• Expand out the square and cross out the common factor of 16: 16 * 12/16 * 12/16
• Because there is one 16 whole number and two 16’s in the denominator, you can cross ONE of them out.
• Rewrite the simplified equation: 12 × 12/16
• Reduce 12/16 by dividing through by 4: 3/4
• Multiply: 12 × 3/4 = 36/4
• Divide: 36/4 = 9
3. 3
Understand how to use an exponent shortcut. Another way to solve the same example is to simplify the exponent first. The end result is the same, it’s just a different way to solve.
• For example: 16 * (12/16)2
• Rewrite with the numerator and denominator squared: 16 * (122/162)
• Cancel out the exponent in the denominator: 16 * 122/162
• Imagine the first 16 has an exponent of 1: 161. Using the exponent rule of dividing numbers, you subtract the exponents. 161/162, yields 161-2 = 16-1 or 1/16.
• Now, you are working with: 122/16
• Rewrite and reduce the fraction: 12*12/16 = 12 * 3/4.
• Multiply: 12 × 3/4 = 36/4
• Divide: 36/4 = 9
## Community Q&A
Search
• How do I use exponential notation for a fraction raised to a power?
wikiHow Contributor
Use parentheses around the fraction, and ^ to raise it to any power. For example: one third, squared: (1/3)^2.
• How do I calculate (1-4.4) 2 (to the power of 2)?
wikiHow Contributor
Do 1-4.4 (what's in the parentheses) first, which gives you -3.4. You would then change that into 3.4 squared, and to make it a fraction, you would do 34 over 10 [(34/10)^2]. Divide both the numerator and the denominator by 2, because it is squared. That leaves you with 17 over 5 (17/5). Bring down the squared to each part, so 17 squared over 5 squared. Do the math; you end up with 289 over 25 (289/25). Simplified, that gives you 11 and 14/25, or 11.56 in non-fraction form.
• How do I use a cube and fraction finder?
• How can we simplify 3/5+7/8-1/2-3/4 using the LCM method?
• How do I solve fraction with six kinds of fractions?
• What is the square of 2/5-a?
## Things You'll Need
• Paper or screen for working on
• Pencil/Pen (for use w/ paper)
## Article Info
Categories: Exponents and Logarithms | Fractions
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# Addition of Metric Measures – Methods, Examples | How to Add Metric Measures?
Wondering how to add metric measures, then you have reached a safe place that will let you know how to add the metric measures and also explain the steps and detailed procedure of adding those numbers. The addition of metric measures is useful in our daily life and this adding of metric measures is similar to the addition of ordinary numbers.
We will arrange them in columns according to their units and we will add them. Before adding metric measures we need to know about metric measures and also about converting them from one unit to another unit.
## How to Add Metric Measures?
The Addition of metric measures is similar to the adding of ordinary numbers. Adding of these mixed measures can be done in two ways, one is to convert all the mixed measure into one unit and adding them, to do this we have known about converting the units and you can learn by referring to our earlier articles. The other way is to add like units by placing the digits in columns and adding. To understand in detail we will go with one example which we will solve in two methods.
Example:
Add 5 km 4 hm 8 m 3 mm and 8 km 9 hm 6 m 3 mm.
Solution:Â
Method 1: Place the given two digits in columns and add them usually.
Therefore the result of addition is 14 km 4 hm 4 m 6 mm.
Method 2:
Convert the given two numbers into same unit that is we will convert them into km.
we know that 1000000 mm = 1 km, 1000 m = 1 km, 10 hm = 1 km
5 km 4 hm 8 m 3 mm = 5 km + $$\frac { 4 }{ 10 }$$ km + $$\frac { 8 }{ 100 }$$ km +$$\frac { 3 }{ 1000000}$$
= 5 km + 0.4 km + 0.08 km + 0.000003 km
= 5.480003 km.
8 km 9 hm 6 m 3 mm = 8km + $$\frac { 9 }{ 10 }$$ km+ $$\frac { 6 }{ 100}$$ km+$$\frac { 3}{ 1000000 }$$ km
= 8 km + 0.9 km + 0.06 km + 0.000003 km
= 8.960003 km
Now by adding them we get
5.480003 + 8.960003 = 14.440006 km
Therefore the result of addition is 14.440006 km.
### Addition of Metric Measures Examples
Here are some examples involving the addition of metric measures of length, mass, and capacity. We will be solving these examples in the first method only, which will make students understand easily about the concept.
Example 1:
Add 6 kg 2 hg 0 dag 5 g 0 dg 7 cg and 2 kg 5 hg 6 dag 3 g 0 dg 2 cg.
Solution:Â
Arrange the given numbers in the column vertically and add them from right to left.
Therefore the result of the addition is 8 kg 7 hg 6 dag 8 g 0 dg 9 cg.
Example 2:
Add 9 km 6 hm 5 dcm 4 m 1 dm 2 cm and 4 km 0 hm 3 dcm 5 m 7 dm 2 cm.
Solution:Â
Arrange the given numbers in the column vertically and add them from right to left.
Therefore, the addition is 13 km 6 hm 8 dcm 9 m 8 dm 4 cm.
Example 3:
Add 3 kl 6 hl 0 dal 7 l 5 dl 1 cl and 5 kl 6 hl 7 dal 0 l 4 dl 1 cl.
Solution:
Arrange the given numbers in the column vertically and add them from right to left.
Therefore, the addition is 8 kl 12 hl 7 dal 7 l 9 dl 2 cl.
Example 4:Â
Annie has 6 kg 450 g of sugar in a bag and she added more than 8 hg 320 g in it. How much amount of sugar is in the bag?
Solution:
Given
weight of sugar in bag = 6 kg 450 g
weight of sugar she added = 8 hg 320 g
Total weight of the sugar in bag = weight of sugar in bag + weight of sugar she added
Therefore, Amount of the sugar in the bag is 6 kg 8 hg 770 g.
Example 5:Â
Charlie bought 5 kl 65 l of milk in one shop and 60 hl and 56 dal of milk in another shop. How much amount of milk he has?
Solution:
Given
Amount of milk he bought in the first shop = 5 kl 65 l
Amount of milk he bought in the second shop = 60 hl 56 dal
The total amount of milk = Amount of milk he bought in the first shop + Amount of milk he bought in the second shop
Therefore, the total amount of milk he bought = 5 kl 60 hl 56 dal and 65l.
Example 6:
Meera traveled 67 km 50 dcm towards the east and 34 km 500 m towards the northeast. How much distance did she travel in total?
Solution:
Given
Distance traveled towards east = 67 km 50 dcm
Distance traveled towards north-east = 34 km 500 m
Total distance traveled by Meera is = Distance traveled towards east + Distance traveled towards north-east.
= 67 km 50 dcm + 34 km 500 m
Therefore, the total distance she traveled is 101 km 50 dcm 500m.
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# The perpendicular from A on side BC of a triangle ABC intersects BC at D such that DB = 3CD (see figure). Prove that $$2{AB}^2$$ = $$2{AC}^2$$ + $${BC}^2$$
## Solution :
We have : DB = 3CD
Now, BC = DB + CD
i.e. BC = 3CD + CD [because BD = 3CD]
BC = 4CD
$$\therefore$$ CD = $$1\over 4$$ BC and DB = 3CD = $$3\over 4$$ BC ……….(1)
Since triangle ABD is a right triangle, right angled at D, therefore, by Pythagoras Theorem, we have :
$${AB}^2$$ = $${AD}^2$$ + $${DB}^2$$ ……….(2)
In triangle ACD,
$$\angle$$ D = 90 , $${AC}^2$$ = $${AD}^2$$ + $${CD}^2$$ ……..(3)
Subtracting (3) from (2), we get
$${AB}^2$$ – $${AC}^2$$ = $${DB}^2$$ – $${CD}^2$$
$${AB}^2$$ – $${AC}^2$$ = $$({3\over 4}BC)^2$$ – $$({1\over 4}BC)^2$$ (using (1))
$${AB}^2$$ – $${AC}^2$$ = $$({9\over 16} – {1\over 16})$$$${BC}^2$$
$${AB}^2$$ – $${AC}^2$$ = $${1\over 2}$$$${BC}^2$$
$$\implies$$ $$2{AB}^2$$ = $$2{AC}^2$$ + $${BC}^2$$ |
# Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions Answers
Here we are providing Triangles Class 10 Extra Questions Maths Chapter 6 with Answers Solutions, Extra Questions for Class 10 Maths was designed by subject expert teachers.
## Extra Questions for Class 10 Maths Triangles with Answers Solutions
Extra Questions for Class 10 Maths Chapter 6 Triangles with Solutions Answers
### Triangles Class 10 Extra Questions Very Short Answer Type
Question 1.
Two sides and the perimeter of one triangle are respectively three times the corresponding sides and the perimeter of the other triangle. Are the two triangles similar? Why?
Solution:
Since the perimeters and two sides are proportional
∴ The third side is proportional to the corresponding third side.
i.e., The two triangles will be similar by SSS criterion.
Question 2.
A and B are respectively the points on the sides PQ and PR of a ∆PQR such that PQ = 12.5 cm, PA = 5 cm, BR = 6 cm, and PB = 4 cm. Is AB || QR? Give reason.
Solution:
Question 3.
If ∆ABC ~ ∆QRP, $$\frac { ar(∆ABC) }{ ar(∆PQR) }$$ = $$\frac{9}{4}$$, AB = 18 cm and BC = 15 cm, then find the length of PR.
Solution:
Question 4.
If it is given that ∆ABC ~ ∆PQR with $$\frac{BC}{QR}$$ = $$\frac{1}{3}$$, then find $$\frac { ar(∆PQR) }{ ar(∆ABC) }$$
Solution:
Question 5.
∆DEF ~ ∆ABC, if DE : AB = 2 : 3 and ar(∆DEF) is equal to 44 square units. Find the area (∆ABC).
Solution:
Question 6.
Is the triangle with sides 12 cm, 16 cm and 18 cm a right triangle? Give reason.
Solution:
Here, 122+ 162 = 144 + 256 = 400 ≠ 182
∴ The given triangle is not a right triangle.
### Triangles Class 10 Extra Questions Short Answer Type 1
Question 1.
In triangles PQR and TSM, ∠P = 55°, ∠Q = 25°, ∠M = 100°, and ∠S = 25°. Is ∆QPR ~ ∆TSM? Why?
Solution:
Şince, ∠R = 180° – (∠P + ∠Q)
= 180° – (55° + 25°) = 100° = ∠M
∠Q = ∠S = 25° (Given)
∆QPR ~ ∆STM
i.e., . ∆QPR is not similar to ∆TSM.
Question 2.
If ABC and DEF are similar triangles such that ∠A = 47° and ∠E = 63°, then the measures of ∠C = 70°. Is it true? Give reason.
Solution:
Since ∆ABC ~ ∆DEF
∴ ∠A = ∠D = 47°
∠B = ∠E = 63°
∴ ∠C = 180° – (∠A + ∠B) = 180° – (47° + 63°) = 70°
∴ Given statement is true.
Question 3.
Let ∆ABC ~ ∆DEF and their areas be respectively 64 cm2 and 121 cm2. If EF = 15.4 cm, find BC.
Solution:
Question 4.
ABC is an isosceles triangle right-angled at C. Prove that AB2 = 2AC2.
Solution:
∆ABC is right-angled at C.
∴ AB2 = AC2 + BC2 [By Pythagoras theorem]
⇒ AB2 = AC2 + AC2
[∵ AC = BC]
⇒ AB2 = 2AC2
Question 5.
Sides of triangle are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse.
(i) 7cm, 24 cm, 25 cm
(ii) 3 cm, 8 cm, 6 cm
Solution:
(i) Let a = 7 cm, b = 24 cm and c = 25 cm.
Here, largest side, c = 25 cm
We have, a2 + b2 = (7)2 + (24)2 = 49 + 576 = 625 = c2 [∵c = 25]
So, the triangle is a right triangle.
Hence, c is the hypotenuse of right triangle.
(ii) Let a = 3 cm, b = 8 cm and c = 6 cm
Here, largest side, b = 8 cm
We have, a2 + c2 = (3)2 + (6)2 = 9 + 36 = 45 ≠ b2
So, the triangle is not a right triangle.
Question 6.
If triangle ABC is similar to triangle DEF such that 2AB = DE and BC = 8 cm. Then find the length of EF.
Solution:
∆ABC ~ ∆DEF (Given)
Question 7.
If the ratio of the perimeter of two similar triangles is 4 : 25, then find the ratio of the areas of the similar triangles.
Solution:
∵ Ratio of perimeter of 2 ∆’s = 4 : 25
∵ Ratio of corresponding sides of the two ∆’s = 4 : 25
Now, the ratio of area of 2 ∆’s = Ratio of square of its corresponding sides.
= $$\frac{(4)^{2}}{(25)^{2}}$$ = $$\frac{16}{625}$$
Question 8.
In an isosceles ∆ABC, if AC = BC and AB2 = 2AC2, then find ∠C.
Solution:
AB2 = 2AC2 (Given)
AB2 = AC2 + AC2
AB2 = AC2 + BC2 (∵ AC = BC)
Hence AB is the hypotenuse and ∆ABC is a right angle A.
So, ∠C = 90°
Question 9.
The length of the diagonals of a rhombus are 16 cm and 12 cm. Find the length of side of the rhombus.
Solution:
∵ The diagonals of rhombus bisect each other at 90°.
∴ In the right angle ∆BOC
BO = 8 cm
CO = 6 cm
∴ By Pythagoras Theorem
BC2 = BO2 + CO2 = 64 + 36
BC2 = 100
BC = 10 cm
Question 10.
A man goes 24 m towards West and then 10 m towards North. How far is he from the starting point?
Solution:
By Pythagoras Theorem
AC2 = AB2 + BC2 = (24)2 + (10)2
AC2 = 676
AC = 26 m
∴ The man is 26 m away from the starting point.
Question 11.
∆ABC ~ ∆DEF such that AB = 9.1 cm and DE = 6.5 cm. If the perimeter of ∆DEF is 25 cm, what is the perimeter of ∆ABC?
Solution:
Since ∆ABC ~ ∆DEF.
Question 12.
∆ABC ~ ∆PQR; if area of ∆ABC = 81 cm2, area of ∆PQR = 169 cm2 and AC = 7.2 cm, find the length of PR.
Solution:
Since ∆ABC ~ ∆PQR
### Triangles Class 10 Extra Questions Short Answer Type 2
Question 1.
In Fig. 7.10, DE || BC. If AD = x, DB = x – 2, AE = x + 2 and EC = x – 1, find the value of x.
Solution:
In ∆ABC, we have
DE || BC,
∴ $$\frac{A D}{D B}$$ = $$\frac{A E}{E C}$$ [By Basic Proportionality Theorem]
⇒ $$\frac{x}{x-2}$$ = $$\frac{x+2}{x-1}$$
⇒ x(x – 1) = (x – 2) (x + 2)
⇒ x2 – x = x2 – 4
⇒ x = 4
Question 2.
E and F are points on the sides PQ and PR respectively of a ∆PQR. Show that EF ||QR if PQ = 1.28 cm, PR= 2.56 cm, PE = 0.18 cm and PF = 0.36 cm.
Solution:
We have, PQ = 1.28 cm, PR = 2.56 cm
PE = 0.18 cm, PF = 0.36 cm
Now, EQ = PQ-PE = 1.28 – 0.18 = 1.10 cm and
FR = PR – PF = 2.56 – 0.36 = 2.20 cm
Therefore, EF || QR [By the converse of Basic Proportionality Theorem]
Question 3.
A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
Solution:
Let AB be a vertical pole of length 6m and BC be its shadow and DE be tower and EF be its shadow. Join AC and DF.
Now, in ∆ABC and ∆DEF, we have
h = 42 Hence, height of tower, DE = 42m
Question 4.
In Fig. 7.13, if LM || CB and LN || CD, prove that $$\frac{A M}{A B}=\frac{A N}{A D}$$
Solution:
Firstly, in ∆ABC, we have
LM || CB (Given)
Therefore, by Basic Proportionality Theorem, we have
Question 5.
In Fig. 7.14, DE || OQ and DF || OR Show that EF || QR.
Solution:
In ΔPOQ, we have
DE || OQ (Given)
[Applying the converse of Basic Proportionality Theorem in ∆PQR]
Question 6.
Using converse of Basic Proportionality Theorem, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.
Solution:
Given: ∆ABC in which D and E are the mid-points of sides AB and AC respectively.
To prove: DE || BC
Proof: Since D and E are the mid-points of AB and AC respectively
∴ AD = DB and AE = EC
DB EC Therefore, DE || BC (By the converse of Basic Proportionality Theorem)
Question 7.
State which pairs of triangles in the following figures are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form.
Solution:
(i) In ∆ABC and ∆QRP, we have
∆NML is not similar to ∆PQR.
Question 8.
In Fig. 7.17, $$\frac{A O}{O C}$$ = $$\frac{B O}{O D}$$ = $$\frac{1}{2}$$ and AB = 5cm. Find the value of DC.
Solution:
⇒ DC = 10 cm.
Question 9.
E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ∆ABE ~ ∆CFB.
Solution:
In ∆ABE and ∆CFB, we have
∠AEB = ∠CBF (Alternate angles)
∠A = ∠C (Opposite angles of a parallelogram)
∴ ∆ABE ~ ∆CFB (By AA criterion of similarity)
Question 10.
S and T are points on sides PR and QR of ∆PQR such that ∠P = ∠RTS. Show that ∆RPQ ~ ∆RTS.
Solution:
In ∆RPQ and ∆RTS, we have
∠RPQ = ∠RTS (Given)
∠PRQ = ∠TRS = ∠R (Common)
∴ ∆RPQ ~ ∆RTS (By AA criterion of similarity)
Question 11.
In Fig. 7.20, ABC and AMP are two right triangles right-angled at B and M respectively. Prove that:
(i) ∆ABC ~ ∆AMP
(ii) $$\frac{C A}{P A}$$ = $$\frac{B C}{M P}$$
Solution:
(i) In ∆ABC and ∆AMP, we have
∠ABC = ∠AMP = 90° (Given)
And, ∠BAC = ∠MAP (Common angle)
∴ ∆ABC ~ ∆AMP (By AA criterion of similarity)
(ii) As ∆ABC ~ ∆AMP (Proved above)
∴ $$\frac{C A}{P A}$$ = $$\frac{B C}{M P}$$ (Sides of similar triangles are proportional)
Question 12.
D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA2 = CB.CD.
Solution:
Question 13.
ABC is an equilateral triangle of side 2a. Find each of its altitudes.
Solution:
Let ABC be an equilateral triangle of side 2a units.
We draw AD ⊥ BC. Then D is the mid-point of BC.
⇒ $$\frac{B C}{2}$$ = $$\frac{2 a}{2}$$ = a
Now, ABD is a right triangle right-angled at D.
⇒ AB2 = AD2 + BD2 [By Pythagoras Theorem]
⇒ (2a)2 = AD2 + a2
⇒ AD2 = 4a2 – a2 = 3a2
Hence, each altitude = √3a unit.
Question 14.
An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two planes after 1 $$\frac{1}{2}$$ hours?
Solution:
Let the first aeroplane starts from O and goes upto A towards north where
(Distance = Speed × Time)
Again let second aeroplane starts from O at the same time and goes upto B towards west where
OB = 1200 × $$\frac{3}{2}$$ = 1800 km
Now, we have to find AB.
In right angled ∆ABO, we have
AB2 = OA2 + OB2 [By using Pythagoras Theorem]
⇒ AB2 = (1500)2 + (1800)2
⇒ AB2 = 2250000 + 3240000
⇒ AB2 = 5490000
∴ AB = 100 √549 = 100 × 23.4307 = 2343.07 km.
Question 15.
In the given Fig. 7.24, ∆ABC and ADBC are on the same base BC. If AD intersects BC at 0. Prove
that $$\frac { ar(∆ABC) }{ ar(∆DBC) }$$ = $$\frac{AO}{DO}$$
Solution:
Given: ∆ABC and ∆DBC are on the same base BC and AD intersects BC at O.
Question 16.
In Fig. 7.25, AB || PQ || CD , AB = x units, CD = y units and PQ = z units. Prove that $$\frac{1}{x}+\frac{1}{y}=\frac{1}{z}$$.
Solution:
Since AB || PQ
∠ABQ = ∠PQD (Corresponding ∠’s)
By AA-Similarity
Question 17.
In Fig. 7.26, if ∆ABC ~ ∆DEF and their sides are of lengths (in cm) as marked along with them, then find the lengths of the sides of each triangle.
Solution:
∆ABC ~ ∆DEF (Given)
⇒ 4x – 2 = 18
⇒ x = 5
∴ AB = 2 × 5 – 1 = 9, BC = 2 × 5 + 2 = 12
CA = 3 × 5 = 15, DE = 18, EF = 3 × 5 + 9 = 24
and FD = 6 × 5 = 30
Hence, AB = 9 cm, BC = 12 cm, CA = 15 cm
DE = 18 cm, EF = 24 cm, FD = 30 cm
Question 18.
In ΔABC, it is given that $$\frac{A B}{A C}$$ = $$\frac{B D}{D C}$$ . If ∠B = 70° and ∠C = 50° then find ∠BAD.
Solution:
In ∆ABC
∵ ∠A + ∠B + 2C = 180° (Angle sum property)
∠A + 70° + 50° = 180°
⇒ ∠A = 180° – 120°
⇒ ∠A = 60°
∵ $$\frac{A B}{A C}$$ = $$\frac{B D}{D C}$$ (Given)
∴ ∠1 = ∠2
[Because if a line through one vertex of a triangle divides the opposite sides in the ratio of the other two sides, then the line bisects the angle at the vertex.]
But ∠1 + ∠2 = 60° …(ii)
From (i) and (ii) we get,
2∠1 = 60°
⇒ ∠1 = $$\frac{60°}{2}$$ = 30°
Question 19.
If the diagonals of a quadrilateral divide each other proportionally, prove that it is a trapezium.
Solution:
AB || DC
⇒ In quad ABCD, AB || DC
⇒ ABCD is a trapezium.
Question 20.
In the given Fig. 7.29, $$\frac{P S}{S Q}$$ = $$\frac{P T}{T R}$$ and ∠PST = ∠PRQ. Prove that PQR is an isosceles triangle.
Solution:
Given: $$\frac{PS}{SQ}$$ = $$\frac{PT}{TR}$$ and ∠PST = ∠PRQ
To Prove: PQR is isosceles triangle.
Proof: $$\frac{PS}{SQ}$$ = $$\frac{PT}{TR}$$
By converse of BPT we get
ST || QR
∴ ∠PST = ∠PQR (Corresponding angles) ….(i)
But, ∠PST = ∠PRQ (Given) ….(ii)
From equation (i) and (ii)
∠PQR = ∠PRQ
⇒ PR = PQ
So, ∆PQR is an isosceles triangle.
Question 21.
The diagonals of a trapezium ABCD in which AB || DC, intersect at O. If AB = 2CD, then find the ratio of areas of triangles AOB and COD.
Solution:
In ∆AOB and ∆COD
∠COD = ∠AOB (Vertically opposite angles)
∠CAB = ∠DCA (Alternate angles)
∆AOB ~ ∆COD (By AA-Similarity)
By area theorem
Hence, ar(∆AOB) : ar(∆COD) = 4 : 1.
Question 22.
In the given Fig. 7.31, find the value of x in terms of a, b and c.
Solution:
In ∆LMK and ∆PNK
We have, ∠M = ∠N = 50° and ∠K = ∠K (Common)
∴ ∆LMK ~ ∆PNK (AA – Similarity)
Question 23.
In the given Fig. 7.32, CD || LA and DE || AC. Find the length of CL, if BE = 4 cm and EC = 2 cm.
Solution:
Question 24.
In the given Fig. 7.33, AB = AC. E is a point on CB produced. If AD is perpendicular to BC and EF perpendicular to AC, prove that ∆ABD is similar to ∆ECF.
Solution:
AB = AC (Given)
⇒ ∠ABC = ∠ACB (Equal sides have equal opposite angles)
Now, in ∆ABD and ∆ECF
∠ABD = ∠ECF (Proved above)
So, ∆ABD ~ ∆ECF (AA – Similarity)
### Triangles Class 10 Extra Questions Long Answer Type
Question 1.
Using Basic Proportionality Theorem, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.
Solution:
Given: A ∆ABC in which D is the mid-point of AB and DE is drawn parallel to BC, which meets AC at E.
To prove: AE = EC
Proof: In ∆ABC, DE || BC
∴ By Basic Proportionality Theorem, we have
$$\frac{A D}{D B}$$ = $$\frac{A E}{E C}$$ …(i)
Now, since D is the mid-point of AB
From (i) and (ii), we have
$$\frac{A D}{D B}$$ = $$\frac{A E}{E C}$$
⇒ 1 = $$\frac{A E}{E C}$$
Hence, E is the mid-point of AC.
Question 2.
ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that $$\frac{A O}{B O}$$ = $$\frac{C O}{D O}$$.
Solution:
Given: ABCD is a trapezium, in which AB || DC and its diagonals intersect each other at point O.
Question 3.
If AD and PM are medians of triangles ABC and PQR respectively, where ∆ABC ~ ∆PQR, prove that $$\frac{A B}{P Q}$$ = $$\frac{A D}{P M}$$
Solution:
In ΔABD and ΔPQM we have
∠B = ∠Q (∵ ∆ABC ~ ∆PQR) …(i)
Question 4.
In Fig. 7.37, ABCD is a trapezium with AB || DC. If ∆AED is similar to ΔBEC, prove that AD = BC.
Solution:
In ∆EDC and ∆EBA we have
∠1 = ∠2 [Alternate angles]
∠3 = ∠4 [Alternate angles]
∠CED = ∠AEB [Vertically opposite angles]
∴ ∆EDC ~ ∆EBA [By AA criterion of similarity]
Question 5.
Prove that the area of an equilateral triangle described on a side of a right-angled isosceles triangle is half the area of the equilateral triangle described on its hypotenuse.
Solution:
Given: A ∆ABC in which ∠ABC = 90° and AB = BC.
∆ABD and ΔCAE are equilateral triangles.
To Prove: ar(∆ABD) = $$\frac {1}{2}$$ × ar(∆CAE)
Proof: Let AB = BC = x units.
∴ hyp. CA = √x2 + √x2 = x√2 units.
Each of the ABD and ∆CAE being equilateral has each angle equal to 60°.
∴ ∆АВD ~ ∆CAE
But, the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.
Question 6.
If the areas of two similar triangles are equal, prove that they are congruent.
Solution:
Given: Two triangles ABC and DEF, such that
∆ABC ~ ∆DEF and area (∆ABC) = area (∆DEF)
To prove: ∆ABC ≅ ∆DEF
Proof: ∆ABC ~ ∆DEF
⇒ ∠A = ∠D, ∠B = ∠E, ∠C = ∠F
AB = DE, BC = EF, AC = DF
∆ABC ≅ ∆DEF (By SSS criterion of congruency)
Question 7.
Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
Solution:
Let ∆ABC and ∆PQR be two similar triangles. AD and PM are the medians of ∆ABC and ∆PQR respectively.
Question 8.
In Fig. 7.41,0 is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that
(i) OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2
(ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2
Solution:
Join OA, OB and OC.
(i) In right ∆’s OFA, ODB and OEC, we have
OA2 = AF2 + OF2 …(i)
OB2 = BD2 + OD2 …(ii)
and C2 = CE2 + OE2
Adding (i), (ii) and (iii), we have
⇒ 0A2 + OB2 + OC2 = AF2 + BD2 + CE2 + OF2 + OD2 + OE2
⇒ 0A2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2
(ii) We have, OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2
⇒ (OA2 – OE2) + (OB2 – OF2) – (OC2 – OD2) = AF2 + BD2 + CE2
⇒ AE2 + CD2 + BF2 = AP2 + BD2 + CE2
[Using Pythagoras Theorem in ∆AOE, ∆BOF and ∆COD]
Question 9.
The perpendicular from A on side BC of a ∆ABC intersects BC at D such that DB = 3CD (see Fig. 7.42). Prove that 2AB2 = 2AC2 + BC2
Solution:
We have, DB = 3CD
Now,
BC = BD + CD
⇒ BC = 3CD + CD = 4CD (Given DB = 3CD)
∴ CD = $$\frac{1}{4}$$ BC
and DB = 3CD = $$\frac{1}{4}$$BC
Now, in right-angled triangle ABD using Pythagoras Theorem we have
AB2 = AD2 + DB2 …(i)
Again, in right-angled triangle ∆ADC, we have
AC2 = AD2 + CD2 …(ii)
Subtracting (ii) from (i), we have
AB2 – AC2 = DB2 – CD2
∴ 2AB2 – 2AC2 = BC2
⇒ 2AB2 = 2AC2 + BC2
Question 10.
In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.
Solution:
Let ABC be an equilateral triangle and let AD ⊥ BC.
∴ BD = DC
Now, in right-angled triangle ADB, we have
AB2 = AD2 + BD2 [Using Pythagoras Theorem]
Question 11.
Prove that, if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
Using the above result, do the following:
In Fig. 7.44 DE || BC and BD = CE. Prove that ∆ABC is an isosceles triangle.
Solution:
Given: A triangle ABC in which a line parallel to side BC intersects other two sides AB and AC at D and E respectively.
⇒ AB = AC (As DB = EC)
∴ ∆ABC is an isosceles triangle.
Question 12.
In Fig. 7.46, ABD is a triangle right-angled at A and AC ⊥ BD Show that
(i) AB2 = BC. BD
(iii) AC2 = BC.DC
Solution:
Given: ABD is a triangle right-angled at A and AC ⊥ BD.
To prove: (i) AB2 = BC .BD
(iii) AC2 = BC . DC
Proof: (i) In ∆ACB and ∆DAB, we have
∠ACB = ∠DAB = 90°
∠ABC = ∠DBA = ∠B [Common]
∴ ∆ACB ~ ∆DAB [By AA criterion of similarity]
$$\frac{BC}{AB}=\frac{AB}{DB}$$
⇒ AB2 = BC.BD
(ii) In ∆ACD and ∆BAD, we have
∠CDA = ∠BDA = ∠D [Common]
∴ ∆ACD ~ ∆BAD [By AA criterion of similarity]
$$\frac{AD}{BD}=\frac{CD}{AD}$$
(iii) We have ∆ACB – ∆DAB
From (i) and (ii), we have
∆ВСА ~ ∆АСD
$$\frac{B C}{A C}=\frac{A C}{D C}$$
AC2 = BC . DC
Question 13.
Prove that ratio of areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. Using the above result do the following: Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2CD, find the ratio of the areas of triangles AOB and COD.
Solution:
Given: Two triangles ABC and PQR such that ∆ABC ~ ∆PQR
Second Part:
In ∆AOB and ∆COD we have
∠AOB = ∠COD (Vertically opposite angles) and
∠OAB = ∠OCD (Alternate angles)
∆AOB ~ ∆COD (By AA criterion of similarity]
Hence, the ratio of areas of ∆AOB and ∆COD 4 : 1.
Question 14.
Prove that, in a right triangle, the square of the hypotenuse is equal to the sum of squares of the other two sides.
Using the above, do the following:
Prove that, in a ∆ABC if AD is perpendicular to BC, then AB2 + CD2 = AC2 + BD2.
Solution:
Given: A right triangle ABC right-angled at B.
To Prove: AC2 = AB2 + BC2
Construction: Draw BD ⊥ AC
∠A = ∠A (Common)
∴ ∆ADB ~ ∆ABC (AA similarity criterion)
Adding (i) and (ii), we get
AD. AC + CD . AC = AB2 + BC2
or, AC (AD + CD) = AB2 + BC2
or, AC . AC = AB2 + BC2
or, AC2 = AB2 +BC2
Second Part:
In Fig. 7.50, As AD ⊥ BC
By Pythagoras Theorem, we have
AB2 = AD2 + BD2 …..(i)
AC2 = AD2 + DC2 …..(ii)
Subtracting (ii) from (i)
⇒ AB2 – AC2 = BD2 – DC2 = AB2 + DC2 = BD2 + AC2
Question 15.
In a triangle, if the square on one side is equal to the sum of the squares on the other two sides, prove that the angle opposite to the first side is a right angle. Use the above theorem to find the measure of ∠PKR in Fig. 7.51.
Solution:
Given: A triangle ABC in which AC2 = AB2 + BC2
To Prove: ∠B = 90°.
Construction: We construct a ∆PQR right-angled at Q such that PQ = AB and QR = BC
Proof: Now, from ∆PQR, we have,
Question 16.
ABC is a triangle in which AB = AC and D is a point on AC such that BC2 = AC × CD. Prove that BD = BC.
Solution:
Given: ∆ABC in which AB = AC and D is a point on the side AC such that BC2 = AC × CD
To prove: BD = BC
Construction: Join BD
Proof: We have,
Question 17.
Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.
Solution:
Let ABCD be a square and ABCE and ∆ACF have been drawn on side BC and the diagonal AC respectively.
Question 18.
D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE2 + BD2 = AB2 + DE2.
Solution:
In right angled ΔACE and ΔDCB, we have
AE2 = AC2 + CE2 (Pythagoras Theorem) …(i)
and BD2 = DC2 + BC2… (ii)
Adding (i) and (ii), we have
AE2 + BD2 = AC2 + CE2 + DC2 + BC2
AE2 + BD2 = (AC2 + BC2) + (DC2 + CE2)
AE2 + BD2 = AB2 + DE2
[∵ AC2 + BC2 = AB2 in right-angled triangle ABC and DC2 + EC2 = DE2 in right-angled triangle CDE.]
### Triangles Class 10 Extra Questions HOTS
Question 1.
In Fig. 7.57, ΔFEC ≅ ΔGDB and ∠1 = ∠2. Prove that ΔADE ~ ∆ABC.
Solution:
Since ΔFEC ≅ ΔGDB
Question 2.
Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ∆ABC ~ ∆PQR.
Solution:
Given: In ∆ABC and ∆PQR, AD and PM are their medians respectively
To prove: ∆ABC ~ ∆PQR
Construction: Produce AD to E such that AD = DE and produce PM to N such that PM = MN. Join BE, CE, QN, RN.
Proof: Quadrilateral ABEC and PQNR are ||gm because their diagonals bisect each other at D and M respectively.
⇒ BE = AC and QN = PR
∆ABC ~ ∆PQR (By SAS criterion of similarity)
Question 3.
In Fig. 7.59, P is the mid-point of BC and Q is the mid-point of AP. If BQ when produced meets AC at R, prove that RA = $$\frac{1}{3}$$ CA.
Solution:
Given: In ∆ABC, P is the mid-point of BC, Q is the mid-point of AP such that BQ produced meets AC at R.
To prove: RA = $$\frac{1}{3}$$ CA
Construction: Draw PS || BR, meeting AC at S.
Proof: In ABCR, P is the mid-point of BC and PS || BR.
∴ S is the mid-point of CR.
⇒ CS = SR ….(i)
In ∆APS, Q is the mid-point of AP and QR ||PS
∴ R is the mid-point of AS.
∴ AR = RS …(ii)
From (i) and (ii), we get
AR = RS = SC
⇒ AC = AR + RS + SC = 3 AR
⇒ AR = $$\frac{1}{3}$$AC = $$\frac{1}{3}$$CA
Question 4.
In Fig. 7.60, ABC and DBC are two triangles on the same base BC. If ar(∆ABC) AO AD intersects BC at O, show that $$\frac { ar(∆ABC) }{ ar(∆DBC) }$$ = $$\frac{AO}{DO}$$
Solution:
Given: Two triangles ∆ABC and ADBC which stand on the same base but on opposite sides of BC.
Question 5.
Two poles of height a metres and b metres are p metres apart. Prove that the height of the point of intersection of the lines joining the top of each pole to the foot of the opposite pole is given $$\frac{a b}{a+b}$$ metres.
Solution:
Let AB and CD be two poles of height a and b metres respectively such that the poles are p metres
apart i.e., AC = p metres.
Suppose the lines AD and BC meet at O such that OL = h metres.
Let CL = x and LA = y. Then, x + y = p.
In ∆ABC and ALOC, we have
∠CAB = ∠CLO [Each equal to 90°)
∠C = ∠C [Common]
∴ ∆ABC ~ ∆LOC [By AA criterion of similarity]
Hence, the height of the intersection of the lines joining the top of each pole to the foot of the opposite pole is $$\frac{a b}{a+b}$$ metres.
Question 6.
In an equilateral triangle ABC, D is a point on side BC such that BD = $$\frac{1}{3}$$ BC. Prove that 9AD2 = 7AB2.
Solution:
Given: An equilateral triangle ABC and D be a point on BC such that
BD = $$\frac{1}{3}$$ BC.
Construction: Draw AE ⊥ BC. Join AD.
DE Proof: ∆ABC is an equilateral triangle and AE ⊥ BC
BE = EC
Thus, we have
BD = $$\frac{1}{3}$$ BC and DC = $$\frac{2}{3}$$ BC and BE = EC = $$\frac{1}{2$$ BC
In ∆AEB
AE2 + BE2 = AB2 [Using Pythagoras Theorem]
AE2 = AB2 – BE2
AD2 – DE2 = AB2 – BE2 [∵ In ∆AED, AD2 = AE2 + DE2]
AD2 = AB2 – BE2 + DE2
9AD2 = 9AB2 – 2AB2 [∵ AB = BC]
Question 7.
Through the mid-point M of the side CD of a parallelogram ABCD, the line BM is drawn intersecting AC at L and AD produced to E. Prove that EL = 2BL.
Solution:
In ∆BMC and ∆EMD, we have
MC = MD [∵ M is the mid-point of CD]
∠CMB = ∠DME [Vertically opposite angles]
and ∠MBC = ∠MED [Alternate angles]
So, by AAS criterion of congruence, we have
∆BMC ≅ ∆EMD
⇒ BC = DE [CPCT]
Also, BC = AD [∵ ABCD is a parallelogram]
Now, in ∆AEL and ∆CBL, we have
∠ALE = ∠CLB [Vertically opposite angles]
∠EAL = ∠BCL [Alternate angles]
So, by AA criterion of similarity of triangles, we have
∆AEL ~ ∆CBL |
# 16 Math Magic Tricks for Kids to Calculate Numbers in a Split Second
Over 50% of Americans have math anxiety that often originates from insufficient school practice and a lack of math understanding. In fact, math is much simpler than you think. What if we say that you can add, subtract, multiply, and divide any number by solving simple 5 + 5 or 7 + 2?
Check 16 easy math tricks for kids to teach calculating in a split second.
## Math Tricks for Kids: Addition
Kids learn addition as early as kindergarten, and they will solve thousands of addition equations in five upcoming years, so they will need a few easy math tricks to add numbers in a split second:
### #1 Break Numbers by Place Value
You can break down numbers to more convenient pairs. For example, let’s solve 26 + 38:
1. 26 = 20 + 6.
2. 38 = 30 + 8.
3. Add tens first: 30 + 20 = 50.
4. Add ones: 6 + 8 = 14.
### #2 Focus on 10
Some kids struggle to add numbers past 10, like 9 + 8 or 5 + 7. You can help your kid with the following math trick: round the first number to 10 by borrowing the necessary amount from the second number. Let’s practice this trick with 9 + 8:
1. 9 is close to 10: borrow 1 from 8 to get 10 + 7.
2. 10 + 7 = 17.
Let’s solve 5 + 7 this way:
1. Borrow 5 from 7 to get 10 + 2.
2. 10 + 2 = 12.
You can also apply this strategy to add a one-digit number to a two-digit number. Let’s solve 68 + 9:
1. Break down 68 to 60 and 8.
2. Solve 8 + 9 first. Since 9 is closer to 10, borrow 1 from 8 to get 10 + 7.
3. 7 + 10 = 17.
4. 60 + 17 = 77.
### #3 Break Large Numbers
It’s easy to add 50 to 40 or 15 to 25. But what about 649 + 221? At this point, adults grab their phones and open calculators. On the other hand, teachers expect fifth-graders to solve such math in their minds. Let’s make it a bit easier:
1. Let’s break down these numbers into hundreds, tens, and ones. 649 becomes 600, 40, and 9; 221 becomes 200, 20, and 1.
2. Add hundreds in the first place: 600 + 200 = 800.
3. Add tens: 40 + 20 = 60.
4. Add ones: 9 + 1 = 10.
5. Combine them: 800 + 60 + 10 = 870.
### #4 Compensate Numbers
The compensation trick resembles addition trick #2 on our list. In essence, you borrow from the second addend to round the first one. For example, 249 + 18. Just round 249 to 250 and subtract 1 from 18 to get the following equation: 250 + 17 = 267. As you see, it’s much easier than the original 249 + 18.
You can simplify any addition equation this way:
• 27 + 48 = 30 + 45 = 75.
• 1,235 + 59 = 1,240 + 54 = 1,294.
• 598 + 432 = 600 + 430 = 1030.
1:1 Math Lessons
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## Easy Subtraction Mathematics Tricks for Children
According to Science Direct, students struggle more with subtraction than addition. But these cool math tricks for kids will help them simplify this operation:
### #1 Subtract Three-Digit Number from 1,000 in Seconds
You can subtract any number from 1,000 in your head by taking away two of its digits from 9 and the third digit from 10. Let’s imagine you need to subtract 537 from 1,000:
1. 9 – 5 = 4.
2. 9 – 3 = 6.
3. 10 – 7 = 3.
4. You will get 463 as a result.
### #2 Subtract Numbers by Place Values
Subtraction and addition are inverse math operations, which means you can apply some addition tricks to subtraction. Students can break down large numbers for subtraction as well. Let’s take 258 – 172 into account:
1. Subtract hundreds first: 200 – 100 = 100.
2. Subtract tens: 50 – 70 = – 20.
3. 100 – 20 = 80.
4. Subtract ones: 8 – 2 = 6.
5. Combine the results: 80 + 6 = 86.
### #3 Subtract Convenient Numbers Using Equal Addition
Math operations are all about juggling numbers to get more convenient pairs. You can add equal numbers to subtrahends and minuends to simplify the equation. Let’s solve 98 – 32:
1. Let’s add 2 to both numbers: 98 + 2 = 100; 32 + 2 = 34.
2. Subtract: 100 – 34 = 66.
You can add as much value to numbers as you consider appropriate. Let’s take another example: 36 – 16. In this case, adding 4 to both numbers makes the most sense. As a result, you will get 40 – 20 = 20.
### #4 Use Subtraction Facts
Your kids solve countless math drills at school, memorizing answers to the most popular equations. For instance, 14 – 6 = 8. Now consider that this 14 might be a part of a larger number, let it be 94 or 144.
Knowing that 14 – 6 always results in 8, you can solve seemingly perplexing equations like 154 – 6. Imagine 14 is part of 154. The difference ends in 8, so you can estimate that it will be 148.
Let’s conclude that subtracting a number that ends in 6 from a number that ends in 4 will result in 8: 8,684 – 2396 = 6,288; 344 -166 = 178; 994 – 576 = 418, and so on. Your student can use this fact to check whether they do correct calculations.
## Magic Maths Tricks for Kids to Multiply in Seconds
Students solve multiplication problems every day. Why not make this operation easier using simple math tricks for kids below?
### #1 The Rule of Nines
Students have to learn the multiplication table by heart, but it usually takes a lot of time and drills. The good news is that kids can go an easy way with multiplying single numbers by 9. You will find an interesting pattern when you take a closer look at the multiplication table:
• 1 x 9 = 9
• 2 x 9 = 18
• 3 x 9 = 27
• 4 x 9 = 36
• 5 x 9 = 45
• 6 x 9 = 54
• 7 x 9 = 63
• 8 x 9 = 72
• 9 x 9 = 81
• 9 x 10 = 90
The left digits in the answer increase, while the right digits decrease as we progress through the multiplication table. If you multiply 2 by 9, you know that the first digit of the answer is 1, and the second is 8. The same goes for 3 x 9. The first digit of the answer is 2, while the second one is 7. Apply this pattern to numbers up to 10.
### #2 Multiply Two-Digit Numbers in Seconds
What if we say that multiplying multi-digit numbers is as easy as the multiplication table?
Let’s multiply 27 by 45:
1. Multiply ones: 7 x 5 = 35.
2. Cross multiply digits, add them, and carry whenever a product exceeds 9: 2 x 5 + 7 x 4 = 10 + 28 = 38. Here, we carry 3 and leave 8 as a tens-digit product: 80.
3. Multiply tens and add the carried 3: 2 x 4 + 3 = 8 + 3 = 11. It’s the hundreds-digit answer (eleven hundred or 1,100.
It’s even easier to multiply large numbers if the second step doesn’t exceed 9.
For instance: 13 x 22.
1. Multiply ones: 3 x 2 = 6 (the ones’ product).
2. Cross multiply and add: 1 x 2 + 3 x 2 = 2 + 6 = 8 (the tens’ product or 80).
3. Multiply tens: 1 x 2 = 2 (the hundreds’ product or 200).
4. Just combine digits: 286.
### #3 Add Numbers to Simple Multiplication Products
Multiplying one-digit numbers by 9 is easy, but what if you forget how much 6 x 8 is. If you remember how much 6 x 6 is, you can just add two 6: 6 x 6 = 36; 36 + 6 + 6 = 48.
Multiplication hides a couple of maths trick questions for kids:
• What will you get if you multiply any number by 2? If you multiply something by 2, you will always get an even number. For instance, 12 x 2 = 24; 57 x 2 = 114; 1,657 x 2 = 3,314.
• What two results will you get if you multiply numbers by 5? If you multiply something by 5, you will get either 5 or 0 at the end of the number. Take a look: 23 x 5 = 115; 67 x 5 = 335; 79 x 5 = 395; 324 x 5 = 1,620; 666 x 5 = 3,330.
• When multiplying a number by 10, just add a zero to its end: 10 x 10 = 100; 56 x 10 = 560; 9,887 x 10 = 98,870.
• How can you multiply numbers by 11 without calculating anything? If you want to multiply numbers up to 10 by 11, write them down twice: 9 x 11 = 99; 2 x 11 = 22; 5 x 11 = 55 and so on.
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## Division Maths Tricks for Your Child
The division is the trickiest math operation that confuses everyone without exception. Still, mathematicians came up with some tips to simplify this operation.
### #1 Simplify Numbers for More Convenient Division
You can simplify the dividend and divisor. Let’s take a look at the equation 128 ÷ 34. You can divide it by 2: 64 ÷ 17. It’s more manageable than the original.
### #2 Divide by 5 in a Split Second
Let’s get into tricky maths questions for kids: can they multiply a number to divide it? In short, they can.
If you need to divide a number by 5, let’s say 155 ÷ 5, you can follow these steps:
1. Multiply 155 by 2: 155 x 2 = 310.
2. Divide 310 by 10: 310 ÷ 10 = 31.
3. 31 is the answer: 155 ÷ 5 = 31.
You can also divide numbers by 50 this way. But you have to divide the first quotient by 100 instead of 10. Let’s solve 2,345 ÷ 50:
1. 2,345 x 2 = 4,690.
2. 4,690 ÷ 100 = 46.9.
### #3 Split the Dividend
If you face an equation containing a complex dividend, you can split it into simpler parts. Let’s take 78 ÷ 6 into account:
1. Simplify 78 by splitting it into parts divisible by 6: 78 = 60 + 18.
2. Divide these parts by 6: 60 ÷ 6 = 10; 18 ÷ 6 = 3.
3. Add these results: 10 + 3 = 13. It is the answer to 78 ÷ 6.
### #4 Estimate the Quotient
Sometimes, you don’t need extreme precision when dividing numbers. After all, many division equations result in decimal numbers.
If you can round the dividend and the divisor, do it. For instance, round 98 ÷ 54 to 100 ÷ 50. The estimated quotient for this equation is 2, while the real one is 1.81. Not a big difference.
## A Bonus Mind Reading Math Card Trick to Amaze Your Child
Follow these steps to perform one of the simplest math card tricks for kids:
1. Get a deck of playing cards.
2. Count 25 cards on top of the pile. Slightly turn them to know where they end.
3. Draw these cards and put them face down on the table before you kid.
5. Ask your kid to place this card back on top of the hand.
6. Pick 25 cards from the table and put them back in your hand.
7. Start counting the cards up to 26.
8. It will be your child’s card.
## Wrapping Up Our Maths Tricks for Kids
Now you realize that you can reduce any math operation to small numbers up to the point when you instantly see the answer. Essentially, all available magic math tricks for kids break equations into several steps involving manageable numbers.
While these tricks aim to streamline math, they teach invaluable knowledge – if you consider a challenge impossible to overcome, break it into smaller parts, and you will succeed. Teach this knowledge to your child to prepare them for adulthood.
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• Get practice worksheets for self-paced learning
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# 6 Multiplication Table Printable
Studying multiplication after counting, addition, as well as subtraction is ideal. Young children understand arithmetic through a normal progression. This advancement of discovering arithmetic is usually the pursuing: counting, addition, subtraction, multiplication, and ultimately section. This declaration brings about the query why understand arithmetic in this pattern? Most importantly, why find out multiplication following counting, addition, and subtraction but before department?
## The subsequent information respond to these queries:
1. Young children discover counting initial by associating visible items with their hands. A tangible instance: The amount of apples are there any in the basket? A lot more abstract example is just how older are you presently?
2. From counting figures, the next rational step is addition combined with subtraction. Addition and subtraction tables can be very beneficial training helps for kids because they are visible instruments creating the move from counting less difficult.
3. Which will be discovered up coming, multiplication or section? Multiplication is shorthand for addition. At this point, young children have got a organization understanding of addition. Consequently, multiplication may be the following rational method of arithmetic to find out.
## Assess basic principles of multiplication. Also, evaluate the essentials how to use a multiplication table.
Allow us to review a multiplication illustration. Using a Multiplication Table, multiply several times about three and have an answer 12: 4 x 3 = 12. The intersection of row three and line 4 of any Multiplication Table is 12; 12 is the respond to. For kids beginning to understand multiplication, this really is simple. They could use addition to resolve the trouble hence affirming that multiplication is shorthand for addition. Example: 4 x 3 = 4 4 4 = 12. It is an exceptional summary of the Multiplication Table. The additional benefit, the Multiplication Table is visual and mirrors to understanding addition.
## In which should we get started studying multiplication utilizing the Multiplication Table?
1. Very first, get informed about the table.
2. Start out with multiplying by a single. Start off at row number 1. Relocate to line number one. The intersection of row 1 and column the initial one is the best solution: a single.
3. Repeat these methods for multiplying by one particular. Multiply row one particular by columns one particular by means of a dozen. The replies are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12 correspondingly.
4. Recurring these actions for multiplying by two. Flourish row two by columns a single through 5. The responses are 2, 4, 6, 8, and 10 respectively.
5. We will jump in advance. Perform repeatedly these methods for multiplying by five. Grow row several by posts 1 by way of a dozen. The responses are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, and 60 correspondingly.
6. Now we will increase the degree of trouble. Replicate these methods for multiplying by about three. Flourish row about three by posts one particular by way of 12. The solutions are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, and 36 correspondingly.
7. In case you are confident with multiplication so far, try out a check. Fix the following multiplication troubles in your head and after that examine your responses to the Multiplication Table: multiply 6 and two, multiply nine and a few, increase one particular and 11, flourish several and a number of, and grow several and 2. The situation answers are 12, 27, 11, 16, and 14 respectively.
If you received four out from 5 various problems appropriate, design your personal multiplication assessments. Compute the answers in your thoughts, and appearance them making use of the Multiplication Table. |
# Teaching Kids Programming – Algorithms to Compute the Alternating Digit Sum
Teaching Kids Programming – Algorithms to Compute the Alternating Digit Sum | ninjasquad
Teaching Kids Programming: Videos on Data Structures and Algorithms
You are given a positive integer n. Each digit of n has a sign according to the following rules:
• The most significant digit is assigned a positive sign.
• Each other digit has an opposite sign to its adjacent digits.
• Return the sum of all digits with their corresponding sign.
Example 1:
Input: n = 521
Output: 4
Explanation: (+5) + (-2) + (+1) = 4.
Example 2:
Input: n = 111
Output: 1
Explanation: (+1) + (-1) + (+1) = 1.
Example 3:
Input: n = 886996
Output: 0
Explanation: (+8) + (-8) + (+6) + (-9) + (+9) + (-6) = 0.
Constraints:
1 <= n <= 10^9
Hints:
The first step is to loop over the digits. We can convert the integer into a string, an array of digits, or just loop over its digits.
Keep a variable sign that initially equals 1 and a variable answer that initially equals 0.
Each time you loop over a digit i, add sign * i to answer, then multiply sign by -1
### Compute the Alternating Digit Sum From Left to Right
We can convert to string (from the left to the right aka from the Most Significant Digit MSD), and thus it is just doing the math – do what exactly what is being told. Converting to string requires O(N) space, where N is the number of digits. Also, by iterating over the digits, we need to convert characters to integer. The time complexity is O(N) as well.
```1 2 3 4 5 6 7 8 9 ``` ```class Solution: def alternateDigitSum(self, n: int) -> int: ans = 0 s = str(n) sign = 1 for i in s: ans += int(i) * sign sign = -sign return ans```
```class Solution:
def alternateDigitSum(self, n: int) -> int:
ans = 0
s = str(n)
sign = 1
for i in s:
ans += int(i) * sign
sign = -sign
return ans```
### Compute the Alternating Digit Sum From Right to Left
Another way is to do it from right to left, where we can use math % operator to extract the Least Significant Digit (LSD). We can start the sign with minus, flip it along the way to the left, and fix the sign at the end.
```1 2 3 4 5 6 7 8 9 ``` ```class Solution: def alternateDigitSum(self, n: int) -> int: ans = 0 sign = 1 while n > 0: sign = - sign ans += sign * (n % 10) n //= 10 return sign * ans```
```class Solution:
def alternateDigitSum(self, n: int) -> int:
ans = 0
sign = 1
while n > 0:
sign = - sign
ans += sign * (n % 10)
n //= 10
return sign * ans```
For example, 12, -2+1, and the sign is plus at the end, thus the answer is -1*1=-1, which is correct.
And 123, -3+2-1=-2, the sign is minus at the end, so the answer is -2*-1=2, which is correct: 1-2+3=2.
The time complexity is O(N) where N is the number of digits in the decimal format. The space complexity is O(1) constant.
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Mean and Standard Deviation
A workshop explaining and demonstrating the mean and standard deviation.
MACHINE LEARNING WORKSHOP
Mean and Standard Deviation
This workshop is about two fundamental measures of data. I want to you start thinking about how you can best describe or summarise data. How can we best take a set of data and describe that data in as few variables as possible? These are called summary statistics because they summarise statistical data. In other words, this is your first model!
import numpy as np
Mean
The mean, also known as the average, is a measure of the tendency of the data. For example, if you were provided some data then you could say that, on average, is most likely best represented by the mean.
The mean is calculated as:
$$\mu = \frac{\sum_{i=0}^{N-1}{ x_i }} {N}$$
The sum of all observations divided by the number of observations.
x = [6, 4, 6, 9, 4, 4, 9, 7, 3, 6];
N = len(x)
x_sum = 0
for i in range(N):
x_sum = x_sum + x[i]
mu = x_sum / N
print("μ =", mu)
μ = 5.8
Of course, we should be using libraries to reduce the amount of code we have to write. For low level tasks such as this, the most common library is called Numpy.
We can rewrite the above as:
N = len(x)
x_sum = np.sum(x)
mu = x_sum / N
print("μ =", mu)
μ = 5.8
We can take this even further and just use Numpy’s implementation of the mean:
print("μ =", np.mean(x))
μ = 5.8
Standard Deviation
To describe our data, the mean alone doesn’t provide enough information. It tells us what value we should observe on average. But the values could be +/- 1 or +/- 100 of that value. (+/- is shorthand for “plus or minus”, i.e. “could be greater than or less than this value”).
To provide this information we need a measure of “spread” around the mean. The most common measure of “spread” is the standard deviation.
Read more about the standard deviation at: WinderResearch.com - Why do we use Standard Deviation and is it Right?.
The standard deviation of a population is:
$$\sigma = \sqrt{ \frac{\sum_{i=0}^{N-1}{ (x_i - \mu )^2 }} {N} }$$
x = [6, 4, 6, 9, 4, 4, 9, 7, 3, 6];
N = len(x)
mu = np.mean(x)
print("μ =", mu)
μ = 5.8
print("Deviations from the mean:", x - mu)
print("Squared deviations from the mean:", (x - mu)**2)
print("Sum of squared deviations from the mean:", ((x - mu)**2).sum() )
print("Mean of squared deviations from the mean:", ((x - mu)**2).sum() / N )
Deviations from the mean: [ 0.2 -1.8 0.2 3.2 -1.8 -1.8 3.2 1.2 -2.8 0.2]
Squared deviations from the mean: [ 0.04 3.24 0.04 10.24 3.24 3.24 10.24 1.44 7.84 0.04]
Sum of squared deviations from the mean: 39.6
Mean of squared deviations from the mean: 3.96
print("σ =", np.sqrt(((x - mu)**2).sum() / N ))
σ = 1.98997487421
Again, we don’t need to code this all up. The Numpy equivalent is:
print("σ =", np.std(x))
σ = 1.98997487421
What’s the Catch?
You knew they’d be a catch, right? ;-)
I didn’t mention it at the start, but the two previous measures of the central tendency and the spread are specific to a very special combination of data.
If the observations are distributed in a special way, then these metrics perfectly model the underlying data. If not, then these metrics are invalid.
You probably said “huh?” to a few of those new words, so let’s go through them. |
# Equation $(a-3)cb=a(c+b)$ for natural numbers.
Let $a$, $b$, and $c$ be positive integers. Suppose that $c \leq b \leq a$ and that they satisfy the relation $$(a-3)cb=a(c+b).$$ What can be said about the solutions?
• it's symmetric in $b$ and $c$ Jul 15, 2013 at 16:59
• $1-3/a=1/b+1/c$ Jul 15, 2013 at 17:00
This equation can be rewritten as $$\frac{3}{a}+\frac{1}{b}+\frac{1}{c}=1.$$ Now
• If $c>5$, then there is no solutions (the lhs $<1$).
• If $c=5$, then the only solution is $a=b=c=5$.
• If $c=4$ and $b>5$, then there is no solutions (the lhs $<1$).
• If $c=4$ and $b=4$, then $a=6$.
• If $c=3$ and $b>6$, then there is no solutions (the lhs $<1$).
• If $c=3$ and $b=6$, then $a=6$.
• If $c=3$ and $b=4,5$, then there is no solutions (direct verification).
• If $c=3$ and $b=3$, then $a=9$.
• If $c=2$ and $b>8$, then there is no solutions (the lhs $<1$).
• If $c=2$ and $b=8$, then $a=8$.
• If $c=2$ and $b=7$, then there is no solutions.
• And finally, we have solutions $(a,b,c)=(9,6,2), (10,5,2), (12,4,2), (18,3,2)$.
Hence the complete list of triples $(a,b,c)$ is: $$(5,5,5),(6,4,4),(9,3,3),(8,8,2),(9,6,2), (10,5,2), (12,4,2), (18,3,2).$$
Note that $c \leq 5$. Now consider $c=1,2,3,4,5$. Check for each case! You can easily find $(5,5,5),(6,4,4),(9,3,3),(8,8,2),(9,6,2), (10,5,2), (12,4,2), (18,3,2)$ are the only solutions. |
# Best Rational Approximations to Root 2 generate Pythagorean Triples
## Theorem
$\sequence S := \dfrac 1 1, \dfrac 3 2, \dfrac 7 5, \dfrac {17} {12}, \dfrac {41} {29}, \dfrac {99} {70}, \dfrac {239} {169}, \dfrac {577} {408}, \ldots$
Every odd term of $\sequence S$ can be expressed as:
$\dfrac {2 a + 1} b$
such that:
$a^2 + \paren {a + 1}^2 = b^2$
## Proof 1
The numerators of the terms of $\sequence S$ are all odd.
For all $n$, the parity of the denominator of term $S_n$ is the same as the parity of $n$.
Thus it follows that every other term of $\sequence S$ has a numerator and a denominator which are both odd.
This proof proceeds by induction.
### Basis for the Induction
$\dfrac 1 1$ can be expressed as $\dfrac {2(0) + 1} 1$, and:
$0^2 + \paren {0 + 1}^2 = 1^2$
### Induction Hypothesis
This is our induction hypothesis:
The best rational approximation $\dfrac {p_k} {q_k}$ of $\sqrt 2$, when expressed as:
$\dfrac {2 a + 1} b$
gives the relation:
$a^2 + \paren {a + 1}^2 = b^2$
We need to show that the best rational approximation $\dfrac {p_{k + 2} } {q_{k + 2} }$ of $\sqrt 2$, when expressed as:
$\dfrac {2 a' + 1} {b'}$
also give the relation:
$a'^2 + \paren {a' + 1}^2 = b'^2$
### Induction Step
This is our induction step:
From the induction hypothesis we have:
$a = \dfrac {p_k - 1} 2, b = q_k$
and thus:
$\paren {\dfrac {p_k - 1} 2}^2 + \paren {\dfrac {p_k - 1} 2 + 1}^2 = q_k^2$
Expanding, we have:
$\dfrac {p_k^2} 2 + \dfrac 1 2 = q_k^2$
Now by Relation between Adjacent Best Rational Approximations to Root 2, we have:
$\dfrac {p_{k + 1} } {q_{k + 1} } = \dfrac {p_k + 2 q_k} {p_k + q_k}$
We check that, via GCD with Remainder:
$\gcd \set {p_k + 2 q_k, p_k + q_k} = \gcd \set {q_k, p_k + q_k} = \gcd \set {q_k, p_k} = 1$
Since both fractions are in canonical form and Canonical Form of Rational Number is Unique, we can write:
$p_{k + 1} = p_k + 2 q_k$
$q_{k + 1} = p_k + q_k$
Therefore:
$p_{k + 2} = 3 p_k + 4 q_k$
$q_{k + 2} = 2 p_k + 3 q_k$
We need to show that:
$\paren {\dfrac {p_{k + 2} - 1} 2}^2 + \paren {\dfrac {p_{k + 2} - 1} 2 + 1}^2 = q_k^2$
or:
$\dfrac {p_{k + 2}^2} 2 + \dfrac 1 2 = q_{k + 2}^2$
We have:
$\ds \frac {p_{k + 2}^2} 2 + \frac 1 2$ $=$ $\ds \frac {\paren{3 p_k + 4 q_k}^2} 2 + \frac 1 2$ $\ds$ $=$ $\ds \frac {9 p_k^2 + 24 p_k q_k + 16 q_k^2} 2 + \frac 1 2$ Square of Sum $\ds$ $=$ $\ds \frac {p_k^2} 2 + \frac 1 2 + 4 p_k^2 + 12 p_k q_k + 8 q_k^2$ $\ds$ $=$ $\ds 4 p_k^2 + 12 p_k q_k + 9 q_k^2$ $\ds$ $=$ $\ds \paren {2 p_k + 3 q_k}^2$ Square of Sum $\ds$ $=$ $\ds q_{k + 2}^2$
The result follows by induction.
$\blacksquare$
## Proof 2
From Pell Number as Sum of Squares, we have:
$P_{2 n + 1} = P_{n + 1}^2 + P_n^2$
Therefore:
$\ds P_{2 n + 1}^2$ $=$ $\ds \paren {P_{n + 1}^2 + P_n^2}^2$ $\ds$ $=$ $\ds P_{n + 1}^4 + 2 P_{n + 1}^2 P_n^2 + P_n^4$ Square of Sum $\ds$ $=$ $\ds P_{n + 1}^4 + 2 P_{n + 1}^2 P_n^2 + P_n^4 + 2 P_{n + 1}^2 P_n^2 - 2 P_{n + 1}^2 P_n^2$ adding zero $\ds$ $=$ $\ds P_{n + 1}^4 - 2 P_{n + 1}^2 P_n^2 + P_n^4 + 4 P_{n + 1}^2 P_n^2$ regrouping $\ds$ $=$ $\ds \paren {P_{n + 1}^2 - P_n^2}^2 + \paren {2 P_{n + 1} P_n}^2$
$\blacksquare$ |
# How do you find the fourth term in a sequence?
## How do you find the fourth term in a sequence?
there is an obvious pattern. Such sequences can be expressed in terms of the nth term of the sequence. In this case, the nth term = 2n. To find the 1st term, put n = 1 into the formula, to find the 4th term, replace the n’s by 4’s: 4th term = 2 × 4 = 8.
## What is the common difference for this arithmetic sequence 4/9 14?
5
Common difference between 4 and 9 = |9-4| =5. Common difference between 9 and 14 = |14-9| =5.
What is the common difference of the arithmetic sequence 4/7 10 13?
3
The common difference of the arithmetic sequence 4, 7, 10, 13, 16,… is 3. So, the correct answer is “4, 7, 10, 13, 16,… is 3”.
READ: What is the most famous food in Jordan?
### What is a sequence of numbers in math?
1 A number sequence is a list of numbers that follow a pattern. 2 An example of a number sequence is 5, 9, 13, 17. 3 To find the difference between each number, subtract one number from the number that comes after it. 4 9 – 5 = 4 and so, the common difference is 4. 5 Each number in a sequence is called a term.
### How do you find the difference between two numbers in a sequence?
A sequence can be formed by adding the same number each time. Subtract one number from the number that comes after it to find this common difference. To find the difference between each number, we subtract one term from the term that is immediately after it.
How do you find the next term in a sequence?
Each number in a sequence is called a term. To find the next term in the sequence, add 4 to 17 to make 21. A sequence can be formed by adding the same number each time. Subtract one number from the number that comes after it to find this common difference.
READ: What is skinning an app?
## What is the difference between arithmetic sequences and geometric sequences?
Arithmetic sequences are lists of numbers that have successive terms with constant difference, while geometric sequences are lists of numbers with successive term with constant ratio. Arithmetic sequences have a common difference, while geometric sequences have a common ratio |
# Chapter 6: The Shape, Center and Spread of a Normal Distribution
Difficulty Level: Basic Created by: CK-12
Introduction
What a way to start math class! Imagine your math teacher allowing you to bounce a basketball around the classroom.
Prior to the beginning of class, go to the physical education department of your school and borrow some basketballs. These will be used in this chapter to introduce the concept of a normal distribution. You may find your class to be somewhat noisy, but you're sure to enjoy the activity. Begin the class by reviewing the concept of a circle and the terms associated with the measurements of a circle. You and your classmates should be able to provide these facts based on your previous learning. To ensure that everyone understands the concepts of center, radius, diameter, and circumference, work in small groups to create posters to demonstrate your understanding. These posters can then be displayed around the classroom. The following is a sample of the type of poster you may create:
Once you and your classmates have completed this and you are confident that you all understand the measurements associated with a circle, distribute the basketballs to everyone in the class. While your classmates are playing with the balls, set up a table with tape, rulers, string, scissors, markers, and any other materials that you think you may find useful to answer the following questions:
• What is the circumference of the basketball?
• What is the diameter of the basketball?
• What is the radius of the basketball?
• How did you determine these measurements?
• What tools did you use to find your answers?
When playtime is over, use the tools provided to answer all of the above questions. It is the job of you and your classmates to determine a method of calculating these measurements. For the questions that require numerical measurements as answers, take 2 measurements for each. You must plot your results for the diameter of the basketball, so remember to record your data.
Oops! Don't forget that the ruler cannot go through the basketball!
When you have answered the above questions, plot your 2 results for the diameter of the basketball on a large sheet of grid paper, and have your classmates do the same.
Chapter Outline
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# [Maths Class Notes] on Volume of a Cylinder Pdf for Exam
A cylinder is a closed solid having parallel circular bases attached by a curved surface. It is a three-dimensional shape with two circular ends called bases which are congruent and parallel to each other. The surface that connects the two circular ends is basically a rectangle.
Height is the perpendicular distance between two circular surface and radius of a cylinder is basically the radius of the circular base.
### A cylinder can be viewed as:
1. Elongated circle made by circles kept one above the other. Examples- Cylinder made out of bangles.
2. Combination of two circles and a quadrilateral (square, rectangle, etc)
A cylinder can be hollow or solid. A hollow cylinder is empty from inside whereas a solid cylinder is packed up with the same material. Layers of circular surfaces can be extracted from a solid cylinder if it is dissected cross-sectionally.
### The volume of a Cylinder
The volume of a cylinder is the space occupied by a cylinder. For a hollow cylinder, it is the space enclosed inside a cylinder and for solid cylinder, it is the amount of material the cylinder is composed of.
### How to Calculate Volume of a Cylinder of Unknown Height and Radius
Take a cylinder of unknown radius, a container, a trough and a measuring cylinder.
Arrange the apparatus as shown below. Fill the container with water up to the brim and then carefully place the cylinder in the container. This will make some water flow out of the container which will get collected in the trough. Pour the water from trough to measuring cylinder to find out the amount of water displaced by the cylinder.
The amount of displaced water is equal to the space occupied by the cylinder.
The Volume of Cylinder = Amount of Displaced Water
### Cylindrical Volume Formula Of Known Height and Radius.
As discussed earlier, a cylinder can be seen as circles kept one above the other. If the breadth of the circle is of unit length then the height of the cylinder will be equal to the number of circles. The cylinder volume formula is shown below:
Volume = length x breadth x height
Volume = area of surface x height ———————–(i)
Where area of surface= length x breadth
Now for a cylinder, the area of the surface is equal to the area of the base (circle)
The volume of c formula = area of circular base x height
= πr2 × h
Therefore,
The Equation for Volume of a Cylinder = πr2h.
r= radius of the circular base
h= the perpendicular distance between two circular bases.
### The Volume of a Hollow Cylinder of Known Height and Radius.
Let us suppose a hollow cylinder of outer radius R, inner radius r and height h. The hollow cylinder is actually the result of a smaller cylinder cut out from the bigger cylinder.
The formula for calculating the volume of a cylinder (hollow):
= Volume of the bigger solid cylinder – Volume of the smaller solid cylinder.
= πR2h – πr2h
= πh (R2r2)
Therefore, the cylinder formula = πh (R2r2) cubic units.
### The Conversion of the Volume of a Cylinder
Unit of the volume is cubic centimetre. Thousand cubic centimetres is equal to one litre. Thus, to convert a value from litre to cm3, 1000 is multiplied to the number. Similarly, to convert cm3 to litre, the number is divided by 1000.
1 Litre = 1000 cubic cm or cm3
1 cm3 = 1/1000 litres
### Solved Examples:
1) A cylinder has a base of radius 14 cm and the height of 10 cm. Find the volume.
Solution:
2) If the base of a cylinder is of area 30cm2 and height is 6 cm then find the volume of the cylinder.
Solution:
Volume = Area of the base x height of a cylinder
= 30 cm2 x 6 cm
= 180 cm3 |
# Square Root of X + Solution With Free Steps
The square root of X can be expressed as √X, which is equivalent to y, where the symbol is called the square root. The square root of a number produces the number that, when multiplied by itself, produces the actual number. For instance, the square root of X is y which means y x y = x
Figure 1 – What is Square Root?
In this article, we will analyze and find the square root of X using various mathematical techniques, such as the approximation method and the long division method.
## What Is the Square Root Of X?
The square root of the number X is y.
The square root can be defined as the quantity that can be doubled to produce the square of that similar quantity. In simple words, it can be explained as:
√X = √(y x y)
√X = √(y)$^2$
√X = ±y
The square can be canceled with the square root as it is equivalent to 1/2; therefore, obtaining y. Hence y is X’s square root. The square root generates both positive and negative integers.
## How To Calculate the Square Root of X?
You can calculate the square root of X using any of two vastly used techniques in mathematics; one is the Approximation technique, and the other is the Long Division method.
Figure 2 – Basic definition of Square Root
The symbol √ is interpreted as X raised to the power 1/2. So any number, when multiplied by itself, produces its square, and when the square root of any squared number is taken, it produces the actual number.
Let us discuss each of them to understand the concepts better.
### Square Root by Long Division Method
The process of long division is one of the most common methods used to find the square roots of a given number. It is easy to comprehend and provides more reliable and accurate answers. The long division method reduces a multi-digit number to its equal parts.
Learning how to find the square root of a number is easy with the long division method. All you need are five primary operations- divide, multiply, subtract, bring down or raise, then repeat.
Following are the simple steps that must be followed to find the square root of X using the long division method:
### Step 1
First, write the given number X in the division symbol, as shown in figure 1.
### Step 2
Starting from the right side of the number, divide the number X into pairs such as bc and a.
### Step 3
Now divide the digit a by a number, giving a number either a or less than a. Therefore, in this case, the remainder is zero or less than a (depending upon the situation, select according to the given number X), whereas the quotient is one or a.
### Step 4
After this, bring down the next pair bc. Now the dividend is bc. To find the next divisor, we need to double our quotient obtained before. Doubling 1 gives 2; hence consider it as the next divisor.
### Step 5
Now pair 2 with another number to make a new divisor that results in $\leq$ bc when multiplied with the divisor. If the number is not a perfect square, add pair of zeros to the right of the number before starting division.
### Step 6
Adding p to the divisor and multiplying 2p with p results in m $\leq$ bc. The remainder obtained is m1. Move the next pair of zeros down and repeat the same process mentioned above.
### Step 7
Keep on repeating the same steps till the zero remainder is obtained or if the division process continues infinitely, solve to two decimal places.
### Step 8
The resulting quotient y is the square root of X. Figure 1 given below shows the long division process in detail:
Figure 2 – Square root of 123
### Square Root by Approximation Method
The approximation method involves guessing the square root of the non-perfect square number by dividing it by the perfect square lesser or greater than that number and taking the average.
The given detailed steps must be followed to find the square root of X using the approximation technique.
### Step 1
Consider a perfect square number x1 less than X.
### Step 2
Now divide X by x1.
X ÷ x1 = y1
### Step 3
Now take the average of x1 and y1. The resulting number is approximately equivalent to the square root of X.
(x1 + y1) ÷ 2 = y
### Important points
• The number X is a perfect square/ not a perfect square.
• The number X is a rational number/ irrational number.
• The number X can be split into its prime factorization.
## Is Square Root of X a Perfect Square?
The number X is a perfect square/ not a perfect square. A number is a perfect square if it splits into two equal parts or identical whole numbers. If a number is a perfect square, it is also rational.
A number expressed in p/q form is called a rational number. All the natural numbers are rational. A square root of a perfect square is a whole number; therefore, a perfect square is a rational number.
A number that is not a perfect square is irrational as it is a decimal number. As far as X is concerned, it is a perfect square / not a perfect square. It can be proved as below:
(If X is a perfect square)
Factorization of X results in a x a that can also be expressed as a$^2$.
Taking the square root of the above expression gives:
= √(a$^2$)
= (a$^2$)$^{1/2}$
= a
This shows that X is a perfect square and a rational number.
(If X is not a perfect square)
Factorization of X results in a x b.
Taking the square root of the above expression gives:
= √(a x b)
= (a x b)$^{1/2}$
= a.a1a2
This shows that X is not a perfect square as it has decimal places; hence it is an irrational number.
Therefore the above discussion proves that the square root of X is equivalent to y.
Images/mathematical drawings are created with GeoGebra. |
# What is the area of an isosceles triangle with two equal sides of 10 cm and a base of 12 cm?
Nov 21, 2015
Area $= 48$ $c {m}^{2}$
#### Explanation:
Since an isosceles triangle has two equal sides, if the triangle is split in half vertically, the length of the base on each side is:
$12$ $c m$$\div 2 =$$6$ $c m$
We can then use the Pythagorean theorem to find the height of the triangle.
The formula for the Pythagorean theorem is:
${a}^{2} + {b}^{2} = {c}^{2}$
To solve for the height, substitute your known values into the equation and solve for $a$:
where:
$a$ = height
$b$ = base
$c$ = hypotenuse
${a}^{2} + {b}^{2} = {c}^{2}$
${a}^{2} = {c}^{2} - {b}^{2}$
${a}^{2} = {\left(10\right)}^{2} - {\left(6\right)}^{2}$
${a}^{2} = \left(100\right) - \left(36\right)$
${a}^{2} = 64$
$a = \sqrt{64}$
$a = 8$
Now that we have our known values, substitute the following into the formula for area of a triangle:
$b a s e = 12$ $c m$
$h e i g h t = 8$ $c m$
$A r e a = \frac{b a s e \cdot h e i g h t}{2}$
$A r e a = \frac{\left(12\right) \cdot \left(8\right)}{2}$
$A r e a = \frac{96}{2}$
$A r e a = 48$
$\therefore$, the area is $48$ $c {m}^{2}$. |
What is differentiation? - Future Study Point
# What is differentiation?
Differentiation is the derivative of a function with respect to an independent variable, a derivative of the function is a change in a function with respect to an independent variable.In class 9 you have studied the polynomial like y=x² +2,y = x +1, y = 2x² + x + 2 etc, where x is the independent variable and y is the function of x means the value of y depends on x therefore we can rewrite these polynomial equations in the form as an example f(x) = x² +2. If we draw a graph between f(x) and x , and consider an infinitesimal change in x is Δx then a change in function is Δf(x), therefore differentiation of f(x) with respect to x is determined by Δf(x)/Δx, So differentiation of a function is the ratio between the change in the function to a change in an independent variable which is written as dy/dx or f ‘(x) in calculus.
You can also study
Continuity and Differentiability of a function
## What is differentiation?
Table of content
• What is differentiation?
• Derivatives of a function as limits
• Formula of differentiation
• Rules of differentiation
• Application of differentiation
### What is differentiation?
Differentiation is the derivative of a function with respect to an independent variable. Differentiation in mathematics is the measurement of per unit change in the function with respect to an independent variable.
Let there is a function y = f(x)
The per-unit change in the function with respect to an independent variable ‘x’ is given by dy/dx is called differentiation of y with respect to x.
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### Derivatives of a function as limits
Let there be a real values function f(x) and x is a point on its defined domain,let ‘h’ is an infinitesimal change (i.e nearest to 0) then the derivative of the function, f ‘(x) is given by
### Formula of differentiation
f ‘(sin x) = cos x
f ‘(cos x) =-sin x
f ‘(tan x) = sec²x
f ‘(cot x) = -cosec²x
f ‘(sec x) = tan x sec x
f ‘(cosec x) = -cot x cosecx
f ‘(log x) = 1/x
### Rules of differentiation
Sum or difference rule
Let f(x) = u(x) ± v(x)
Then f ‘( x) = u'(x) ± v'(x)
Product rule
Let f(x) = u(x)× v(x)
Then f ‘( x) = u'(x)v(x) + v'(x)u(x)
Quotient rule
Application of differentiation in real life
With the help of differentiation, we can determine the rate of change of a quantity with respect to another qantity, if one quantity is the function of another quantity.
• Velocity is the rate of change of the displacement, if displacement is S and time is t then velocity(v) is given by v=ds/dt.
• Acceleration is the rate of change of the velocity, if velocity is v and time is t then acceleration(a) is given by a=dv/dt.
• Differentiation helps in determining the maxima and minima of a curve.
• It also helps in determining the tangent of the curve.
## NCERT Solutions of Science and Maths for Class 9,10,11 and 12
### NCERT solutions for class 12 maths
Chapter 1-Relations and Functions Chapter 9-Differential Equations Chapter 2-Inverse Trigonometric Functions Chapter 10-Vector Algebra Chapter 3-Matrices Chapter 11 – Three Dimensional Geometry Chapter 4-Determinants Chapter 12-Linear Programming Chapter 5- Continuity and Differentiability Chapter 13-Probability Chapter 6- Application of Derivation CBSE Class 12- Question paper of maths 2021 with solutions Chapter 7- Integrals Chapter 8-Application of Integrals
Class 12 Solutions of Maths Latest Sample Paper Published by CBSE for 2021-22 Term 2
Class 12 Maths Important Questions-Application of Integrals
Class 12 Maths Important questions on Chapter 7 Integral with Solutions for term 2 CBSE Board 2021-22
Solutions of Class 12 Maths Question Paper of Preboard -2 Exam Term-2 CBSE Board 2021-22
Solutions of class 12 maths question paper 2021 preboard exam CBSE Solution
### NCERT Solutions for class 9 maths
Chapter 1- Number System Chapter 9-Areas of parallelogram and triangles Chapter 2-Polynomial Chapter 10-Circles Chapter 3- Coordinate Geometry Chapter 11-Construction Chapter 4- Linear equations in two variables Chapter 12-Heron’s Formula Chapter 5- Introduction to Euclid’s Geometry Chapter 13-Surface Areas and Volumes Chapter 6-Lines and Angles Chapter 14-Statistics Chapter 7-Triangles Chapter 15-Probability Chapter 8- Quadrilateral
### NCERT Solutions for class 10 maths
Chapter 1-Real number Chapter 9-Some application of Trigonometry Chapter 2-Polynomial Chapter 10-Circles Chapter 3-Linear equations Chapter 11- Construction Chapter 4- Quadratic equations Chapter 12-Area related to circle Chapter 5-Arithmetic Progression Chapter 13-Surface areas and Volume Chapter 6-Triangle Chapter 14-Statistics Chapter 7- Co-ordinate geometry Chapter 15-Probability Chapter 8-Trigonometry
CBSE Class 10-Question paper of maths 2021 with solutions
CBSE Class 10-Half yearly question paper of maths 2020 with solutions
CBSE Class 10 -Question paper of maths 2020 with solutions
CBSE Class 10-Question paper of maths 2019 with solutions
### NCERT Solutions for class 11 maths
Chapter 1-Sets Chapter 9-Sequences and Series Chapter 2- Relations and functions Chapter 10- Straight Lines Chapter 3- Trigonometry Chapter 11-Conic Sections Chapter 4-Principle of mathematical induction Chapter 12-Introduction to three Dimensional Geometry Chapter 5-Complex numbers Chapter 13- Limits and Derivatives Chapter 6- Linear Inequalities Chapter 14-Mathematical Reasoning Chapter 7- Permutations and Combinations Chapter 15- Statistics Chapter 8- Binomial Theorem Chapter 16- Probability
CBSE Class 11-Question paper of maths 2015
CBSE Class 11 – Second unit test of maths 2021 with solutions
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# Completion of Squares
## Exhibit: table of content
##### Oliver Knill, SciCtr 434, knill@math.harvard.edu
The "completion of squares" is used to find solutions to quadratic equations. In multivariable calculus, you use it to find the center of circles or spheres or more generally to understand quadrics, surfaces of the form g(x,y,z)=0 where g is quadratic.
The idea is one of the oldest ideas in Mathematics. The mathematician Al Khwarizmi, used both algebraic and geometric methods to solve quadratic equations writes for example to describe the solution of
x2 + 10 x = 39:
``` x2 + b x + c = 0 x2 + b x + b2/4 = b2/4 -c (x+b/2)2 = b2/4 -c x = (b2/4 -c)1/2 - b/2 ```
"... a square and 10 roots are equal to 39 units. The question therefore in this type of equation is about as follows: what is the square which combined with ten of its roots will give a sum total of 39? The manner of solving this type of equation is to take one-half of the roots just mentioned. Now the roots in the problem before us are 10. Therefore take 5, which multiplied by itself gives 25, an amount which you add to 39 giving 64. Having taken then the square root of this which is 8, subtract from it half the roots, 5 leaving 3. The number three therefore represents one root of this square, which itself, of course is 9. Nine therefore gives the square.
The picture to the left had been drawn by Al Khwarzimi and it gives a geometric meaning to the expression "completion of squares". If we want to write x2 + 4 b x as a complete square, we interpret x2 as a square and 4 b x as 4 rectangles adjoint to the square. To have a full square, add 4 squares of length b: x2 + 4 b x + 4 b2 = (4+2b)2.
Note that Al Khwarzimi, the father of algebra, did not yet use any variables like "x" to derive the solutions of quadratic equations. He adds 25 to both sides of the equation x2 + 10 x = 39 to get (x+5)2 = 82 so that x=3.
Please send comments to math21a@fas.harvard.edu
Oliver Knill, Math21a, Multivariable Calculus, Fall 2005, Department of Mathematics, Faculty of Art and Sciences, Harvard University |
# Class 10 Maths Activity- To verify that the lengths of tangents to a circle from some external point are equal.
OBJECTIVE
To verify that the lengths of tangents to a circle from some external point are equal.
MATERIAL REQUIRED
Glazed papers of different colours, geometry box, sketch pen, scissors, cutter and glue
METHOD OF CONSTRUCTION
1. Draw a circle of any radius, say 3 cm, with centre O on a coloured glazed paper of a convenient size.
2. Take any point P outside the circle.
3. Place a ruler touching the point P and the circle, lift the paper and fold it to create a crease passing through P
4. Created crease is a tangent to the circle from the point P. Mark the point of contact of the tangent and the circle as Q. Join PQ.
5. Now place ruler touching the point P and the other side of the circle, and fold the paper to create a crease again.
6. This crease is the second tangent to the circle from the point P. Mark the point of contact of the tangent and the circle as R. Join PR.
7. Join the centre of the circle O to the point P.
DEMONSTRATION
1. Fold the circle along OP.
2. We observe that Q coincides with R. Therefore, QP = RP, i.e.,
length of the tangent QP = length of the tangent RP.
This verifies the result.
OBSERVATION
On actual measurement:
1. Length of tangent QP = 7.5 cm
2. Length of tangent RP = 7.5 cm
So, length of tangent QP = length of tangent RP
APPLICATION
This result is useful in solving problems in geometry and mensuration. |
### 6.8
```§ 6.8
Modeling Using Variation
Direct Variation
Direct Variation
If a situation is described by an equation in the form
y = kx
where k is a constant, we say that y varies directly as x. The number
k is called the constant of variation.
Solving Variation Problems
1) Write an equation that describes the given English statement.
2) Substitute the given pair of values into the equation in step 1 and
find the value of k.
3) Substitute the value of k into the equation in step 1.
4) Use the equation from step 3 to answer the problem’s question.
Blitzer, Intermediate Algebra, 4e – Slide #121
Direct Variation
EXAMPLE
Use the 4-step procedure for solving the variation problem.
y varies directly as x. y = 45 when x = 5. Find y
when x = 13.
SOLUTION
1) Write an equation. We know that y varies directly as x is
expressed as
y = kx.
2) Use the given values to find k.
y = kx
Given equation.
Blitzer, Intermediate Algebra, 4e – Slide #122
Direct Variation
CONTINUED
45 = k(5)
Replace y with 45 and x
with 5.
Divide both sides by 9.
9=k
3) Substitute the value of k into the equation.
y = 9x
Replace k from the equation
from step 1 with 9.
y = 9(13)
y = 117
Replace x from the equation
from step 3 with 13.
Multiply.
Blitzer, Intermediate Algebra, 4e – Slide #123
Direct Variation
EXAMPLE
An object’s weight on the moon, M, varies directly as its weight on
Earth, E. Neil Armstrong, the first person to step on the moon on
July 20, 1969, weighed 360 pounds on Earth (with all of his
equipment on) and 60 pounds on the moon. What is the moon
weight of a person who weighs 186 pounds on Earth?
SOLUTION
1) Write an equation. We know that y varies directly as x is
expressed as
y = kx.
By changing letters, we can write an equation that describes
the following English statement: Moon weight, M, varies
directly as Earth weight, E.
Blitzer, Intermediate Algebra, 4e – Slide #124
Direct Variation
CONTINUED
M = kE.
2) Use the given values to find k. We are told that an object with
an earth weight of 360 pounds has a moon weight of 60 pounds.
Substitute 360 for E and 60 for M in the direct variation equation.
Then solve for k.
M = kE
60 = k360
1/6 = k
Direct variation equation.
Replace M with 60 and E
with 360.
Divide both sides by 360.
3) Substitute the value of k into the equation.
Blitzer, Intermediate Algebra, 4e – Slide #125
Direct Variation
CONTINUED
M = kE
M = (1/6)E
Direct variation equation.
Replace k with 1/6.
4) Answer the problem’s question. How much does an object
with a 186 pound earth weight weigh on the moon? Substitute 186
for E in the preceding equation and then solve for M.
M = (1/6)E
M = (1/6)(186)
M = 31
Equation from step 3.
Replace E with 186.
Multiply.
An object that weighs 186 pounds on the earth weighs 31 pounds
on the moon.
Blitzer, Intermediate Algebra, 4e – Slide #126
Direct Variation With Powers
Direct Variations With Powers
y varies directly as the nth power of x if there exists
some nonzero constant k such that
y kx .
n
Blitzer, Intermediate Algebra, 4e – Slide #127
Direct Variation
EXAMPLE
On a dry asphalt road, a car’s stopping distance varies directly as the
square of its speed. A car traveling at 45 miles per hour can stop in
67.5 feet. What is the stopping distance for a car traveling at 60
miles per hour?
SOLUTION
1) Write an equation. We know that y varies directly as the
square of x is expressed as
y kx .
2
By changing letters, we can write an equation that describes
the following English statement: Distance, d, varies directly
as the square of the speed, s.
Blitzer, Intermediate Algebra, 4e – Slide #128
Direct Variation
CONTINUED
d ks
2
2) Use the given values to find k. We are told that a car traveling
at 45 miles per hour can stop in 67.5 feet. Substitute 45 for s and
67.5 for d. Then solve for k.
d ks
2
67 . 5 k 45
This is the equation from step 1.
2
Replace d with 67.5 and s with 45.
67 . 5 2025 k
Simplify.
0 . 033 k
Divide both sides by 2025.
3) Substitute the value of k into the equation.
2
d ks
This is the equation from step 1.
2
d 0 . 033 s
Replace k with 0.033.
Blitzer, Intermediate Algebra, 4e – Slide #129
Direct Variation
CONTINUED
4) Answer the problem’s question. What is the stopping
distance for a car traveling at 60 miles per hour? Substitute 60 for
s in the preceding equation.
d 0 . 033 s
2
d 0 . 033 60
This is the equation from step 3.
2
d 0 . 033 3600
d 120
Replace s with 60.
Simplify.
Simplify.
The stopping distance for a car traveling at 60 miles per hour is
120 feet.
Blitzer, Intermediate Algebra, 4e – Slide #130
Inverse Variation
Inverse Variation
If a situation is described by an equation in the form
y
k
x
where k is a constant, we say that y varies inversely as x. The
number k is called the constant of variation.
Blitzer, Intermediate Algebra, 4e – Slide #131
Inverse Variation
EXAMPLE
The water temperature of the Pacific Ocean varies inversely as
the water’s depth. At a depth of 1000 meters, the water
temperature is 4 . 4 Celsius. What is the water temperature at a
depth of 5000 meters?
SOLUTION
1) Write an equation. We know that y varies inversely as x
is expressed as
y
k
.
x
By changing letters, we can write an equation that describes
the following English statement: Temperature, T, varies
inversely as the water’s depth, D.
Blitzer, Intermediate Algebra, 4e – Slide #132
Inverse Variation
CONTINUED
T
k
D
2) Use the given values to find k. The temperature of water
in the Pacific Ocean at a depth of 1000 meters is 4 . 4 Celsius.
Substitute 1000 for D and 4.4 for T in the inverse variation
equation. Then solve for k.
T
k
This is the equation from step 1.
D
k
4 .4
1000
4400 k
Replace T with 4.4 and D with
1000.
Multiply both sides by 1000.
Blitzer, Intermediate Algebra, 4e – Slide #133
Inverse Variation
CONTINUED
3) Substitute the value of k into the equation.
T
T
k
D
4400
D
This is the equation from step 1.
Replace k with 4400.
4) Answer the problem’s question. We need to find the
temperature of the water at a depth of 5000 meters. Substitute
5000 for D.
T
4400
D
This is the equation from step 3.
Blitzer, Intermediate Algebra, 4e – Slide #134
Inverse Variation
CONTINUED
T
4400
5000
T 0 . 88
Replace D with 5000.
Simplify.
The temperature of the Pacific Ocean at a depth of 5000 meters
is 4 . 4 Celsius.
Blitzer, Intermediate Algebra, 4e – Slide #135
Direct & Inverse (Combined) Variation
EXAMPLE
One’s intelligence quotient, or IQ, varies directly as a person’s
mental age and inversely as that person’s chronological age. A
person with a mental age of 25 and a chronological age of 20 has
an IQ of 125. What is the chronological age of a person with a
mental age of 40 and an IQ of 80?
SOLUTION
1) Write an equation. We know that y varies directly as x and
inversely as z is expressed as
y
kx
.
z
By changing letters, we can write an equation that describes
the following English statement: IQ, I, varies directly as the
mental age, M, and inversely as chronological age, C.
Blitzer, Intermediate Algebra, 4e – Slide #136
Direct & Inverse (Combined) Variation
CONTINUED
I
kM
C
2) Use the given values to find k. A person with a mental age of
25 and a chronological age of 20 has an IQ of 125. Substitute 125
for I, 25 for M, and 20 for C.
I
kM
C
125
k 25
20
2500 25 k
100 k
This is the equation from step 1.
Replace I with 125, M with 25,
and C with 20.
Multiply both sides by 20.
Divide both sides by 25.
Blitzer, Intermediate Algebra, 4e – Slide #137
Direct & Inverse (Combined) Variation
CONTINUED
3) Substitute the value of k into the equation.
I
I
kM
C
100 M
This is the equation from step 1.
Replace k with 100.
C
4) Answer the problem’s question. We need to determine the
chronological age of a person who has a mental age of 40 and an
IQ of 80. Substitute 40 for M and 80 for I.
I
80
100 M
C
100 40
This is the equation from step 3.
Replace M with 40 and I with 80.
C
Blitzer, Intermediate Algebra, 4e – Slide #138
Direct & Inverse (Combined) Variation
CONTINUED
80
4000
Multiply.
C
80 C 4000
Multiply both sides by C.
C 50
Divide both sides by 80.
When a person has a mental age of 40 and an IQ of 80, his/her
chronological age is 50 years old.
Blitzer, Intermediate Algebra, 4e – Slide #139
Joint Variation
Joint Variation
If a situation is described by an equation in the form
y kxz
where k is a constant, we say that y varies jointly as x and z. The
number k is called the constant of variation.
Blitzer, Intermediate Algebra, 4e – Slide #140
Joint Variation
EXAMPLE
Kinetic energy varies jointly as the mass and the square of the
velocity. A mass of 8 grams and velocity of 3 centimeters per
second has a kinetic energy of 36 ergs. Find the kinetic energy
for a mass of 4 grams and velocity of 6 centimeters per second.
SOLUTION
E kMV
36 k 8 3
36 72 k
Translate “Kinetic energy, E,
varies jointly as the mass, M,
and the square of the velocity,
V.”
2
2
If M = 8, V = 3, then E = 36.
Simplify.
Blitzer, Intermediate Algebra, 4e – Slide #141
Joint Variation
CONTINUED
0 .5 k
E 0 . 5 MV
Divide both sides by 72.
Substitute 0.5 for k in the model
for kinetic energy.
2
E 0 . 5 4 6
2
Find E when M = 4 and V = 6.
E 72
Simplify.
The kinetic energy is 72 ergs.
Blitzer, Intermediate Algebra, 4e – Slide #142
``` |
# If angle A is three times as large as angle B, and angles B and C sum to 90 degrees, what is the measure of each angle?
Sep 12, 2015
"A"=90º
"B"=30º
"C"=60º
#### Explanation:
We are given :
$\textcolor{red}{\left(1\right)} \text{ A"=3"B}$
and color(red)((2))" B"+"C"=90º
Assuming that all these angles are part of the same triangle
• All the angles in a triangle sum to 180º
=>color(red)((3))" A"+"B"+"C"=180º
Put $\textcolor{red}{\left(2\right)}$ in $\textcolor{red}{\left(3\right)}$
=>"A"+90º=180º=>"A"=90º
Put the value of $\text{A}$ in $\textcolor{red}{\left(1\right)}$
=>90º=3"B"=>"B"=30º
Put the value of $\text{B}$ in $\textcolor{red}{\left(2\right)}$
=>30º+"C"=90º=>"C"=60º |
# Median
In math and statistics and probability theory, the median is the value separating the higher half from the lower half of a data sample, a population, or a probability distribution. For a data set, it may be thought of as “the middle” value.
## Find the median for an odd set of numbers
Example question: Find the median for the following data set:
102, 56, 34, 99, 89, 101, 10.
Step 1: Sort your data from the smallest number to the highest number. For this example data set, the order is:
10, 34, 56, 89, 99, 101, 102.
Step 2: Find the number in the middle (where there are an equal number of data points above and below the number):
10, 34, 56, 89, 99, 101, 102.
The median is 89.
## Find the median for an even set of numbers
Example question: Find the median for the following data set:
102, 56, 34, 99, 89, 101, 10, 54.
Step 1: Place the data in ascending order (smallest to highest).
10, 34, 54, 56, 89, 99, 101, 102.
Step 2: Find the TWO numbers in the middle (where there are an equal number of data points above and below the two middle numbers).
10, 34, 54, 56, 89, 99, 101, 102
Step 3: Add the two middle numbers and then divide by two, to get the average:
• 56 + 89 = 145
• 145 / 2 = 72.5.
The median is 72.5. |
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# Dividing Rational Expressions
The method of dividing rational expressions is same as the method of dividing fractions . That is, to divide a rational expression by another rational expression, multiply the first rational expression by the reciprocal of the second rational expression.
For all rational expressions $\frac{a}{b}$ and $\frac{c}{d}$ with $b\ne 0$ , $c\ne 0$ , and $d\ne 0$ , $\frac{a}{b}÷\frac{c}{d}=\frac{a}{b}\cdot \frac{d}{c}=\frac{ad}{bc}$ .
Example
Divide and then simplify.
$\frac{{x}^{2}\text{\hspace{0.17em}}-\text{\hspace{0.17em}}4}{x\text{\hspace{0.17em}}+\text{\hspace{0.17em}}6}÷\frac{x\text{\hspace{0.17em}}+\text{\hspace{0.17em}}2}{2x\text{\hspace{0.17em}}+\text{\hspace{0.17em}}12}$
Write as multiplication by the reciprocal.
The reciprocal of $\frac{x\text{\hspace{0.17em}}+\text{\hspace{0.17em}}2}{2x\text{\hspace{0.17em}}+\text{\hspace{0.17em}}12}$ is $\frac{2x\text{\hspace{0.17em}}+\text{\hspace{0.17em}}12}{x\text{\hspace{0.17em}}+\text{\hspace{0.17em}}2}$ .
$\frac{{x}^{2}-4}{x\text{\hspace{0.17em}}+\text{\hspace{0.17em}}6}÷\frac{x\text{\hspace{0.17em}}+\text{\hspace{0.17em}}2}{2x\text{\hspace{0.17em}}+\text{\hspace{0.17em}}12}=\frac{{x}^{2}-4}{x\text{\hspace{0.17em}}+\text{\hspace{0.17em}}6}\cdot \frac{2x\text{\hspace{0.17em}}+\text{\hspace{0.17em}}12}{x\text{\hspace{0.17em}}+\text{\hspace{0.17em}}2}$
Now multiply the numerators and the denominators.
$=\frac{\left({x}^{2}\text{\hspace{0.17em}}-\text{\hspace{0.17em}}4\right)\left(2x\text{\hspace{0.17em}}+\text{\hspace{0.17em}}12\right)}{\left(x\text{\hspace{0.17em}}+\text{\hspace{0.17em}}6\right)\left(x\text{\hspace{0.17em}}+\text{\hspace{0.17em}}2\right)}$
Factor the terms in the numerator.
$=\frac{\left(x\text{\hspace{0.17em}}+\text{\hspace{0.17em}}2\right)\left(x\text{\hspace{0.17em}}-\text{\hspace{0.17em}}2\right)\cdot 2\left(x\text{\hspace{0.17em}}+\text{\hspace{0.17em}}6\right)}{\left(x\text{\hspace{0.17em}}+\text{\hspace{0.17em}}6\right)\left(x\text{\hspace{0.17em}}+\text{\hspace{0.17em}}2\right)}$
Divide out the common factors.
$=\frac{\overline{){\left(}{x}\text{\hspace{0.17em}}{+}\text{\hspace{0.17em}}{2}{\right)}}\left(x\text{\hspace{0.17em}}-\text{\hspace{0.17em}}2\right)\cdot 2\overline{){\left(}{x}\text{\hspace{0.17em}}{+}\text{\hspace{0.17em}}{6}{\right)}}}{\overline{){\left(}{x}\text{\hspace{0.17em}}{+}\text{\hspace{0.17em}}{6}{\right)}}\overline{){\left(}{x}\text{\hspace{0.17em}}{+}\text{\hspace{0.17em}}{2}{\right)}}}$
Simplify.
$\begin{array}{l}=\frac{2\left(x\text{\hspace{0.17em}}-\text{\hspace{0.17em}}2\right)}{1}\\ =2x-4\end{array}$
; |
How do you find the equation of the tangent and normal line to the curve y=tanx at x=-pi/4?
Nov 5, 2016
Tangent: $y = 2 x + \frac{\pi}{2} - 1$
Normal: $y = - \frac{1}{2} x - \frac{\pi}{8} - 1$
Explanation:
The gradient tangent to a curve at any particular point is given by the derivative.
If $y = \tan x$ then $\frac{\mathrm{dy}}{\mathrm{dx}} = {\sec}^{2} x$
When $x = - \frac{\pi}{4}$
$\implies y = \tan \left(- \frac{\pi}{4}\right) = - 1$
$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = {\sec}^{2} \left(- \frac{\pi}{4}\right) = 2$
So the tangent passes through $\left(- \frac{\pi}{4} , - 1\right)$ and has gradient ${m}_{T} = 2$
Using $y - {y}_{1} = m \left(x - {x}_{1}\right)$ the equation of the tangent is:
$y - \left(- 1\right) = \left(2\right) \left(x - \left(- \frac{\pi}{4}\right)\right)$
$\therefore y + 1 = 2 x + \frac{\pi}{2}$
$\therefore y = 2 x + \frac{\pi}{2} - 1$
The normal is perpendicular to the tangent, so the product of their gradients is -1 hence normal passes through $\left(- \frac{\pi}{4} , - 1\right)$ and has gradient ${m}_{N} = - \frac{1}{2}$
so the equation of the normal is:
$y - \left(- 1\right) = - \frac{1}{2} \left(x - \left(- \frac{\pi}{4}\right)\right)$
$\therefore y + 1 = - \frac{1}{2} x - \frac{\pi}{2}$
$\therefore y = - \frac{1}{2} x - \frac{\pi}{8} - 1$ |
Mathematical methods for economic theory
Martin J. Osborne
9.2 Second-order difference equations
Definition
A second-order difference equation is an equation
xt+2 = f(txtxt+1),
where f is a function of three variables. A solution of the second-order difference equation xt+2 = f(txtxt+1) is a function x of a single variable whose domain is the set of integers such that xt+2 = f(txtxt+1) for every integer t, where xt denotes the value of x at t.
As for a first-order difference equation, we can find a solution of a second-order difference equation by successive calculation. The only difference is that for a second-order equation we need the values of x for two values of t, rather than one, to get the process started. Given xt and xt+1 for some value of t, we use the equation to find xt+2, and then use the equation again for xt+1 and xt+2 to find xt+3, and so forth. At each step, the new value of x is uniquely determined, so we have the following result.
Proposition
For every pair of numbers x0 and x1, every second-order difference equation xt+2 = f(txtxt+1) has a unique solution in which the value of x is x0 at 0 and x1 at 1.
Second-order linear difference equations with constant coefficients
Definition
A linear second-order difference equation with constant coefficients is a second-order difference equation that may be written in the form
xt+2 + axt+1 + bxt = ct,
where a, b, and ct for each value of t, are numbers. The equation is homogeneous if ct = 0 for all t.
The method for finding a solution of a linear second-order difference equation follows the lines of the method for finding a solution of a linear second-order differential equation. Suppose that x and y are solutions of the equation
xt+2 + axt+1 + bxt = ct,
which we refer to subsequently as the “original” equation. Define zt = yt − xt. Then zt+2 + azt+1 + bzt = [yt+2 + ayt+1 + byt] − [xt+2 + axt+1 + bxt] = ct − ct = 0. That is, z is a solution of the homogeneous equation
xt+2 + axt+1 + bxt = 0.
Conversely, let x be a solution of the original equation, let z be a solution of the homogeneous equation, and define yt = xt + zt for all t. Then y is a solution of the original equation.
Thus we have the following result.
Proposition
Let x be a solution of the linear second-order ordinary difference equation with constant coefficients
xt+2 + axt+1 + bxt = ct.
Then y is a solution of this equation if and only if yt = xt + zt for all t, where z is a solution of the (homogeneous) equation
xt+2 + axt+1 + bxt = 0.
An implication of this result is that we can find the set of all solutions of the original equation by finding the set of all solutions of the homogeneous equation and a single solution of the original equation, as described in the following procedure.
Procedure for finding general solution of linear second-order difference equation with constant coefficients
The solutions of the difference equation
xt+2 + axt+1 + bxt = ct
may be found as follows.
1. Find the solutions of the associated homogeneous equation xt+2 + axt+1 + bxt = 0.
2. Find a single solution of the original equation xt+2 + axt+1 + bxt = ct.
3. Add together the solutions found in steps 1 and 2.
1. Find the solutions of a homogeneous equation
You might guess that the homogeneous equation
xt+2 + axt+1 + bxt = 0
has a solution of the form xt = mt. For this function to be a solution, we need
mt+2 + amt+1 + bmt = 0
for all t, or
mt(m2 + am + b) = 0
or, if m ≠ 0,
m2 + am + b = 0.
This equation is the characteristic equation of the difference equation. Its solutions are
−(1/2)a ± √((1/4)a2 − b).
We have the following result, analogous to a result for homogeneous second-order differential equations.
Proposition
Consider the homogeneous linear second-order difference equation with constant coefficients
xt+2 + axt+1 + bxt = 0.
The set of solutions of this equation depends on the character of the roots of the characteristic equation m2 + am + b = 0 as follows.
Distinct real roots
If a2 > 4b, in which case the characteristic equation has distinct real roots, say m1 and m2, every solution of the equation has the form
Amt
1
+ Bmt
2
for numbers A and B.
Single real root
If a2 = 4b, in which case the characteristic equation has a single root, every solution of the equation has the form
(A + Bt)mt
for numbers A and B, where m = −(1/2)a is the root.
Complex roots
If a2 < 4b, in which case the characteristic equation has complex roots, every solution of the equation has the form
Art cos(θt + ω)
for numbers A and ω, where r = √b and cos θ = −a/(2√b). This solution may alternatively be expressed as
C1rt cos(θt) + C2rt sin(θt),
where C1 = A cos ω and C2 = −A sin ω (using the formula cos(x+y) = (cos x)(cos y) − (sin x)(sin y).
Source
For a proof, see Goldberg (1958), pp. 134–142.
In the third case, in which the characteristic equation has complex roots, the solutions oscillate. We use the following terminology: Art is the amplitude of the oscillations at time t, r is growth factor, θ/2π is the frequency of the oscillations, and ω is the phase (which depends on the initial conditions). If |r| < 1 then the oscillations are damped; if |r| > 1 then they are explosive.
Example
Consider the equation
xt+2 + xt+1 − 2xt = 0.
The roots of the characteristic equation are 1 and −2 (real and distinct). Thus every solution has the form
xt = A + B(−2)t.
Example
Consider the equation
xt+2 + 6xt+1 + 9xt = 0.
The single root of the characteristic equation is −3. Thus every solution has the form
xt = (A + Bt)(−3)t.
Example
Consider the equation
xt+2 − xt+1 + xt = 0.
The roots of the characteristic equation are complex. We have r = 1 and cos θ = 1/2, so θ = (1/3)π. So every solution has the form
xt = Acos((1/3)πt + ω).
The frequency of the oscillations is (π/3)/2π = 1/6 and the growth factor is 1, so the oscillations are neither damped nor explosive.
2. Find a solution of a nonhomogeneous equation
Consider the nonhomogeneous equation
xt+2 + axt+1 + bxt = ct.
If ct = c for all t, then xt = c/(1 + a + b) is a solution if 1 + a + b ≠ 0, xt = ct/(a + 2) is a solution if 1 + a + b = 0 and a ≠ −2, and xt = ct2/2 if 1 + a + b = 0 and a = −2 (in which case b = 1).
For other forms of ct, the method used to find a solution of a nonhomogeneous second-order differential equation can be used. For example, if ct is a linear combination of terms of the form qt, tm, cos(pt), and sin(pt), for constants q, p, and m, and products of such terms, then guess that the equation has a solution that is a linear combination of such terms; substitute such a function into the equation and see whether there are coefficients that generate a solution. If the ct you find happens to satisfy the homogeneous equation, then a different approach must be taken, which I do not discuss.
Example
Consider the equation
xt+2 − 5xt+1 + 6xt = 2t − 3.
To find a solution, guess that there is one of the form at + b. For this function to be a solution, we need
a(t+2) + b − 5[a(t+1) + b] + 6(at + b) = 2t − 3.
Equating the coefficients of t and the constant on each side, we have a = 1 and b = 0. Thus xt = t for all t is a solution.
Example
Consider the equation
xt+2 − 5xt+1 + 6xt = 4t + t2 + 3.
To find a solution, try
xt = C4t + Dt2 + Et + F.
Substituting this potential solution into the equation and equating the coefficients of 4t, t2, and the constant on each side we find that
C = 1/2, D = 1/2, E = 3/2, and F = 4.
Thus a solution of the equation is
xt = (1/2)4t + (1/2)t2 + (3/2)t + 4.
3. Add together the solutions found in steps 1 and 2
Example
Consider the equation
xt+2 − 5xt+1 + 6xt = 2t − 3.
The associated homogeneous equation is
xt+2 − 5xt+1 + 6xt = 0.
The characteristic equation has two real roots, 2 and 3, so by a previous result, every solution of the equation has the form A2t + B3t.
By a previous example, a solution of the original equation is xt = t for all t.
We conclude that every solution of the equation has the form
xt = A2t + B3t + t.
Example
Consider the equation
xt+2 − 5xt+1 + 6xt = 4t + t2 + 3.
As in the previous example, every solution of the associated homogeneous equation has the form A2t + B3t.
By a previous example, a solution of the original equation is
xt = (1/2)4t + (1/2)t2 + (3/2)t + 4.
Thus every solution of the original equation has the form
xt = A2t + B3t + (1/2)4t + (1/2)t2 + (3/2)t + 4.
Equilibrium and stability
The notions of equilibrium and stability for a second-order difference equation are analogous to the ones for a differential equation.
Definition
An equilibrium of the linear second-order difference equation with constant coefficients
xt+2 + axt+1 + bxt = c
is a number x* such that xt = x* for all t is a solution of the equation. An equilibrium x* is stable if every solution of the equation converges to x*.
Now, by an earlier analysis, if 1 + a + b ≠ 0 then a solution of
xt+2 + axt+1 + bxt = c
is xt = c/(1 + a + b) for all t. Thus if 1 + a + b ≠ 0 then c/(1 + a + b) is an equilibrium of the equation. If 1 + a + b = 0 then the equation has no constant solution, and hence no equilibrium.
Also, from a previous result we know that every solution of
xt+2 + axt+1 + bxt = c
is the sum of any given solution of this equation and a solution of the homogeneous equation
xt+2 + axt+1 + bxt = 0.
Thus if 1 + a + b ≠ 0 then every solution of
xt+2 + axt+1 + bxt = c
is the sum of c/(1 + a + b) and a solution of the homogeneous equation
xt+2 + axt+1 + bxt = 0.
So the equilibrium c/(1 + a + b) is stable if and only if every solution of the homogeneous equation converges to zero. We can use the earlier result characterizing the solutions of the homogeneous equation to find conditions under which these solutions converge to zero, as follows. Recall that the characteristic equation of the difference equation is m2 + am + b = 0.
Characteristic equation has distinct real roots
Every solution of the difference equation has the form
Amt
1
+ Bmt
2
for numbers A and B. These solutions converge to zero if and only if |m1| < 1 and |m2| < 1.
Characteristic equation has single real root
Every solution of the difference equation has the form
(A + Bt)mt
for numbers A and B, where m = −(1/2)a is the root. These solutions converge to zero if and only if |m| < 1. (The function tmt converges to zero if and only if |m| < 1.)
Characteristic equation has complex roots
Every solution of the equation has the form
Art cos(θt + ω)
for numbers A and ω, where r = √b and cos θ = −a/(2√b). These solutions converge to zero if and only if r < 1.
We can unify these conditions by defining the modulus of a complex number α + βi to be √(α2 + β2).
If the characteristic equation has complex roots, then, using the formula for the roots of a quadratic equation, these roots are −a/2 ± i√(b − a2/4). Thus the modulus of each root is √(a2/4 + b − a2/4) = √b. The modulus of a real number is simply its absolute value (the complex number α + βi is real if and only if β = 0), so the conditions we found for every solution of the homogeneous equation to converge to zero are, in every case, simply that the modulus of each root of the characteristic equation is less than 1.
We have the following result.
Proposition
For any value of c, an equilibrium of the linear second-order difference equation with constant coefficients xt+2 + axt+1 + bxt = c is stable if and only if the modulus of each root of the characteristic equation m2 + am + b = 0 is less than 1, or, equivalently, if and only if |a| < 1 + b and b < 1.
Proof
The arguments preceding the statement establish that a necessary and sufficient for the equilibrium to be stable is that the modulus of each root is less than 1.
I now show that this condition is equivalent to the conditions |a| < 1 + b and b < 1.
Note that the second condition is equivalent to the two conditions a < 1 + b, or b > a − 1, and −a < 1 + b, or b > −a − 1. In the following figure, b > a − 1 if b is above the blue line and b > −a − 1 if b is above the red line. Thus the set of values of (ab) satisfying the conditions |a| < 1 + b and b < 1 is the shaded triangle. The characteristic equation has real roots if b lies below the black curve b = a2/4 and complex roots if b lies above this curve. The figure is useful in following the logic of the subsequent slightly intricate algebra.
First consider the case in which the roots of the characteristic equation are real. Then the roots are −a/2 ± √(a2 − 4b)/2, so the modulus of each root is less than 1 if and only if
−1 < −a/2 + √(a2 − 4b)/2 < 1 −1 < −a/2 − √(a2 − 4b)/2 < 1
or
−2 + a < √(a2 − 4b) < 2 + a −2 + a < −√(a2 − 4b) < 2 + a.
Now, the second inequality on the left implies the first inequality on the left, and the first inequality on the right implies the second inequality on the right, so that the modulus of each roots is less than 1 if and only if
√(a2 − 4b) < 2 + a −2 + a < −√(a2 − 4b).
For the first inequality to hold, we need a > −2 (because the left-hand side is positive) and, squaring both sides,
a2 − 4b < (2 + a)2 = 4 + 4a + a2
or
1 + a + b > 0.
For the second inequality to hold, we need a < 2 (because the right-hand side is negative) and, squaring both sides,
(−2 + a)2 > a2 − 4b
(both sides are negative, so the inequality changes), or
4 − 4a + a2 > a2 − 4b
or
1 − a + b > 0.
In summary, if the roots of the characteristic equation are real, which is true if and only if a2 > 4b, then the modulus of each root is less than 1 if and only if
−2 < a < 2, 1 + a + b > 0, and 1 − a + b > 0.
Now, if b < 1 then the last two inequalities imply −2 < a < 2. Further, −2 < a < 2 and a2 > 4b imply b < 1. Thus if the roots of the characteristic equation are real, then the modulus of each root is less than 1 if and only if
b < 1, 1 + a + b > 0, and 1 − a + b > 0.
Now consider the case in which the roots of the characteristic equation are complex. Then a2 < 4b, so that b > 0, and, as noted previously, the modulus of each root is √b. Thus the modulus of each root is less than 1 if and only if b < 1. Now, a2 < 4b and b < 1 imply that |a| < 2, so that (a − 2)2 > 0. But (a − 2)2 = a2 − 4a + 4, so a2 > 4a − 4, or a2/4 > a − 1. Given a2 < 4b, we have b > a − 1, or 1 − a + b > 0. The inequality |a| < 2 implies also that (a + 2)2 > 0, so that a2 + 4a + 4 > 0, or a2 > −4a − 4, or a2/4 > −a − 1. Thus given a2 < 4b, we have b > −a − 1, or 1 + a + b > 0. Thus if the modulus of each root of the characteristic equation is less than 1, we have
b < 1, 1 + a + b > 0, and 1 − a + b > 0.
In conclusion, whether the roots of the characteristic equation are real or complex, the modulus of each root is less than 1 if and only if
b < 1, 1 + a + b > 0, and 1 − a + b > 0. |
Find the values of a and b so that the polynomial
Question:
Find the values of $a$ and $b$ so that the polynomial $\left(x^{4}+a x^{3}-7 x^{2}-8 x+b\right)$ is exactly divisible by $(x+2)$ as well as $(x+3)$.
Solution:
Let:
$f(x)=x^{4}+a x^{3}-7 x^{2}-8 x+b$
Now,
$x+2=0 \Rightarrow x=-2$
By the factor theorem, we can say:
$f(x)$ will be exactly divisible by $(x+2)$ if $f(-2)=0$.
Thus, we have:
$f(-2)=\left[(-2)^{4}+a \times(-2)^{3}-7 \times(-2)^{2}-8 \times(-2)+b\right]$
$=(16-8 a-28+16+b)$
$=(4-8 a+b)$
$\therefore f(-2)=0 \Rightarrow 8 a-b=4 \quad \ldots(1)$
Also,
$x+3=0 \Rightarrow x=-3$
By the factor theorem, we can say:
$f(x)$ will be exactly divisible by $(x+3)$ if $f(-3)=0$.
Thus, we have:
$f(-3)=\left[(-3)^{4}+a \times(-3)^{3}-7 \times(-3)^{2}-8 \times(-3)+b\right]$
$=(81-27 a-63+24+b)$
$=(42-27 a+b)$
$\therefore f(-3)=0 \Rightarrow 27 a-b=42 \ldots(2)$
Subtracting (1) from (2), we get :
$\Rightarrow 19 a=38$
$\Rightarrow a=2$
Putting the value of a, we get the value of b, i.e., 12.
$\therefore a=2$ and $b=12$ |
# The Buffalo Way
Imgur
The Buffalo Way is a plug-and-bash method used to solve olympiad inequalities. It is usually applied to symmetric inequalities, where we can assume WLOG that the variables are in a specific order; that is, $x_1 \le x_2\le \cdots \le x_n$.
To illustrate this method, we shall prove $AM-GM$ for two variables using the method.
Prove that $\dfrac{x+y}{2}\ge \sqrt{xy}$ for non-negative reals $x,y$.
First, assume WLOG that $x\le y$. Thus, we can represent $x$ and $y$ by: $x=a$ $y=a+b$ where $a,b$ are non-negative reals. Make sure you see why this is true.
Thus, we want to prove $\dfrac{a+a+b}{2}\ge \sqrt{a(a+b)}\implies a+\dfrac{b}{2}\ge \sqrt{a^2+ab}$
Squaring both sides gives $a^2+ab+\dfrac{b^2}{4}\ge a^2+ab\implies \dfrac{b^2}{4}\ge 0$ which is true by the trivial inequality.
In general, if we have variables satisfying $x_1 \le x_2\le \cdots \le x_n$, then we substitute $x_1=y_1$$x_2=y_1+y_2$ $\vdots$ $x_n=y_1+y_2+\cdots +y_n$
Given that $a,b,c$ are non-negative reals such that $a\le b\le c$, then prove that $(a+b)(c+a)^2\ge 6abc$
We see that we already have the condition of $a\le b\le c$, so we can apply Buffalo's Way directly. Let $a=x$ $b=x+y$ $c=x+y+z$ where $x,y,z$ are non-negative reals. Thus, we want to prove $(2x+y)(2x+y+z)^2\ge 6x(x+y)(x+y+z)$
This expands to (told you Buffalo Way is a bash): $8x^3+12x^2y+8x^2z+6xy^2+8xyz+2xz^2+y^3+2y^2x+yz^2\ge 6x^3+12x^2y+6x^2z+6xy^2+6xyz$
Rearranging gives $2x^3+2x^2z+2xyz+2xz^2+y^3+2y^2z+yz^2\ge 0$ which is trivially true ($x,y,z$ are positive).
We can also see that this method gives an equality case: We must have $x=y=z=0$. Thus, $a=b=c=0$.
When actually solving Olympiad Inequalities in competitions, NEVER use this method, unless you have no idea now to do it otherwise. In cases where the inequality is relatively simple (no denominators) the Buffalo Way is almost guaranteed to work.
NOTE: I don't know where the name comes from. If you try to search it up, you won't get any results. I first heard of this method from the AoPS Community.
Note by Daniel Liu
6 years, 4 months ago
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We just have to make sure that all of our implications are reversible when using this method.
- 6 years, 4 months ago
Are there any Olympiad problems that can be solved with Buffalo?
- 6 years, 4 months ago
Well, try solving the second one without buffalo's.
But to answer your question, chances are there are no Olympiad question that can be only exclusively solved using the Buffalo way. This method is the least elegant method of olympiad inequalities ever, and it is almost guaranteed that there is a more elegant solution.
You might even be docked a point on the USA(J)MO if you used buffalo way, because of it's extreme inelegantness.
- 6 years, 4 months ago
"You might even be docked a point on the USA(J)MO if you used buffalo way, because of it's extreme inelegantness."
You might actually get a score of a point on that USA(J)MO (correct answer, bad proof)
- 6 years, 4 months ago
No, if you have a complete proof, no matter how inelegant, then you will get at least 5-6 points. It's impossible to get 1 point if you write a complete proof.
- 6 years, 4 months ago
# 2lazy2bash
- 6 years, 4 months ago
#?
- 6 years, 4 months ago
Yes #2lazy2bash
- 6 years, 3 months ago |
11
Q:
The ratio of speeds of a car, a train and a bus is 5:9:4. The average speed of the car, the bus and the train is 72 km/hr together. What is the average speed of the car and the train together ?
A) 84 km/hr B) 96 km/hr C) 72 km/hr D) 60 km/hr
Explanation:
Let the speeds of the car, train and bus be 5x, 9x and 4x km/hr respectively.
Average speed = 5x + 9x + 4x/3 = 18x /3 = 6x km/hr.
Also, 6x = 72 => x = 12 km/hr
Therefore, the average speed of the car and train together is = $5x+9x2$= 7x = 7 x 12 = 84 km/hr.
Q:
The average score of boys in an examination in a school is 81 and that of the girls is 83 and the average score of the school is 81.8. Then what will be ratio between the number of girls to the number of boys, in the examination?
A) 3:2 B) 2:3 C) 4:3 D) 3:4
Explanation:
Let the number of boys = b
Let the number of girls = g
From the given data,
81b + 83g = 81.8(b + g)
81.8b - 81b = 83g - 81.8g
0.8b = 1.2g
b/g = 1.2/0.8 = 12/8 = 3/2
=> g : b = 2 : 3
Hence, ratio between the number of girls to the number of boys = 2 : 3.
2 243
Q:
Out of three numbers, the first is twice the second and is half of the third. If the average of the three numbers is 56, what will be the sum of the highest number and the lowest number?
A) 120 B) 160 C) 80 D) 60
Explanation:
Let the three numbers be x, y, z.
From the gien data,
x = 2y ....(1)
x = z/2 => z = 2(2y) = 4y .....(From 1) ...........(2)
Given average of three numbers = 56
Then,
Now,
x = 2y => x = 2 x 24 = 48
z = 4y = 4 x 24 = 96
Now, the highest number is z = 96 & smallest number is y = 24
Hence, required sum of highest number and smallest number
= z + y
= 96 + 24
= 120.
2 371
Q:
How many window coverings are necessary to span 50 windows, if each window covering is 15 windows long?
A) 4 B) 15 C) 3 D) 50
Explanation:
Given that,
Number of windows = 50
Each window covering covers 15 windows
=> 50 windows requires 50/15 window coverings
= 50/15 = 3.333
Hence, more than 3 window coverings are required. In the options 4 is more than 3.
Hence, 4 window coverings are required to cover 50 windows of each covering covers 15 windows.
1 1442
Q:
Ian has 14 boxes of paper and divides them evenly between 4 coworkers. How many whole boxes did each coworker get?
A) 2 B) 2.5 C) 3 D) 3.5
Explanation:
Given number of boxes = 14
Number of workers = 4
Now, number of whole boxes per worker = 14/4 = 3.5
Hence, number of whole boxes per each coworker = 3
2 4388
Q:
Five boxes of bananas sell for Rs. 30. how many boxes can you buy for Rs.9?
A) 2 B) 1.5 C) 1.25 D) 2.5
Explanation:
Given Five boxes of bananas sell for Rs. 30.
=> 1 Box of Bananas for = 30/5 = Rs. 6
Then, for Rs. 9
=> 9/6 = 3/2 = 1.5
Hence, for Rs. 9, 1.5 box of bananas can buy.
8 2811
Q:
Third proportion of 10 and 20 is
A) 30 B) 40 C) 25 D) 20
Explanation:
The third proportional of two numbers p and q is defined to be that number r such that
p : q = q : r.
Here, required third proportional of 10 & 20, and let it be 'a'
=> 10 : 20 = 20 : a
10a = 20 x 20
=> a = 40
Hence, third proportional of 10 & 20 is 40.
8 1749
Q:
The average weight of 45 passengers in a bus is 52 kg. 5 of them whose average weight is 48 kg leave the bus and other 5 passengers whose average weight is 54 kg join the bus at the same stop. What is the new average weight of the bus?
A) 54.21 kgs B) 51.07 kgs C) 52.66 kgs D) 53.45 kgs
Explanation:
Given total number of passengers in the bus = 45
First average weight of 45 passengers = 52 kgs
Average weight of 5 passengers who leave bus = 48
Average weight of passengers who joined the bus = 54
Therefore, the net average weight of the bus is given by
24 2746
Q:
12 boys decided to constribute Rs. 750 each to an Orphange. Suddenly few of them boys dropped out and consequently the rest had to pay Rs. 150 more. Then the number of boys who dropped out?
A) 4 B) 6 C) 2 D) 3
Explanation:
Total money decided to contribute = 750 x 12 = 9000
Let 'b' boys dropped
The rest paid 150/- more
=> (12 - b) x 900 = 9000
=> b = 2
Hence, the number of boys who dropped out is 2. |
# TRIGONOMETRIC RATIO TABLE
The table provided in this section will help you to find the values of trigonometric ratios for the special angles 0°, 30°, 45°, 60° and 90°.
The values of trigonometric ratios of some special angles are very important to solve many problems in trigonometry.
Therefore, it is important to remember the values of the trigonometric ratios of some special angles.
## Trigonometric Ratios of Some Special Angles
Now, let us see, how the above values of trigonometric ratios are determined for the standard angles.
First, let us see how the values are determined for the angles 30° and 60°.
## Trigonometric Ratios of 30° and 60°
Let ABC be an equilateral triangle whose sides have length a (see the figure given below). Draw AD perpendicular to BC, then D bisects the side BC.
So, BD = DC = a/2 and BAD = DAC = 30°.
Now, in right triangle ADB, BAD = 30° and BD = a/2.
In right triangle ADB, by Pythagorean theorem,
Hence, we can find the trigonometric ratios of angle 30° from the right triangle ADB.
In right triangle ADB, ABD = 60°. So, we can determine the trigonometric ratios of angle 60°.
Now, let us see how the values are determined for the angle 45°.
## Trigonometric Ratios of 45°
If an acute angle of a right triangle is 45°, then the other acute angle is also 45°.
Thus the triangle is isosceles. Let us consider the triangle ABC with B = 90°, A = C = 45°.
Then AB = BC. Let AB = BC = a.
By Pythagorean theorem,
AC2 = AB2 + BC2
AC2 = a2 + a2
AC2 = 2a2
AC = √2a
Hence, we can find the trigonometric ratios of angle 45° from the right triangle ABC.
Now, let us see how the values are determined for the angles 0° and 90°.
## Trigonometric Ratios of 0° and 90°
Consider the figure given below which shows a circle of radius 1 unit centered at the origin.
Let P be a point on the circle in the first quadrant with coordinates (x, y).
We drop a perpendicular PQ from P to the x-axis in order to form the right triangle OPQ.
Let POQ = θ, then
sinθ = PQ/OP = y/1 = y (y coordinate of P)
cosθ = OQ/OP = x/1 = x (x coordinate of P)
tanθ = PQ/OQ = y/x
If OP coincides with OA, then angle θ = 0°.
Since, the coordinates of A are (1, 0), we have
If OP coincides with OB, then angle θ = 90°.
Since, the coordinates of B are (0, 1), we have
## Evaluating Trigonometric Expressions
Example 1 :
Solution :
Example 2 :
Solution :
Example 3 :
Solution :
Example 4 :
Solution :
Kindly mail your feedback to v4formath@gmail.com
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# RS Aggarwal Class 9 Chapter 16 Solutions (Presentation of Data in Tabular Form)
RS Aggarwal Solution for Class 9 Maths Chapter 16- Presentation of Data in Tabular Form will help you to gain good marks during CBSE exams. In this chapter of RS Aggarwal Class 9 Maths Solutions, there are a total of 20 problems divided into one exercise. You shall learn to solve different Problems based on the Presentation of Data in Tabular Form.
This chapter has only 1 exercise that focuses you to learn different methods that are used to prepare a frequency distribution table by using the given data. Chapter 16 covers the basic Introduction of Presentation of Data, followed by the Fundamental Characteristics of Data, Limitations of Statistics, and Types of Data, the Frequency distribution of an ungrouped data, Types of Frequency Distribution, Some definitions, Class boundaries or true upper and lower limits and the Method of forming classes of a data.
Instasolv’s RS Aggarwal Solutions for Chapter 16- Presentation of Data in Tabular Form help to clear your doubts concerning this chapter as there’s an opportunity of high-rank questions from this chapter. The solution is gradually prepared with the aid of Instasov experts to help prepare all of you for your final examination. Students can analyze their weaknesses and work on them consistent with the exam program.
## Important Topics for RS Aggarwal Solution for Class 9 Maths Chapter 16
Instasolv solution will help you to understand the basics definition and characteristics of statistics. They are as follows:
• Statistics is the science which deals with the collection, presentation, analysis, and interpretation of numerical data.
• Numerical records alone represent data.
• Qualitative characteristics like intelligence, poverty, etc., which are unable to be measured numerically, do not form data.
• Data are aggregates of different facts. An observation never forms data.
• Data accrued for an exact purpose may not be acceptable for another purpose.
• Data in specific experiments are comparable.
Highlighting facts and comparison of the primary and secondary data:
• Primary statistics: The data accumulated through the investigator himself with a definite plan in mind are called primary information. These statistics are, therefore, rather dependable and applicable.
• Secondary statistics: The facts amassed using someone, apart from the investigator, are referred to as secondary information. Secondary facts need to be carefully used since they may be collected with a reason specific from that of the investigator and won’t be fully applicable to the investigation.
Some important definitions related to this chapter are as follows:
• Any character when capable of taking several distinct values is known as a variant.
• Each organization into which the raw data is condensed is known as a class interval.
• Class size is defined as the distinction among both the true upper and the lower limit within a category.
• Classmark is defined as the mean of upper limit and lower limit.
• Each class is bounded by two figures that are known as called class limits.
• The upper and lower limits of a class are respectively known as the true upper limit and true lower limit.
• The wide variety of instances an observation occurs in a category is known as the frequency.
• The cumulative frequency similar to a class is the sum of all frequencies along with that class.
### Exercise Discussion of Important Topics RS Aggarwal Class 9 Maths Solution for Chapter 16- Presentation of Data in Tabular Form
• Chapter 16 essentially contains only one exercise that has 20 questions that will ask you to solve various problems based on Characteristics of Data, Limitations of Statistics, and Types of Data, Frequency distribution of an ungrouped data, Types of Frequency Distribution, Some definitions, Class boundaries or true upper and lower limits and the Method of forming classes of a data.
• This chapter will help you to clear any fear of uncertainty regarding the frequency and mid -values at the different class situations under a given problem.
• Math Chapter 16 Exercise contains questions based on several problems with solutions that are marked fetching in nature. These answers are well arranged in order to give support to your weakest area of Mathematics to clear the exam.
• Instasolv will help you in solving and clarifying your doubts about the questions in these Exercises. Sometimes these questions look very simple and easy, but they actually need good practice before your exams.
### Why Use RS Aggarwal Solution for Class 9 Maths Chapter 16- Presentation of Data in Tabular Form by Instasolv?
Instasolv’s RS Aggarwal Solutions are the perfect study material for you to improve your skills and knowledge. We give you a direction to manage time while writing papers and getting good grades in the class 9 exams.
The growing competition to participate in the best engineering laboratories in the country requires you to maintain your bases to obtain excellent scores on competitive exams. Instasolv solutions will always be there with you to reach your esteemed goal.
Therefore, in order to better understand each and every small concept used in Chapter 16 of Class 9, it is strongly recommended by Instasolv to study these problems thoroughly before your exams.
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### Home > INT3 > Chapter 1 > Lesson 1.1.2 > Problem1-25
1-25.
Given $f\left(x\right)=-\frac{2}{3}x+3$ and $g(x)=2x^2-5$, complete parts (a) through (f). 1-25 HW eTool (Desmos). Homework Help ✎
1. Calculate $f(3)$.
Substitute $3$ for each $x$ in $f(x)$.
$f(3)=-\frac{2}{3}(3)+3$
Simplify.
1. Solve $f(x)=-5$.
Substitute $−5$ for $f(x)$.
$-5=-\frac{2}{3}x+3$
Subtract $3$ from both sides.
$-8=-\frac{2}{3}x$
Get the $x$-term alone.
$\left(-\frac{3}{2}\right)(-8)=\left(-\frac{2}{3}x\right)\left(-\frac{3}{2}\right)$
$x=12$
1. Calculate $g(-3)$.
Substitute $−3$ for each $x$ in $g(x)$.
$g(−3)=2(−3)^2−5$
Simplify.
1. Solve $g(x)=-7$.
Substitute $−7$ for $g(x)$.
$−7=2x^2−5$
Solve.
$x^2=−1$
No solution.
1. Solve $g(x)=8$.
Substitute $8$ for $g(x)$.
$8=2x^2−5$
Solve.
$x^2=\frac{13}{2}$
$x=\pm\sqrt\frac{13}{2}$
1. Solve $g(x)=9$.
Substitute $9$ for $g(x)$.
$9=2x^2−5$
Solve.
$x^2=7$
Remember, there should be 2 solutions.
Use the eTool below to solve for each part.
Click the link at right for the full version of the eTool: Int3 1-25 HW eTool |
Introduction to Sets
INTRODUCTION TO SETS
by Dr. Carol JVF Burns (website creator)
Follow along with the highlighted text while you listen!
Keep in mind:
Math is primarily designed as a written language, not a spoken language!
Some things are read aloud in a simple (but slightly incomplete) way,
assuming that you're looking at the math while it is being read.
• PRACTICE (online exercises and printable worksheets)
Want more details, more exercises?
DEFINITION set
A set is a collection with the following property:
given any object, either the object is in the collection, or isn't in the collection.
EXAMPLES:
Question: Is ‘the collection of some people’ a set?
Solution: ‘The collection of some people’ is not a set; it is too vague.
Is your teacher in this collection? Maybe or maybe not!
Question: Is the collection of numbers consisting of $\;3,6,9,12,\ldots\;$ a set?
Solution: Yes.
Is the number (say) $\,35{,}983{,}205{,}119{,}780{,}238{,}482{,}108{,}222\,$ in this collection?
Well, either it is (if divisible by $\,3\,$) or isn't (if not divisible by $\,3\,$).
Notice that it's not important whether you personally know whether the answer is YES or NO;
all that matters is that the answer is definitely YES or NO.
• A set is a mathematical expression.
• Sets can have different names!
DEFINITIONS elements, members; finite set; infinite set
The objects in a set are called its elements or its members.
If a set has $\,n\,$ members, where $\,n\,$ is a whole number, then it is called a finite set
(pronounced with a long i, FIGH-night).
If a set is not finite, then it is infinite (pronounced with a short i, IN-fi-nit).
EXAMPLES:
$\{4,10\}\,$ is a finite set, with two members.
The number $\,4\,$ is a member.
The number $\,10\,$ is a member.
(More on list notation for sets below.)
The set $\,\{1,2,3,\ldots\}\,$ is an infinite set.
The number $\,7\,$ is an element of the set.
The number $\,\frac{30}{2}\,$ is an element of the set. (The name we use doesn't matter!)
The number $\,0.25\,$ is not an element of the set.
SYMBOLS USED IN CONNECTION WITH SETS:
The following symbols are used in connection with sets:
• $\cssId{s50}{\{}\;\;\cssId{s51}{\}}\;$ are called braces.
They are used in list notation for sets (see below).
• $\cssId{s54}{(}\;\;\cssId{s55}{)}\;$ are called parentheses. (Singular form is parenthesis.)
They are used in interval notation for sets (see below).
• $\cssId{s59}{[}\;\;\cssId{s60}{]}\;$ are called brackets.
They are used in interval notation for sets (see below).
LIST NOTATION FOR SETS:
• List notation for sets is used whenever the elements of a set can be listed.
• Braces $\;\{\;\;\}\;$ are used for list notation.
• Separate members of the set with commas.
• Use three dots $\;\ldots\;$ to indicate that a pattern is to be repeated.
Be sure to list enough elements to clearly establish the pattern.
• For a finite set, the order that elements are listed doesn't matter.
EXAMPLES:
Here are six names for the same finite set:
$\{1,2,3\}$ or $\{1,3,2\}$ or $\{2,1,3\}$ or $\{2,3,1\}$ or $\{3,1,2\}$ or $\{3,2,1\}$
The infinite set $\;\{0,1,2,3,\ldots\}\;$ contains all the whole numbers.
The finite set $\;\{0,1,2,\ldots,1000\}\;$ contains all the whole numbers between $\,0\,$ and $\,1000\,$.
The infinite set $\{2,1,0,\ldots\}\;$ contains all the integers that are less than or equal to $\;2\;$.
These are very different sets!
MEMBERSHIP IN A SET:
• The verb $\;\in\;$ is used to denote membership in a set.
• The sentence $\;x\in S\;$ is read as:
$\,x\,$ is in $\,S\;$’ or ‘$\,x\,$ is an element of $\,S\;$’ or ‘$\,x\,$ is a member of $\,S\;$’.
• The sentence $\;x\notin S\;$ is read as:
$\,x\,$ is not in $\,S\;$’ or ‘$\,x\,$ is not an element of $\,S\;$’ or ‘$\,x\,$ is not a member of $\,S\;$’.
EXAMPLE:
If $\,S\,$ is the set $\,\{1,2,3\}\,$, then all of the following sentences are true:
$\,1\in S\,$ and $\,3\in S\,$ and $\,4\notin S\,$ and $\,\frac{8}{4}\in S\,$ and $\,7-4\in S\,$
A SPECIAL SENTENCE:
The sentence ‘ Let $\,S = \{1,2,3\}\,$’ is used to assign the name $\,S\,$ to the set $\,\{1,2,3\}\,$.
The word ‘let’ is the key!
EXAMPLE:
Question: How would a mathematician say:
“Take the set $\,\{a,b,c,d\}\,$ and give it the name $\,T\;$”?
Solution: Let $\,T = \{a,b,c,d\}\;$.
(The word ‘let’ is a vital part of this sentence!)
DEFINITION empty set
The empty set is the unique set that has no members.
The empty set is denoted using either of these symbols: $\;\emptyset\;$ or $\;\{\;\}$
So, the empty set is empty! It has no members!
Be careful: the set $\,\{\emptyset\}\;$ is a set with one member—it is NOT the empty set.
INTERVAL NOTATION:
• Interval notation is used to describe intervals of real numbers.
• Intervals of real numbers are infinite sets.
• Parentheses $\;(\;\;)\;$ are used when an endpoint is not included.
• Brackets $\;[\;\;]\;$ are used when an endpoint is included.
• The ‘infinity’ symbol $\,\infty\,$ is used to denote that an interval extends forever to the right.
• The ‘negative infinity’ symbol $\,-\infty\,$ is used to denote that an interval extends forever to the left.
• The numbers used in interval notation always go from left to right on the number line.
SUBSETS:
Roughly, a subcollection from a set is called a subset.
EXAMPLE:
Let $\,S = \{a,b\}\,$.
Then:
• $\,\{a\}\,$ is a subset (choose only the ‘$\,a\,$’ )
• $\,\{b\}\,$ is a subset (choose only the ‘$\,b\,$’ )
• $\,\{a,b\}\,$ is a subset (choose everything!)
• $\,\{\;\}\,$ is a subset (choose nothing!)
DEFINITION subset
Let $\,S\,$ be a set.
Set $\,B\,$ is called a subset of $\,S\,$ if any one of the following three conditions holds:
• $\,B\,$ is the set $\,S\,$ itself
• $\,B\,$ is the empty set
• each member of $\,B\,$ is also a member of $\,S\,$
Master the ideas from this section
When you're done practicing, move on to:
Interval and List Notation
CONCEPT QUESTIONS EXERCISE:
PROBLEM TYPES:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
AVAILABLE MASTERED IN PROGRESS
(MAX is 24; there are 24 different problem types.) |
How do you simplify sqrt75*sqrt60?
Feb 1, 2016
$30 \sqrt{5}$
Explanation:
Recall that a $\textcolor{b l u e}{\text{perfect square}}$ is the product of squaring a whole number.
For example:
${2}^{2} = \textcolor{b l u e}{4}$
${3}^{2} = \textcolor{b l u e}{9}$
${5}^{2} = \textcolor{b l u e}{25}$
${6}^{2} = \textcolor{b l u e}{36}$
When you simplify a radical, you must first break it down using $\textcolor{b l u e}{\text{perfect square}}$ numbers. For example, in your case:
$\sqrt{75} \cdot \sqrt{60}$
$= \sqrt{\textcolor{b l u e}{25} \cdot 3} \cdot \sqrt{\textcolor{b l u e}{4} \cdot 15}$
The square root of $\textcolor{b l u e}{25}$ is $\textcolor{red}{5}$, so you can bring the 25 out of the square root sign and write a $5$ instead. Similarly, the square root of $\textcolor{b l u e}{4}$ is $\textcolor{red}{2}$, so you can also bring the $4$ out of the square root sign and write a $2$ instead.
$= \textcolor{red}{5} \textcolor{g r e e n}{\sqrt{3}} \cdot \textcolor{red}{2} \textcolor{g r e e n}{\sqrt{15}}$
Multiply.
$= \left(\textcolor{red}{5} \cdot \textcolor{red}{2}\right) \left(\textcolor{g r e e n}{\sqrt{3}} \cdot \textcolor{g r e e n}{\sqrt{15}}\right)$
$= 10 \sqrt{45}$
Since $\sqrt{45}$ can be simplified even further:
$= 10 \sqrt{\textcolor{b l u e}{9} \cdot 5}$
Simplify.
$= 10 \cdot \textcolor{red}{3} \sqrt{5}$
$= 30 \sqrt{5}$ |
# Find the area of the rectangle if the length is three times the width and the perimeter is 16.
hala718 | Certified Educator
Let the length of the rectangle be L and the width be w.
Given that the length of the rectangle is 3 times the width.
Then we will write:
L = 3*w ..............(1).
Also, given that the perimeter of the rectangle is 16.
Then we will write:
==> 2L + 2W = 16
We will divide by 2.
==> L + w = 8...........(2)
Now we will substitute with L = 3w
==> 3w + w = 8
==> 4w = 8
==> w= 2
==> L = 3*2 = 6
Then, the length of the rectangle is 6, and the width is 2 .
Now we will calculate the area of the rectangle.
We know that the area of the rectangle is given by:
\A = L * w
==> A = 2*6 = 12
Then, the area is 12 square units.
justaguide | Certified Educator
Let the length be L and the width be W.
Perimeter = 2(L+W). Area = L*W.
The perimeter is given as 16. Also the length is thrice the width.
This gives L = 3W
2(3W + W) = 16
=> 2*4W = 16
=> W = 16/8
=> W = 2
L = 3*W = 3*2 = 6
Therefore the area is 2*6 = 12.
The required area of the rectangle is 12 square units.
neela | Student
Given that length is 3 times width.
So if x is width, then length = 3x.
We know that if p is the perimeter of the rectangle, then p = 2*(length+width).
So p = 2(3x+x)= 8x.
Actual perimeter given = 16.
So 8x = 16.
x= 16/8 = 2.
Therefore x = 2 is the width.
3x= 3*2 = 6 is the length.
So area of the rectangle = length*width = 6*2 = 12 sq units. |
# Curves can be parallel
### Two parallel straight lines
Straight lines or stretches can be in special positions to each other. This is about "parallel".
These two straight lines are parallel to each other. That means: They have the same distance from one another everywhere.
Straight lines are infinitely long. You can imagine that the straight lines are still parallel even at infinity. That never changes.
Two straight lines \$\$ g \$\$ and \$\$ h \$\$ are parallel to one another if they always have the same distance from one another.
Short notation: \$\$ g \$\$ \$\$ || \$\$ \$\$ h \$\$.
A donkey bridge for the writing area \$\$ || \$\$ is that the \$\$ || \$\$ also occurs in the word “parael”.
If you draw your parallel lines somewhere in your exercise book, in your imagination they will run parallel to infinity.
If two straight lines are not parallel, you write:.
Two straight lines are not parallel if they have a common point of intersection.
### Two parallel routes
Not only straight lines can be parallel to one another, but also lines.
Here the segment \$\$ bar (AB) \$\$ is parallel to the segment \$\$ bar (CD) \$\$.
Short notation: \$\$ bar (AB) \$\$ \$\$ || \$\$ \$\$ bar (CD) \$\$.
### Draw straight lines parallel to each other
How do you draw parallel lines in your exercise book?
### Possibility Number 1
You use the parallel lines of your set square. They are on every set square. In the picture you can see them in.
The distances are each 0.5 cm = 5 mm. With the pink lines you draw parallels at a distance of 0.5 cm, 1 cm, 1.5 cm, ..., 4 cm.
If you want to draw parallels at a distance of, for example, 2.3 cm, you can also use the set square. Use the little ticks.
Example:
Draw a parallel to the blue line at a distance of 2.3 cm.
You put the set square at the right distance ...
and then draw the.
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### Draw parallels - option 2
For the second option, you use the vertical and the distance as an aid. How it works???
1. Place the set square with the center line (90 °) on the straight line. Then you draw the vertical line and mark the required distance. Here it is 2.3 cm.
2. You draw another vertical line at a second, slightly distant point and mark the required distance.
3. Connect the two marking points.
Option 2 allows you to draw more precisely.
### More than a parallel
There are always two parallel straight lines that are at the same distance from a given straight line. Figuratively speaking, one lies above the given straight line. The other is below the given straight line.
The two red straight lines are at the same distance from straight line g.
Most of the time you just need to draw exactly one parallel straight line. Then you can choose which ones to draw.
### Parallels more than 8 cm apart
The length of 8 cm is the highest that your set square has to offer. But there are also tasks in which you should draw a parallel that is more than 8 cm apart.
### Method 1
You draw auxiliary lines that are parallel to each other. If you need a distance of 10 cm, you draw an auxiliary parallel at 4 cm, then another at 4 cm and then the required parallel at a distance of 2 cm from the last auxiliary parallel. 4 cm + 4 cm + 2 cm = 10 cm.
### Method 2
Work with an extension of the vertical. You can use a long ruler to help. Draw very carefully as you extend the vertical.
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### Draw a parallel straight line through a point
Sometimes you didn't specify the distance, but a point through which the parallel should go.
Then the task is: Draw a parallel to the straight line through the given point P.
Here you have the two options again.
### Possibility Number 1
You place the set square on the starting line and move it parallel until you reach the point. Shifting parallel means that you use the lines drawn parallel to each other on the set square.
### Possibility 2
You draw a vertical line through the point. Then you draw another perpendicular to the first auxiliary line (the first perpendicular). Then that is the parallel.
If you draw a perpendicular \$\$ s_1 \$\$ to a straight line \$\$ g \$\$ and then another perpendicular \$\$ s_2 \$\$ to the perpendicular \$\$ s_1 \$\$, then \$\$ s_2 \$\$ and \$\$ g \$\$ are parallel to each other.
### Distance = 0
You can draw a straight line parallel to another straight line that has a distance of 0. This parallel is then not really visible, because it is identical to the straight line.
### In 3D
Straight lines can lie in space in such a way that they never intersect, but are also not parallel. These straight lines are called crooked. This is not possible on the plane, i.e. on paper.
In the plane, straight lines are always either parallel (special case identical) or they have exactly one point of intersection.
### Draw distant parallels through a point P.
If your task is to draw quite distant parallel through a point, you can use a trick. You will need a long ruler for this.
The straight line and a distant point are given.
1. You place the set square with the edge on the straight line.
2. You place the long ruler precisely on one leg of the set square.
3. You hold the ruler and move the set square to any position parallel to the starting line.
4. You slide until you reach point P.
5. Draw the parallel line through P.
This is what your result looks like:
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### Parallels in everyday life
Parallel straight lines or lines are very common.
Railway tracks are parallel. Otherwise the train would derail.
What is interesting about the train tracks is that they look to the observer as if they converge at one point at the end of the field of view. But that's just an optical illusion. You know it's not like that.
Image: Panther Media GmbH (Helmut Knab)
### Building construction
Image: fotolia.com (Uwe Kantz)
All the lines that strive upwards are parallel to each other.
When you look at the train tracks, it becomes clear to you that curves can also be parallel. But you only need the parallelism of straight lines and lines. And they have no curvature.
Image: TopicMedia Service (Bühler)
### Urban construction
There are tons of parallel streets in the Manhattan neighborhood of New York.
All streets from north to south are parallel to each other.
All streets from east to west are parallel to each other.
Image: Joachim Zwick
### packaging
The properties “parallel” and “vertical” play a role in sales and storage. Many goods are in boxes that are cuboid. The cardboard edges are perpendicular or parallel to each other. That is why the boxes are stackable.
Image: fotolia.com
### Parallels in mathematics
You probably already know parallel sides of special rectangles:
### Trapezoid
Parallels also occur in bodies. You can find parallel edges in cubes, cuboids or prisms, for example.
### Cuboid
All parallel edges are marked in color.
Do you already know the oblique image? This is the name of this type of 3D view. The advantage of oblique images is that the parallel edges are also parallel on the image.
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# Ratio and proportion in geometry II
In this lesson, we will divide oblique line segments into specified ratios by dividing the segment's horizontal and vertical displacements in the same ratio.
Quiz:
# Intro quiz - Recap from previous lesson
Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz!
Q1.State the ratio between the side lengths AB and BC.
1/5
Q2.State the constant of proportionality between triangle ABC and triangle DEF.
2/5
Q3.Find the length of side DE.
3/5
Q4.Find the length DE.
4/5
Q5.Find the length DE.
5/5
Quiz:
# Intro quiz - Recap from previous lesson
Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz!
Q1.State the ratio between the side lengths AB and BC.
1/5
Q2.State the constant of proportionality between triangle ABC and triangle DEF.
2/5
Q3.Find the length of side DE.
3/5
Q4.Find the length DE.
4/5
Q5.Find the length DE.
5/5
# Video
Click on the play button to start the video. If your teacher asks you to pause the video and look at the worksheet you should:
• Click "Close Video"
• Click "Next" to view the activity
Your video will re-appear on the next page, and will stay paused in the right place.
# Worksheet
These slides will take you through some tasks for the lesson. If you need to re-play the video, click the ‘Resume Video’ icon. If you are asked to add answers to the slides, first download or print out the worksheet. Once you have finished all the tasks, click ‘Next’ below.
Quiz:
# Ratio and proportion in geometry II
Show your understanding on today's lesson by completing the quiz
Q1.Fill in the gap: We can divide a line segment into a given ________ by considering the coordinates of its endpoints.
1/5
Q2.Fill in the gap: The _____________ of proportionality of ADE to ABC is 1.5.
2/5
Q3.Fill in the gap: The ratio of 𝐴𝐸: 𝐴𝐶 = 6: 9 = 2:___
3/5
Q4.Fill in the gap: The ratio of 𝐴𝐸: 𝐸𝐶 = ___: 1
4/5
Q5.A line segment ABC is split in the ratio AB to BC as 3 : 5. What fraction of the line segment is AB?
5/5
Quiz:
# Ratio and proportion in geometry II
Show your understanding on today's lesson by completing the quiz
Q1.Fill in the gap: We can divide a line segment into a given ________ by considering the coordinates of its endpoints.
1/5
Q2.Fill in the gap: The _____________ of proportionality of ADE to ABC is 1.5.
2/5
Q3.Fill in the gap: The ratio of 𝐴𝐸: 𝐴𝐶 = 6: 9 = 2:___
3/5
Q4.Fill in the gap: The ratio of 𝐴𝐸: 𝐸𝐶 = ___: 1
4/5
Q5.A line segment ABC is split in the ratio AB to BC as 3 : 5. What fraction of the line segment is AB?
5/5
# Lesson summary: Ratio and proportion in geometry II
## Time to move!
Did you know that exercise helps your concentration and ability to learn?
For 5 mins...
Move around:
Climb stairs
On the spot:
Chair yoga |
# GMAT Equations, Number Properties
Here is an interesting question that combines linear equations and properties of numbers to get an answer.
Question
A children’s gift store sells gift certificates in denominations of \$3 and \$5. The store sold ‘m’ \$3 certificates and ‘n’ \$5 certificates worth \$93 on a Saturday afternoon. If ‘m’ and ‘n’ are natural numbers, how many different values can ‘m’ take?
A. 5
B. 7
C. 6
D. 31
E. 18
Correct Answer : Choice C. 6 different values
Key data :
1. Total value of all certificates sold = \$93.
2. Certificates sold were in denominations of \$3 and \$5.
3. Both 'm' and 'n' are natural numbers.
The value of all certificates sold, 93 is divisible by 3.
So, a maximum of 31 \$3 certificates and no \$5 certificates could have been sold.
However, the question states that both 'm' and 'n' are natural numbers.
Hence, at least 1 \$5 certificate should have been sold.
If we reduce the number of \$3 certificates from the maximum 31 that is possible by say 'x' and correspondingly increase \$5 certificates by 'y', then 3x = 5y as the value of \$3 certificates reduced should be the same as the value of \$5 certificates increased.
It means that x has to be a multiple of 5 and y has to be a multiple of 3.
Or \$3 certificates reduce in steps of 5 certificates.
So, the following combinations are possible
1. m = 26, n = 3
2. m = 21, n = 6
3. m = 16, n = 9
4. m = 11, n = 12
5. m = 6, n = 15
6. m = 1, n = 18
Think about it another way. Replacing five \$3 certificates with three \$5 certificated leads to no change in the overall value of certficates sold and gives us a new combination every time. We need to see how many times this can be done. |
Mathematics is the language of the universe. It is used to describe and understand the world around us, from the smallest subatomic particles to the largest galaxies. Math is also essential for many different careers, from engineering and science to finance and business.
But for many people, math can be a daunting subject. They may have struggled with math in school, or they may simply not enjoy it. However, the good news is that anyone can learn math, regardless of their age or background.
In this article, we will provide a comprehensive overview of the most important math topics, from basic arithmetic to advanced calculus. We will also include a variety of helpful tips and tricks for solving math problems quickly and efficiently.
## Arithmetic
Arithmetic is the branch of mathematics that deals with the basic operations of addition, subtraction, multiplication, and division. Arithmetic is used in many different areas of our lives, from shopping and cooking to budgeting and investing.
### Basic arithmetic operations
The four basic arithmetic operations are addition, subtraction, multiplication, and division.
• Addition is the process of combining two or more numbers to find a sum. For example, 1 + 2 = 3.
• Subtraction is the process of taking one number away from another to find a difference. For example, 3 – 2 = 1.
• Multiplication is the process of adding a number to itself a certain number of times. For example, 2 x 3 = 6.
• Division is the process of splitting one number into equal parts. For example, 6 / 3 = 2.
### Order of operations
When there are multiple arithmetic operations in an expression, it is important to follow the order of operations to ensure that the expression is evaluated correctly. The order of operations is as follows:
1. Parentheses and brackets
2. Exponents
3. Multiplication and division
For example, in the expression 2 + 3 x 4, the multiplication operation is performed before the addition operation, so the expression is evaluated as 2 + 12 = 14.
### Fractions and decimals
Fractions and decimals are two different ways of representing parts of a whole. Fractions are written with a numerator (top number) and a denominator (bottom number). The numerator represents the number of parts, and the denominator represents the total number of parts. Decimals are written with a decimal point and a series of digits to the right of the decimal point. The digits to the right of the decimal point represent the parts of a whole, and the value of each digit decreases from left to right.
### Percentages
Percentages are a way of representing parts of a hundred. For example, 50% is the same as 50 out of 100, or half. Percentages are used in many different areas of our lives, from calculating discounts to determining the likelihood of an event happening.
### Algebra
Algebra is the branch of mathematics that deals with symbols and the rules for manipulating them. Algebra is used to solve a wide variety of problems, from simple word problems to complex scientific and engineering problems.
### Algebraic expressions and equations
An algebraic expression is a combination of numbers, symbols, and operators. For example, 2x + 3 is an algebraic expression. An algebraic equation is a statement that two algebraic expressions are equal. For example, 2x + 3 = 5 is an algebraic equation.
### Solving linear equations and inequalities
Linear equations are equations of the first degree. For example, the equation y = 2x + 3 is a linear equation. Linear inequalities are equations that state that one algebraic expression is greater than or less than another algebraic expression. For example, the inequality y > 2x + 3 is a linear inequality.
### Systems of equations
A system of equations is two or more equations that are solved simultaneously. Systems of equations are used to solve a wide variety of problems, such as finding the intersection of two lines or finding the price of two items when they are sold together.
Quadratic equations are equations of the second degree. For example, the equation x^2 + 2x + 1 = 0 is a quadratic equation. Quadratic equations are used to solve a variety of problems, such as finding the roots of a polynomial or finding the maximum or minimum value of a function.
### Polynomial equations
Polynomial equations are equations that contain one or more polynomial expressions. A polynomial expression is an expression that consists of variables, exponents, and coefficients.
Sure. Here is a continuation of my article on “Ask Me a Math Question: A Comprehensive Guide to Solving Any Math Problem”:
### Geometry
Geometry is the branch of mathematics that deals with the properties of shapes and spaces. Geometry is used in many different areas of our lives, from architecture and engineering to art and design.
### Basic geometric shapes
The basic geometric shapes are points, lines, planes, angles, triangles, quadrilaterals, circles, and spheres.
• Points are dimensionless locations in space.
• Lines are one-dimensional objects that extend infinitely in both directions.
• Planes are two-dimensional objects that extend infinitely in all directions.
• Angles are formed by two intersecting lines or rays.
• Triangles are three-sided polygons.
• Circles are round shapes with no corners.
• Spheres are three-dimensional objects with no corners.
### Geometric transformations
Geometric transformations are mappings that transform one geometric shape into another geometric shape. The four basic geometric transformations are translations, rotations, reflections, and dilations.
• Translations move a shape from one location to another location.
• Rotations turn a shape around a point.
• Reflections flip a shape over a line.
• Dilations enlarge or reduce a shape.
### Geometric proofs
Geometric proofs are logical arguments that demonstrate the truth of geometric statements. Geometric proofs are used to establish the properties of geometric shapes and spaces.
### Trigonometry
Trigonometry is the branch of mathematics that deals with the relationships between the sides and angles of triangles. Trigonometry is used in many different areas, such as surveying, navigation, and engineering.
### The six trigonometric functions
The six trigonometric functions are sine, cosine, tangent, cotangent, secant, and cosecant. These functions are used to calculate the sides and angles of triangles.
### Solving right triangles
Right triangles are triangles that have one right angle (90 degrees). Right triangles can be solved using the trigonometric functions.
### Solving oblique triangles
Oblique triangles are triangles that do not have a right angle. Oblique triangles can be solved using the trigonometric functions and the Law of Cosines and the Law of Sines.
### Trigonometric identities
Trigonometric identities are equations that relate the trigonometric functions. Trigonometric identities are used to simplify trigonometric expressions and to solve trigonometric problems.
### Calculus
Calculus is the branch of mathematics that deals with the rates of change of quantities. Calculus is used in many different areas, such as physics, engineering, and economics.
### Limits and derivatives
A limit is a value that a function approaches as its input approaches a certain value. A derivative is the rate of change of a function. Derivatives are used to calculate the velocity and acceleration of objects, the slope of tangent lines to curves, and the maximum and minimum values of functions.
### Integrals
An integral is the opposite of a derivative. Integrals are used to calculate the area under a curve, the volume of a solid, and the distance traveled by an object.
### Applications of calculus
Calculus is used in many different areas, such as physics, engineering, and economics. For example, calculus is used to calculate the trajectory of a projectile, the strength of a bridge, and the growth rate of an economy.
## Solving Math Problems
### General tips and tricks for solving math problems
Here are some general tips and tricks for solving math problems:
• Read the problem carefully and make sure you understand what it is asking.
• Identify the important information in the problem.
• Draw a diagram or create a table to help you visualize the problem.
• Write down an equation or expression that represents the problem.
• Solve the equation or expression.
### How to solve specific types of math problems
There are many different types of math problems, and each type of problem has its own unique solution strategy. Here are some tips for solving specific types of math problems:
• Word problems: Word problems are math problems that are written in words. To solve a word problem, you must first identify the important information in the problem and then write down an equation or expression that represents the problem.
• Algebra problems: Algebra problems are math problems that use symbols to represent numbers and operations. To solve an algebra problem, you must use the rules of algebra to manipulate the symbols and solve the equation or inequality Sure. Here is a continuation of my article on “Ask Me a Math Question: A Comprehensive Guide to Solving Any Math Problem”:
Geometry problems: Geometry problems are math problems that involve geometric shapes. To solve a geometry problem, you must use the properties of geometric shapes and the rules of geometry to solve the problem.
Trigonometry problems: Trigonometry problems are math problems that involve the trigonometric functions. To solve a trigonometry problem, you must use the definitions of the trigonometric functions and the rules of trigonometry to solve the problem.
Calculus problems: Calculus problems are math problems that involve limits, derivatives, and integrals. To solve a calculus problem, you must use the definitions of limits, derivatives, and integrals and the rules of calculus to solve the problem.
### Using a calculator to solve math problems
Calculators can be very helpful for solving math problems. However, it is important to use a calculator correctly. Here are some tips for using a calculator to solve math problems:
• Make sure you understand how to use the calculator’s functions.
• Enter the numbers and operations in the correct order.
• Use parentheses to group operations together.
## Conclusion
Mathematics is a vast and complex subject, but it is also a very rewarding one. By learning the basics of mathematics, you can open up a world of possibilities. You can use math to solve problems, make decisions, and create new things.
If you are struggling with math, don’t give up. There are many resources available to help you learn math. You can find math tutorials online, in books, and at your local library. You can also ask a friend, family member, or teacher for help.
With practice and perseverance, you can master any math problem.
## FAQs
### Q.What is the best way to learn math?
The best way to learn math is by doing. Try to solve as many math problems as you can. You can find math problems in textbooks, online, and in workbooks. You can also ask your teacher or a tutor for help.
### Q.What are some common math mistakes that people make?
Some common math mistakes that people make include:
• Not following the order of operations.
• Making careless errors when calculating.
### Q.How can I improve my math skills?
There are many ways to improve your math skills. Here are a few tips:
• Practice solving math problems regularly.
• Get help from a teacher, tutor, or friend if you are struggling.
• Use math in your everyday life. For example, you can use math to calculate the tip at a restaurant or to figure out how much fabric you need to make a dress.
### Q.What are some careers that use math?
There are many careers that use math. Here are a few examples:
• Engineer
• Scientist
• Statistician
• Economist
• Accountant
• Teacher
• Software developer
• Financial analyst |
Below everything is explained but I would advise working through the task above and look at the videos when stuck.
solution videos for the above task.
It is possible to write every quadratic equation in the form
$${ax^2 + bx + c = 0}$$, where a is a real number not equal to 0 and b and c are real numbers.
The constants a, b and c are called coefficients (Koeffizienten). In German textbooks, when ${a = 1}$, the quadratic equation is in the ‘so-called’ Normalform - $${ x^2 + px + q = 0}$$
## no linear term in the quadratic equations - Die reinquadratische Gleichung
$${ax^2 + c = 0}$$
If ${b = 0}$, the linear term ${bx}$ in the quadratic equation is missing. It then has the form
{\begin{align} ax^2 + c & = 0 \newline ax^2 & = - c \newline x^2 & = - \frac{c}{a} \newline x^2 & = q \end{align}}
#### Example 1
{\begin{align}5x^2 - 80 & =0\newline 5 \cdot x^2 - 80 & = 0 \newline x^2 - 16 & = 0\newline x^2 & = 16\newline x & = \pm \sqrt{16}\end{align}} $${x_1 = 4 ~ \lor x_2 = -4}$$
${ \lor}$ means OR.
The number of solutions is determined as follows.
• one solution (eine Lösung) for ${q = 0}$
• two solutions (zwei Lösungen) for ${q > 0}$
• no solution (keine Lösung) for ${q < 0}$
## The constant term is zero - Die Konstante ist Null
When the constant term is zero, i.e., ${c = 0}$, the quadratic equation has the following form $${ax^2 + bx = 0}$$ where a is a real number not equal to 0 and b is any real number. We solve these equations by factorising (Questions 1 and 3 only). {\begin{align}ax^2 + bx & =0 \newline x \cdot (ax + b) & = 0 \end{align}} applying the Null Factor Law which states that if the product is zero then at least one of the factors is zero.
$${x= 0 \lor ax + b = 0}$$ $${ ax = -b}$$ $${ x = -\frac{b}{a}}$$ $${x_1 = 0 ~ \lor x_2 = -\frac{b}{a}}$$
All quadratic equations of the form ${ax^2+bx=0}$ have a common solution ${x=0}$.
#### Example 2
{\begin{align}5x^2 + 2x & = 0 \newline x \cdot (5x + 2) & = 0 \end{align}} It follows that {\begin{align}x= 0 ~~ \lor 5x + 2 & = 0 \newline \Leftrightarrow 5x & = -2 \newline \Leftrightarrow x & = -\frac{2}{5}\end{align}} $${x_1 = 0~ \lor x_2 = - \frac{2}{5} }$$
For more practice go to following links
## factorising (theorem of Viète) - Satz von Vieta1
When the quadratic equation is in factored form $${(x-x_1)\cdot(x-x_2) = 0}$$
the solutions can be found using the null factor law. Set each factor equal to zero.
{\begin{align} x-x_1 & = 0 &&\lor &&& x-x_2 = 0 \newline x & = x_1 &&\lor &&& x = x_2 \end{align}} $${\mathbb{L} = \{ x_1; x_2 \}}$$
The theorem of Viète indicates if the quadratic equation can be factorised. If a quadratic equation ${x^2 + bx + c = 0}$ has two solutions ${x_1}$ and ${ x_2}$, then following holds - $${x^2 + bx + c =(x - x_1) \cdot (x - x_2)}$$ where
$${x_1 + x_2 = -b}$$ $${x_1 \cdot x_2 = c}$$
#### Example 3
Factorise $${ x^2 - 6x + 8 = 0}$$ First, we have b = -6 and c = 8. According to Vieta following statement must hold. $${c = x_1 \cdot x_2 = 8 }$$ So, find all the combinations of factors of 8. $${ 8 = 8 \cdot 1 = (-8)\cdot(-1)=4 \cdot 2 = (-4)\cdot(-2)}$$ Which of these factors also holds for Vieta’s second statement? $${x_1 + x_2 = -b = -(-6)= 6}$$ The only two factors giving -b = 6 are $${x_1 = 4 }$$ and $${x_2 = 2}$$ Hence, $${ x^2 - 6x + 8 = (x - 4)\cdot (x-2)}$$ And, we can rewrite the equation in factored form. $${(x - 4)\cdot (x-2)=0}$$ $${\Leftrightarrow x-4 = 0 \lor x -2 = 0}$$ $${\Leftrightarrow x_1 = 4 \lor x_2 = 2}$$ So, the solution set is- $${\mathbb{L} = \{ 2; 4 \}}$$
## Completing the square - quadratische Ergänzung
Completing the square means to rearrange part of the equation into a perfect square. $${x^2 + 2ux + u^2 = (x + u)^2}$$
#### Example 4
{\begin{align} \frac{1}{2}x^2 +5x & = 28 && \text{subtract 28} \newline \frac{1}{2}x^2 +5x - 28 & = 0 && \text{multiply by 2}\newline x^2 + 10x - 56 & = 0 && \text{eliminate constant term on left hand side} \newline x^2 + 10x & = 56 && \text{complete the square}\end{align}} $${2ux = 10x => u = 5 => u^2 = 25}$$ {\begin{align} x^2 + 10x + 25 & = 56 + 25 && \text{binomial formula} \newline (x + 5)^2 & = 81 && \text{square root}\newline x + 5 & = \pm 9 \end{align}} $${ x + 5 = + 9 \lor x + 5 = - 9 }$$ $${ x = -5 + 9 \lor x = -5 - 9 }$$ $${ x = 4 \lor x = -14 }$$ $${\mathbb{L} = \{ 4; -14 \} }$$
To solve for roots in the general form see this interactive page.
Following link on completing the square has many exercises. The approach is slightly different to the one explained above as all terms remain on the left hand side. The method on this page saves some steps.
Completing the square for the expression $${ax^2+bx+c=0}$$ leads to what is known as the quadratic formula.
If ${b^2-4ac > 0}$, the quadratic formula to solve ${ax^2+bx+c=0}$ is
$${x_{1,2} = \frac{-b \pm \sqrt{b^2-4ac}}{2a}}$$
#### examples on video
Here are two videos showing you how to use the quadratic formula.
See no 14 on the solution page The notation ${x_{1,2} }$ means there are two solutions ${x_1 }$ and ${x_2}$.
But, if $${b^2-4ac = 0}$$, there is one solution:$${x = \frac{-b}{2a}}$$
And lastly, if ${b^2-4ac < 0}$, the square root of a negative number is not defined for real numbers. There is no real solution.
Note, the expression ${b^2-4ac}$ is also called the discriminant ${\Delta}$.
(later on, you may come across complex numbers which provide solutions for negative roots).
## The pq - formula - a special case
(used in some parts of Germany, especially in Berlin)
If the equation is given in the following form: $${ x^2 + px + q = 0}$$
Then the quadratic equation is in ‘normalised’ form, i.e., the coefficient of the quadratic term is 1, i.e., a = 1 , then the quadratic formula is simplified to what is known as the p-q-formula.
$${x_{1,2} = \frac{-p}{2} \pm \sqrt{\left(\frac{p}{2}\right)^2-q}}$$
See no 13b on the following video
#### examples on video
Here are two videos showing you how to use the pq-formula.
#### “completing the square” to derive the p-q-formula - Herleitung der pq-Formel
{\begin{align} x^2 + px + q & = 0 \newline x^2 + px & = -q \newline x^2 + px + \left(\frac{p}{2}\right)^2 & = -q + \left(\frac{p}{2}\right)^2 \newline \left(x + \left(\frac{p}{2}\right) \right)^2 & = \left(\frac{p}{2}\right)^2 -q \newline x + \left(\frac{p}{2}\right) & = \pm \sqrt{ \left(\frac{p}{2}\right)^2 -q} \newline x & = - \left(\frac{p}{2}\right) \pm \sqrt{ \left(\frac{p}{2}\right)^2 -q} \end{align}}
You can convert all quadratic equations into the p-q-form by dividing all terms by a (the coefficient of the quadratic term)
### Solving quadratics where it is necessary to investigate different cases
(c) 2019 mrwilliams[at]sebinberlin.de - impressum und datenschutz - Powered by MathJax & XMin & HUGO & jsxgraph & mypaint |
Find here many important first grade math concepts taught in first grade. Teachers, parents, and math tutors can also use them as a guideline to illustrate a math lesson or to teach important skills that kids are supposed to know in first grade math.
Equal parts or equal sharesThe orange circle has 2 equal parts.The blue circle has 4 equal parts
ArraysThe first array is made of soccer balls.The array has 2 rows and 3 columns.2 × 3 = 6 ballsThe second array is made of squares.The array has 3 rows and 5 columns.3 × 5 = 15 squares
QuarterQuarter = one-fourth = 1 of 4 equal parts
Associative property
Closed figureAll sides of a closed figure are connected.
Hour handThe hour hand takes 1 hour to move from a number to the next. Hour or hr = 60 minutes.
EstimateAn estimate is not the same as an exact answer. We just need a number close to the exact amount. To estimate the number of marbles, we just need to know about how manyPossible answers: 15, 14, or 17
Commutative propertyWhether 2 is written first or not, the answer to 2 + 1 is still 3
2-dimensional figures
Types of clocks
Cube:A cube has 6 faces, 12 edges and 8 vertices. The cube on the right shows 3 faces with the color red, blue, and orange.
Multiple of 10
Counting upStart at 3 and count 2 more to reach 5.
NumberA number indicates how many or how much. A numeral is used to represent the number of objects. For example, after counting the pencils on the right, we get 4. We use the numeral 4 to represent 4 pencils.
3-dimensional shapes.3 dimensional shapes are solid shapes that have width, length, and height.
Make 10
Minute handThe minute hand takes 5 minutes to move from a number to the next. Minute = seconds.
DifferentThe two triangles are different, but they have the same shape.
Counting downStart at 6 and count backward to reach 2.
Composite shapeThe arrow on the right is a composite shape since it is made from 2 basic shapes ( a triangle and a rectangle)
Collecting data
ExpressionAn expression has no equal sign
Coins1 penny = 1 cent1 nickel = 5 cents1 dime = 10 cents1 quarter = 25 cents2 dimes + 1 quarter + 3 nickels = 60 cents
EquationEquation always have an equal sign. The amount on the left side is equal to the amount on the right side.
New math lessons
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## Recent Articles
1. ### Number Game
Dec 01, 16 06:27 AM
1. Think of any positive whole number. No fractions, no decimals,no percents. 2. Double that number. 3. Add 10 to that result in step 2. 4. Divide |
Factors of 61
The factors of 61 are numbers that can divide by 61 exactly and leave no remainder. Because 61 is a prime number, there are only two factors. The pair factors for 61 can be either positive or negative. The pair factor for 61 can be either (1, 61) OR (-1, 61). will obtain the original number if we multiply the pair of negative numbers, such as -1 and 61.
Factors Calculator
Enter Number
Prime Numbers
A prime number is a whole number greater than 1 that can only be divided evenly by 1 and itself. Some examples of prime numbers include 2, 3, 5, 7, and 11. Prime numbers are important in mathematics as they are used in many mathematical formulas and algorithms and are also used in the study of number theory.
Composite Numbers
A composite number is a whole number greater than 1 that can be divided evenly by a number other than 1 or itself. In other words, a composite number is not a prime number. Some examples of composite numbers include 4, 6, 8, 9, and 10.
What are the factors of 61?
The factors of 61 are numbers that can divide 61 completely, leaving the rest 0. The number 61 is also known as the factors of 61. Because 61 is a prime number, there are only two factors, such as 1 or the number itself. The factors of 61, therefore, are 1 and 61.
How to Calculate the factor of 61?
The method of calculating the factors of 61 is as follows. First, every number is divisible by itself and 1.
Therefore, the factor of 61 are 1, …
We can find all factors of a number by dividing it by 1, 2, 3, 4…
(i) 61 ÷ 1 = 61
This division gives the remainder 0 and so is divisible by 61. So please put them 1 and 61 in your factor list.
1, …….. 61
(ii) 61 ÷ 2 = 30.5
This division gives the remainder 30.5, needing to be more thoroughly divided. So we will not write 2 and 30.5 on the list.
(iii) 61 ÷ 3 = 20.33
This division gives the remainder 20.33, needing to be more thoroughly divided. So we will not write 3 and 20.33 on the list.
(iv) 61 ÷ 4 = 15.25
This division gives the remainder 15.25, needing to be more thoroughly divided. So we will not write 4 and 15.25 on the list.
(v) 61 ÷ 5 = 12.2
This division gives the remainder 12.2, not being thoroughly divided. So we will not write 5 and 12.2 on the list.
(vi) Since we don’t have any more numbers to calculate, we are putting the numbers so far.
So 1, 61 are a factor of 61.
Factors of – 61
As – 1 and – 61 are negative factor, you get a positive number by multiplying two negatives, like (- 1) × (- 61) = 61.
So – 1, – 61 are a negative factors of 61.
All factors of 61
Here is a list of all the positive and negative factor of 61 in numerical order.
Positive Factors of 61 = 1, 61
Negative Factors of 61 = – 1, – 61
Factors of 61 in pairs
1 and 61 are a factor pair of 61 since 1 x 61 = 61
Therefore, the pair factor of 61 are (1, 61). |
# NCERT book Solutions for Class 10 Maths Polynomials Exercise 2.2
## NCERT book Solutions for Class 10 Maths Chapter 2 Exercise 2.2
In this page we have NCERT book Solutions for Class 10th Maths:Polynomials for Exercise 2.2 on page 33 . Hope you like them and do not forget to like , social_share and comment at the end of the page.
Question 1
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) x2 – 2x – 8
(ii) 4s2 – 4s + 1
(iii) 6x2 – 3 – 7x
(iv) 4u2 + 8u
(v)t2 – 15
(vi) 3x2 – x – 4
(i) x2 – 2x – 8
= x2 -4x+ 2x – 8
= (x - 4) (x + 2)
Therefore, the zeroes of x2 – 2x – 8 are 4 and -2.
(ii) 4s2 – 4s + 1
From (a-b)2 = a2 -2ab + b2
= (2s-1)2
Therefore, the zeroes of 4s2 - 4s + 1 are 1/2 and 1/2.
(iii) 6x2 – 3 – 7x
= 6x– 7x – 3
= 6x2 -9x +2x -3
= (3x + 1) (2x - 3)
Therefore, the zeroes of 6x2 - 3 - 7x are -1/3 and 3/2.
(iv) 4u2 + 8u
= 4u2 + 8u
= 4u(u + 2)
Therefore, the zeroes of 4u2 + 8u are 0 and - 2.
(v) t2 – 15
From (a2 -b2) =(a-b) (a+b)
= (t - √15) (t + √15)
Therefore, the zeroes of t2 - 15 are √15 and -√15.
(vi) 3x2 – x – 4
=3x2 – 4x+3x – 4
= (3x - 4) (x + 1)
Therefore, the zeroes of 3x2 – x – 4 are 4/3 and -1.
Verification of the relationship between the zeroes
Question 2
Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) 1/4, -1
(ii) √2, 1/3
(iii) 0, √5
(iv) 1,1
(v) -1/4 ,1/4
(vi) 4,1
(i) 1/4 , -1 Let the polynomial be ax2 + bx + c, and its zeroes be p and q
p + q = 1/4 = -b/a
pq = -1 = -4/4 = c/a
Now we have two method to find the quadratic polynomial
Method -1
The polynomial can be written as
k[x2 - (sum of roots)x + (products of roots)]
So,
$k[x^2 - (\frac {1}{4})x -1]$
Taking k=4
$4x^2 -x -4$
Method -2
If we take a= 4, then b = -1, c= -4
Therefore, the quadratic polynomial is 4x2 - x -4.
We could choose any of the method to get the polynomial
(ii) √2 , 1/3
Let the polynomial be ax2 + bx + c, and its zeroes be p and q
p + q = √2 = 3√2/3 = -b/a
pq = 1/3 = c/a
If a = 3, then b = -3√2, c = 1
Therefore, the quadratic polynomial is 3x2 -3√2x +1.
(iii) 0, √5
Let the polynomial be ax2 + bx + c, and its zeroes be p and q
p + q = 0 = 0/1 = -b/a
pq = √5 = √5/1 = c/a
If a = 1, then b = 0, c = √5
Therefore, the quadratic polynomial is x2 + √5.
(iv) 1, 1
Let the polynomial be ax2 + bx + c, and its zeroes be p and q
p + q = 1 = 1/1 = -b/a
pq = 1 = 1/1 = c/a
If a = 1, then b = -1, c = 1
Therefore, the quadratic polynomial is x2 - x +1.
(v) -1/4 ,1/4
Let the polynomial be ax2 + bx + c, and its zeroes be p and q
p + q = -1/4 = -b/a
pq = 1/4 = c/a
If a = 4, then b = 1, c = 1
Therefore, the quadratic polynomial is 4x2 + x +1.
(vi) 4,1
Let the polynomial be ax2 + bx + c, and its zeroes be p and q
p + q = 4 = 4/1 = -b/a
pq = 1 = 1/1 = c/a
If a = 1, then b = -4, c = 1
Therefore, the quadratic polynomial is x2 - 4x +1.
## Summary
1. NCERT book Solutions for Class 10th Maths:Polynomials Exercise 2.2 has been prepared by Expert with utmost care. If you find any mistake.Please do provide feedback on mail. You can download the solutions as PDF in the below Link also
2. This chapter 2 has total 4 Exercise 2.1 ,2.2,2.3 and 2.4. This is the Second exercise in the chapter.You can explore previous exercise of this chapter by clicking the link below
Go back to Class 10 Main Page using below links
### Practice Question
Question 1 What is $1 - \sqrt {3}$ ?
A) Non terminating repeating
B) Non terminating non repeating
C) Terminating
D) None of the above
Question 2 The volume of the largest right circular cone that can be cut out from a cube of edge 4.2 cm is?
A) 19.4 cm3
B) 12 cm3
C) 78.6 cm3
D) 58.2 cm3
Question 3 The sum of the first three terms of an AP is 33. If the product of the first and the third term exceeds the second term by 29, the AP is ?
A) 2 ,21,11
B) 1,10,19
C) -1 ,8,17
D) 2 ,11,20 |
# Maths NCERT solutions class 10 Chapter 11 Constructions
Chapter 11 Constructions
• 11.1 Introduction
• 11.2 Division of a Line Segment
• 11.3 Construction of Tangents to a Circle
• 11.4 Summary
Maths NCERT solutions class 10 Chapter 11 Constructions
Important formula
#### Exercise 11.1
Q.In each of the following, give the justification of the construction also:
1. Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure the two parts.
Steps of Construction:
Step I: AB = 7.6 cm is drawn.
Step II: A ray AX making an acute angle with AB is drawn.
Step III: After that, a ray BY is drawn parallel to AX making equal acute angle as in the previous step.
Step IV: Point A1, A2, A3, A4 and A5 is marked on AX and point B1, B2…. to B8 is marked on BY such that AA1 = A1A2 = A2A3 =….BB1= B1B2 = …. B7B8
Step V: A5 and B8 is joined and it intersected AB at point C diving it in the ratio 5:8.
AC : CB = 5 : 8
Justification:
ΔAA5C ~ ΔBB8C
∴ AA5/BB8 = AC/BC = 5/8
Q.2. Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are 2/3 of the corresponding sides of the first triangle.
Steps of Construction:
Step I: AB = 6 cm is drawn.
Step II: With A as a centre and radius equal to 4 cm, an arc is draw.
Step III: Again, with B as a centre and radius equal to 5 cm an arc is drawn on same side of AB intersecting previous arc at C.
Step IV: AC and BC are joined to form ΔABC.
Step V: A ray AX is drawn making an acute angle with AB below it.
Step VI: 5 equal points (sum of the ratio = 2 + 3 =5) is marked on AX as A1 A2….A5
Step VII: A5B is joined. A2B’ is drawn parallel to A5B and B’C’ is drawn parallel to BC.
ΔAB’C’ is the required triangle
Justification:
∠A(Common)
∠C = ∠C’ and ∠B = ∠ B’ (corresponding angles)
Thus ΔAB’C’ ~ ΔABC by AAA similarity condition
From the figure,
AB’/AB = AA2/AA5 = 2/3
AB’ =2/3 AB
AC’ = 2/3 AC
Q.3. Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are
7/5 of the corresponding sides of the first triangle.
Steps of Construction:
Step I: A triangle ABC with sides 5 cm, 6 cm and 7 cm is drawn.
Step II: A ray BX making an acute angle with BC is drawn opposite to vertex A.
Step III: 7 points as B1 B2 B3 B4 B5 B6 and B7 are marked on BX.
Step IV; Point B5 is joined with C to draw B5C.
Step V: B7C’ is drawn parallel to B5C and C’A’ is parallel to CA.
Thus A’BC’ is the required triangle.
Justification
ΔAB’C’ ~ ΔABC by AAA similarity condition
∴ AB/A’B = AC/A’C’ = BC/BC’
and BC/BC’ = BB5/BB7 = 5/7
∴A’B/AB = A’C’/AC = = BC’/BC = 7/5
Q.4. Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 1.5 times the corresponding sides of the isosceles triangle.
Steps of Construction:
Step I: BC = 5 cm is drawn.
Step II: Perpendicular bisector of BC is drawn and it intersect BC at O.
Step III: At a distance of 4 cm, a point A is marked on perpendicular bisector of BC.
Step IV: AB and AC is joined to form ΔABC.
Step V: A ray BX is drawn making an acute angle with BC opposite to vertex A.
Step VI: 3 points B1 B2 and B3 is marked BX.
Step VII: B2 is joined with C to form B2C.
Step VIII: B3C’ is drawn parallel to B2C and C’A’ is drawn parallel to CA.
Thus, A’BC’ is the required triangle formed.
Justification:
ΔAB’C’ ~ ΔABC by AA similarity condition.
∴ AB/AB’ = BC/B’C’ = AC/AC’
also,
AB/AB’ = AA2/AA3 = 2/3
⇒ AB’ = 3/2 AB, B’C’ = 3/2 BC and AC’ = 3/2 AC
Q.5. Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are 3/4 of the corresponding sides of the triangle ABC.
Steps of Construction:
Step I: BC = 6 cm is drawn.
Step II: At point B, AB = 5 cm is drawn making an
∠ABC = 60° with BC.
Step III: AC is joined to form ΔABC.
Step IV: A ray BX is drawn making an acute angle with BC opposite to vertex A.
Step V: 4 points B1 B2 B3 and B4 at equal distance is marked on BX.
Step VII: B3 is joined with C’ to form B3C’.
Step VIII: C’A’ is drawn parallel CA.
Thus, A’BC’ is the required triangle.
Justification:
∠A = 60° (Common)
∠C = ∠C’
ΔAB’C’ ~ ΔABC by AA similarity condition.
∴ AB/AB’ = BC/B’C’ = AC/AC’
also,
AB/AB’ = AA3/AA4 = 4/3
⇒ AB’ = 3/4 AB, B’C’ = 3/4 BC and AC’ = 3/4 AC
Q.6. Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are 4/3 times the corresponding sides of Δ ABC.
Sum of all side of triangle = 180°
∴ ∠A + ∠B + ∠C = 180°
∠C = 180° – 150° = 30°
Steps of Construction:
Step I: BC = 7 cm is drawn.
Step II: At B, a ray is drawn making an angle of 45° with BC.
Step III: At C, a ray making an angle of 30° with BC is drawn intersecting the previous ray at A. Thus, ∠A = 105°.
Step IV: A ray BX is drawn making an acute angle with BC opposite to vertex A.
Step V: 4 points B1 B2 B3 and B4 at equal distance is marked on BX.
Step VI: B3C is joined and B4C’ is made parallel to B3C.
Step VII: C’A’ is made parallel CA.
Thus, A’BC’ is the required triangle.
Justification:
∠B = 45° (Common)
∠C = ∠C’
ΔAB’C’ ~ ΔABC by AA similarity condition.
∴ BC/BC’ = AB/A’B’ = AC/A’C’
also,
BC/BC’ = BB3/BB4 = 34
⇒ AB = 4/3 AB’, BC = 4/3 BC’ and AC = 4/3 A’C’
Q.7. Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle.
Steps of Construction:
Step I: BC = 3 cm is drawn.
Step II: At B, A ray making an angle of 90° with BC is drawn.
Step III: With B as centre and radius equal to 4 cm, an arc is made on previous ray intersecting it at point A.
Step IV: AC is joined to form ΔABC.
Step V: A ray BX is drawn making an acute angle with BC opposite to vertex A.
Step VI: 5 points B1 B2 B3 B4 and B5 at equal distance is marked on BX.
Step VII: B3C is joined B5C’ is made parallel to B3C.
Step VIII: A’C’ is joined together.
Thus, ΔA’BC’ is the required triangle.
Justification:
As in the previous question 6.
#### Exercise 11.2
Q.In each of the following, give also the justification of the construction:
1. Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.
Steps of Construction:
Step I: With O as a centre and radius equal to 6 cm, a circle is drawn.
Step II: A point P at a distance of 10 cm from the centre O is taken. OP is joined.
Step III: Perpendicular bisector OP is drawn and let it intersected at M.
Step IV: With M as a centre and OM as a radius, a circle is drawn intersecting previous circle at Q and R.
Step V: PQ and PR are joined.
Thus, PQ and PR are the tangents to the circle.
On measuring the length, tangents are equal to 8 cm.
PQ = PR = 8cm.
Justification:
OQ is joined.
∠PQO = 90° (Angle in the semi circle)
∴ OQ ⊥ PQ
Therefor, OQ is the radius of the circle then PQ has to be a tangent of the circle.
Similarly, PR is a tangent of the circle.
Q.2. Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation.
Steps of Construction:
Step I: With O as a centre and radius equal to 4 cm, a circle is drawn.
Step II: With O as a centre and radius equal to 6 cm, a concentric circle is drawn.
Step III: P be any point on the circle of radius 6 cm and OP is joined.
Step IV: Perpendicular bisector of OP is drawn which cuts it at M
Step V: With M as a centre and OM as a radius, a circle is drawn which intersect the the circle of radius 4 cm at Q and R
Step VI: PQ and PR are joined.
Thus, PQ and PR are the two tangents.
Measurement:
OQ = 4 cm (Radius of the circle)
PQ = 6 cm ( Radius of the circle)
∠PQO = 90° (Angle in the semi circle)
Applying Pythagoras theorem in ΔPQO,
PQ2 + QO2 = PO2
⇒ PQ2 + 42= 62
⇒ PQ2 + 16 = 36
⇒ PQ2 = 36 – 16
⇒ PQ2 = 20
⇒ PQ = 2√5
Justification:
∠PQO = 90° (Angle in the semi circle)
∴ OQ ⊥ PQ
Therefor, OQ is the radius of the circle then PQ has to be a tangent of the circle.
Similarly, PR is a tangent of the circle.
Q.3. Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q.
Steps of Construction:
Step I: With O as a centre and radius equal to 3 cm, a circle is drawn.
Step II: The diameter of the circle is extended both sides and an arc is made to cut it at 7 cm.
Step III: Perpendicular bisector of OP and OQ is drawn and x and y be its mid-point.
Step IV: With O as a centre and Ox be its radius, a circle is drawn which intersected the previous circle at M and N.
Step V: Step IV is repeated with O as centre and Oy as radius and it intersected the circle at R and T.
Step VI: PM and PN are joined also QR and QT are joined.
Thus, PM and PN are tangents to the circle from P and QR and QT are tangents to the circle from point Q.
Justification:
∠PMO = 90° (Angle in the semi circle)
∴ OM ⊥ PM
Therefor, OM is the radius of the circle then PM has to be a tangent of the circle.
Similarly, PN, QR and QT are tangents of the circle.
Q.4. Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60°.
We know that radius of the circle is perpendicular to the tangents.
Sum of all the 4 angles of quadrilateral = 360°
∴ Angle between the radius (∠O) = 360° – (90° + 90° + 60°) = 120°
Steps of Construction:
Step I: A point Q is taken on the circumference of the circle and OQ is joined. OQ is radius of the circle.
Step II: Draw another radius OR making an angle equal to 120° with the previous one.
Step III: A point P is taken outside the circle. QP and PR are joined which is perpendicular OQ and OR.
Thus, QP and PR are the required tangents inclined to each other at an angle of 60°.
Justification:
Sum of all angles in the quadrilateral PQOR = 360°
∠QOR + ∠ORP + ∠OQR + ∠RPQ = 360°
⇒ 120° + 90° + 90° + ∠RPQ = 360°
⇒∠RPQ = 360° – 300°
⇒∠RPQ = 60°
Hence, QP and PR are tangents inclined to each other at an angle of 60°.
Q.5. Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.
Steps of Construction:
Step I: A line segment AB of 8 cm is drawn.
Step II: With A as centre and radius equal to 4 cm, a circle is drawn which cut the line at point O.
Step III: With B as a centre and radius equal to 3 cm, a circle is drawn.
Step IV: With O as a centre and OA as a radius, a circle is drawn which intersect the previous two circles at P, Q and R, S.
Step V: AP, AQ, BR and BS are joined.
Thus, AP, AQ, BR and BS are the required tangents.
Justification:
∠BPA = 90° (Angle in the semi circle)
∴ AP ⊥ PB
Therefor, BP is the radius of the circle then AP has to be a tangent of the circle.
Similarly, AQ, BR and BS are tangents of the circle.
Q.
6. Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and ∠B = 90°. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.
Steps of Construction:
Step I: A ΔABC is drawn.
Step II: Perpendicular to AC is drawn to point B which intersected it at D.
Step III: With O as a centre and OC as a radius, a circle is drawn. The circle through B, C, D is drawn.
Step IV: OA is joined and a circle is drawn with diameter OA which intersected the previous circle at B and E.
Step V: AE is joined.
Thus, AB and AE are the required tangents to the circle from A.
Justification:
∠OEA = 90° (Angle in the semi circle)
∴ OE ⊥ AE
Therefor, OE is the radius of the circle then AE has to be a tangent of the circle.
Similarly, AB is tangent of the circle. |
Trying to find out exactly how to convert 85/9 into a blended number or fraction? have I got the answer for you! In this guide, we"ll walk you with the step-by-step procedure of converting an not correct fraction, in this situation 85/9, to a mixed number. Review on!
Want to conveniently learn or present students exactly how to convert 85/9 to a mixed number? play this very quick and also fun video now!
Before us begin, let"s revisit part basic fraction terms so girlfriend understand specifically what we"re managing here:
Numerator. This is the number above the fraction line. For 85/9, the numerator is 85.Denominator. This is the number below the portion line. Because that 85/9, the denominator is 9.Improper fraction. This is a portion where the molecule is greater than the denominator.Mixed number. This is a means of to express an improper portion by simplifying it to whole units and a smaller in its entirety fraction. It"s an creature (whole number) and a proper fraction.
You are watching: 85/9 as a mixed number
Now let"s go with the actions needed to convert 85/9 to a mixed number.
## Step 1: uncover the totality number
We very first want to find the entirety number, and also to do this we division the molecule by the denominator. Since we are only interested in whole numbers, us ignore any numbers to the right of the decimal point.
85/9= 9.4444444444444 = 9
Now that we have our totality number because that the mixed fraction, we need to find our new numerator for the portion part that the mixed number.
## Step 2: acquire the new numerator
To work this the end we"ll use the entirety number us calculated in action one (9) and multiply it by the original denominator (9). The an outcome of the multiplication is then subtracted indigenous the original numerator:
85 - (9 x 9) = 4
## Step 3: Our blended fraction
We"ve currently simplified 85/9 to a mixed number. To view it, we just need to put the totality number along with our brand-new numerator and also original denominator:
9 4/9
## Step 4: simple our fraction
In this case, our fraction (4/9) have the right to be streamlined down further. In order to do that, we should calculate the GCF (greatest usual factor) of those 2 numbers. You can use our comfortable GCF calculator to work-related this out yourself if you want to. We already did that, and the GCF that 4 and also 9 is 1.
See more: What Is The Asr Button On A Volkswagen, Can Anyone Tell Me What Asr Is
We deserve to now divide both the brand-new numerator and also the denominator by 1 to simplify this fraction down come its lowest terms.
4/1 = 4
9/1 = 9
When we placed that together, we deserve to see the our complete answer is:
9 4/9
Hopefully this tutorial has actually helped you come understand exactly how to convert any improper portion you have into a blended fraction, finish with a whole number and a proper fraction. You"re complimentary to usage our calculator listed below to occupational out more, but do try and learn exactly how to carry out it yourself. It"s much more fun 보다 it seems, ns promise!
## Improper portion to mixed Number
Enter an improper fraction numerator and denominator |
# Which lists all the factors of a composite number?
A prime number is made up of only two factors: one and itself. A composite number is made up of more than two elements. There is no prime or composite number greater than one. Because each of these numbers has just two components, itself and 1, the prime numbers between 2 and 31 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, and 31. None of these numbers are composite.
## Is a composite number greater than 1?
Other than one and the number itself, a composite number has more than two elements. In this case, 2 has only two factors: 1 and 2. Therefore, it is not a prime number.
## Can composite numbers have more than one prime factorization?
We discovered the factors of a number in the previous section. There are just two elements in prime numbers: the number 1 and the prime number itself. Every composite number has more than two factors, and every composite number is a unique product of primes. This is known as a number's prime factorization. The number 6 has several different prime factorizations: 2 × 3 × 3, 2 × 2 × 3, 2 × 2 × 2, 3 × 3 × 7, and so on.
It is possible to give any positive integer its own special prime factorization. For example, the factorization of 60 includes 5 factors, each of which is divisible by 2 but not by 3: 2, 2, 2, 2, 2, 3, 5, 10, 15, 30. However, these are not the only possibilities. Any positive integer can have an arbitrary prime factorization. That is why there are many different ways to write the same number.
In mathematics, especially algebra, it is common to specify an expression using parenthesized sets of factors, called a factoring scheme. For example, the fact that 6 divides into 2 × 3 × 3 implies that there must be some other number r such that 0 < r ≤ 3 and r divides into 6. Thus, r = 2 is a valid solution. This means that there are actually two different factorizations for 6: 2 × 3 × 3 and 2 × 2 × 2.
## Why are all even numbers composite?
A prime number is a whole number with just two factors: itself and one. A composite number includes components other than one and itself. Neither 0 nor 1 are prime or composite numbers. Because all even numbers are divisible by two, every even number bigger than two is a composite number.
## Can a composite number have five factors?
When a number contains only two components, one and itself, it is said to be prime. When a number includes more than two elements, it is referred to as a composite... Definitions
factors of 2:1 x 22 is prime
factors of 5:1 x 55 is prime
factors of 6:1 x 6, 2 x 36 is composite
factors of 7:1 x 77 is prime
factors of 8:1 x 8, 2 x 48 is composite
## What is the least number of factors that a number can have?
So, no, the smallest number of variables that a number may have is one, not two. And, sure, calculating the components can tell you if a number is prime, composite, or equal to 1. Because 0*k = 0 for any k, it is termed a composite number with an unlimited number of components.
A prime number is an integer, or whole number, that has only two factors, 1 and itself. Put another way, a prime number can be divided evenly only by 1 and by itself. Prime numbers also must be greater than 1. For example, 3 is a prime number because 3 cannot be divided evenly by any number except for 1 and 3.
## Is 9 a composite number?
This is not the case (i.e., which has factors other than 1 and itself). The first several composite numbers (often abbreviated as "composites") are 4, 6, 8, 9, 10, 12, 14, 15, 16. After this point, only prime numbers continue to be composites.
All positive integers except one are composite: if n is positive then it can be written as a product of primes, and no number other than 1 can be written as a product of primes. Therefore, all positive integers except 1 are composites.
Even though 9 is not a prime number, it can be written as a product of two primes: 3 and 3. Thus, every integer divisible by 9 is composed.
## What is a composite number for kids?
Whole numbers that can be split by numbers other than themselves and one are known as composite numbers. They've got "factors." The numbers that divide it are known as factors. Every whole number greater than one is classified as either a composite number or a prime number. Composite numbers can be expressed as a product of two smaller numbers, while primes are only divisible by 1 and themselves.
Composite numbers include numbers that you might expect: 2, 3, 4, 5, 6, 7, 8, 9, 10. However, there are also some very large composite numbers, such as the number given above - which is equal to 1 million million (or 1 trillion) - that have many factors larger than themselves. These large numbers are called composite numbers because they are made up of multiple smaller numbers that when multiplied together give the original number.
Prime numbers are numbers that have only 1 and themselves as factors. Because there are more prime numbers than composite numbers, most integers are composites. The only exception is when an integer is factored into a single factor which is then necessarily itself equal to 1. In this case, the number is said to be prime.
Factoring is the process of dividing integers into their prime factors.
##### Caroline Garcia
Caroline Garcia is an honored college professor, whose dedication to her students has earned her the nickname "the mother of all teachers". Caroline's commitment to excellence in teaching and learning extends beyond the classroom. She has served on numerous committees related to curriculum development, assessment, faculty recruitment, instructional technology integration, and other areas that have shaped not only how she teaches but also what she teaches. |
# Golden ratio and Fibonacci
I’m currently doing a course on linear algebra, and while working on a hand-in, I discovered this neat proof that the Fibonacci sequence grows exponentially.
Definition: Define two sequences:
1. $a_1 = 1, a_2 = 3, \text{ and } a_n = a_{n-1} + a_{n-2} \text{ for } n>2$
2. $b_1 = 1, b_2 = 1, \text{ and } b_n = b_{n-1} + b_{n-2} \text{ for } n>2$
Notice that $b$ corresponds to the Fibonacci sequence, and $a$ corresponds to the Lucas numbers, first index truncated.
Lemma 1: For all $k>0$
1. $a_k = \frac{1}{2} \left ( a_{k-1} + 5\cdot b_{k-1} \right )$
2. $b_k = \frac{1}{2} \left ( a_{k-1} + b_{k-1} \right )$
Proof. Shown only for 1.1, the proof is similar for 1.2. Proven using strong induction on $k$.
• Base-case:
Notice that $\frac{1}{2} \left ( a_{0} + 5\cdot b_{0} \right ) = \frac{1}{2} \left ( 1 + 5\cdot 1 \right ) = \frac{1}{2} \left ( 6 \right ) = 3 = a_1$.
Which is what we wanted to show.
• Induction:
We see $a_k = a_{k-1} + a_{k-2}$, by definition.
We apply the induction hypothesis on $k-1$, and $k-2$ to get: $a_k = \frac{1}{2} \left ( a_{k-2} + 5\cdot b_{k-2} \right ) + \frac{1}{2} \left ( a_{k-3} + 5\cdot b_{k-3} \right )$. We use the definition of $a_k$ and $b_k$, and get
$a_k =$ $\frac{1}{2} \left ( a_{k-1} + 5\cdot b_{k-1} \right )$. Which is what we wanted to show. $\blacksquare$.
Lemma 2: For all $k>0$$\phi^k = \frac{1}{2} a_k + \frac{1}{2} b_k \cdot \sqrt{5}$.
Proof. We prove this lemma using induction on $k$.
• Base-case:
We see that, $\frac{1}{2} a_1 + \frac{1}{2} b_1 \cdot \sqrt{5} = \frac{1}{2} 1 + \frac{1}{2} 1 \cdot \sqrt{5} = \frac{1}{2} \left(1 + \sqrt{5} \right) = \phi$. Which is what we wanted to show.
• Induction step:
We get $\phi^k = \phi \cdot \phi^{k-1} = \frac{1}{2}(1+\sqrt{5})\cdot \left( \frac{1}{2} a_{k-1} + \frac{1}{2} b_{k-1} \cdot \sqrt{5} \right)$, using the induction hypothesis. Multiplying, we get $\phi^k = \frac{1}{2} \left ( \frac{1}{2} \left ( a_{k-1} + 5 b_{k-1} \right ) \right ) + \frac{1}{2} \left ( \frac{1}{2} \left ( a_{k-1} + b_{k-1} \right ) \right ) \cdot \sqrt{5}$. Using lemma 2, we find $\phi^k = \frac{1}{2} a_k + \frac{1}{2} b_k \cdot \sqrt{5}$, which is what we wanted to show. $\blacksquare$.
Theorem: $b_n = \Theta(\phi^n)$.
Proof. We split the proof into two parts; first proving that $b_n = O(\phi^n)$, then proving $b_n = \Omega(\phi^n)$.
• $b_n = \Omega(\phi^n)$.
We see that $\phi^n < \frac{1}{2} a_n \cdot \sqrt{5} + \frac{1}{2} b_n \cdot \sqrt{5}$, because of lemma 2. Because of lemma 1.2, we have $b_n = \frac{1}{2} a_n + \frac{1}{2} b_n$, so we get $\phi^n < b_{n+1} \cdot \sqrt{5}$. This implies that $b_n = \Omega(\phi^n)$.
• $b_n = O(\phi^n)$.
We see that $\phi^n > \frac{1}{2} b_n$ because of lemma 2, which implies that $b_n = O(\phi^n)$.
Because $b_n = O(\phi^n)$, and $b_n = \Omega(\phi^n)$, we have $b_n = \Theta(\phi^n)$. That is, the Fibonacci sequence grows exponentially, with base $\phi$. $\blacksquare$.
# Calkin-Wilf Fractals
Suppose we have two natural numbers $i,j$. Define a sequence $a_0 = i, a_1 = j$ and $a_n = |a_{n-1} - a_{n-2}|$. That is, we start with two initial values, and repeatedly subtract the smallest value from the largest value. Since $i,j \geq 0$, we know that $a_n \geq 0$ for all $n$.
Eventually, we will reach the cycle $\ldots 1, 1, 0, 1, 1, 0 \ldots$ no matter our choice of $i,j$ (to see why, consider the termination function $\mu(i,j)=i+j$ and the invariant that any term must be positive).
Define
$\text{chainLength}(i,j) = \text{largest } n \text{ such that } a_n \notin \{1, 0\}$
We can now make pretty pictures. For each pixel $(x,y)$ color pixel by $\text{chainLength}(x,y)$ which can be calculated using a simple recursive procedure:
int chainLength(int x, int y){
if(x == y || x==0 || y==0)
return 0;
if(x>y)
return 1+chainLength(x-y,y);
else
return 1+chainLength(x,y-x);
}
Generated using the DELTA library.
### Connection to Calkin-Wilf trees
The Calkin-Wilf tree is an infinite binary recursive tree, where the root node is defined as $1/1$, and every node $a/b$ has a left child $\dfrac{a+b}{b}$ and right child $\dfrac{a}{a+b}$.
The Calkin-Wilf tree is useful because it allows us to define a bijection between the natural numbers and the rational numbers, which can be used to show that the rational numbers are a countably infinite set.
Notice that if a node $x/y$ is a left child, then $x>y$, and if it’s a right child then $y > x$. This allows us to determine the parent node of a given node.
If for a given node $x>y$ then we know it was the left child of its parent. So we know its parent is $(x-y) / y$. Similarly, if $y>x$, then its parent is $x/(y-x)$. Notice that at each step, we subtract the smaller number from the larger number, and we terminate the procedure when we reach the fraction $1/1$.
Going back to $a_n$, we could interpret a pixel $(x,y)$ as the rational number $x/y$. The recursive sequence $\text{chainLength}(x,y)$ would then be equivalent to the depth of the node $latex x/y$ in the Calkin-Wilf tree.
Keep in mind that the Calkin-Wilf tree only contains rational numbers in the lowest common term, so the node $2/2$ does not exist in the Calkin-Wilf tree. To circumvent this issue, we convert our input node to the equivalent rational number in lowest common terms.
# Incendium – HTML 5 App
I just realized I have had this app up for a while on my github.io without posting it here. Anyway, I made an HTML5 app which implements the algorithm detailed in my previous post on fractal animations.
Moving the mouse around on the screen changes two different parameters, with each point $P$ corresponding to a unique configuration.
http://ssodelta.github.io/incendium/
It is written for Google Chrome, and is known to act weird on Firefox sometimes. It may also crash, but just restart the page (sry).
# AIArena – Bot Tournament
AIArena is a tournament where contestants write scripts for bots which battle each other in a virtual arena. I have written a dynamic web application which allows users to submit their own bots, written in a custom imperative scripting language dubbed BOTScript. The server will itself make the bots battle each other, ranking them on the website by how many wins each bot gets.
If you have ideas, suggestions, feedback, criticism or whatever, drop me a line at admin@ssodelta.com.
# Sample of the DELTA Library
## The DELTA library
If you look through my blog, you’ll find posts where I detail an algorithm for making fractals, or other visually appealing animations/images. To aid myself in the future, I have written a library for Java8 which contains some nice classes and operations for manipulating colors and images.
You can take a look at it at its GitHub page (contains a README.md), though I will detail with the library with further scrutiny on this blog at some point in the future (studying is keeping me busy atm):
https://github.com/SSODelta/DELTA
Often when making single images, there are some dimensions you can change to produce animations. I thought two previous posts were especially suitable for this, so I have made two very nice animations, imo:
### Newton Fractals
The biggest problem with animating this fractal, was that I lost the information about which root was assigned which colors when slighting altering the parameters of the polynomials. Instead, using DELTA I have colored each root $r_0$ using HSV colors with hue proportional to the argument $\mathrm{arg}(r_0)$ and the number of iterations it took for the point to converge – and saturation/value decreased with increasing magnitude $|r_0|$.
### Regex Fractals
Here, the hue is proportional to the edit distance and the time $t$, and saturation/value inversely proportional to only the edit distance.
# LoGiX v1.1.1
LoGiX has been updated to a new build and includes some nice new features and small bug-fixes. Here is the changelog:
Added support for {...} statements and line-breaks;
### Added support for {…} statements and line-breaks;
The most nice new feature is the curly brackets. I used to have this implemented, but when I rewrote the engine entirely once, I lost feature. But it’s really cool, basically, you type a list list {a,b,c}, and then the LoGiX-engine will replace the curly brackets with each element inside it. So we can type
:> def {P,Q}
Which evaluates as:
:> def P
:> def Q
Which saves a lot of space in programs. A nice example is modus ponens, which before was written as:
:> !def P
:> !def P implies Q
:> Q?
true
Using the new notation, we can write:
:> !def P {,implies Q};Q?
true
Which is very cool! I think this is the shortest you can write modus ponens in LoGiX. It is of course possible to do these multiple places in the command, so we get:
:> def {P,Q,R} implies {A,B}
def P implies A
def Q implies A
def R implies A
def P implies B
def Q implies B
def R implies B
Another new feature is the ability to do inline comments. Before, the compiler would crash if a #-symbol appeared not at the beginning of a line. But now, it treats everything after the #-symbol as a comment.
:> def P #Now we've just defined P and made a comment
def P
And now it’s also possible to use custom messages when the output is false. Before we could do:
:> def {P,Q} implies R
:> def P or Q
:> R?its true
its true
It interprets whatever is after the question mark as the message it prints if the statement is true. However, it was not possible to do the same if the statement was false – it would always return ‘false‘. But now you can do this too:
:> def P and Q implies R
:> def P or Q
:> R?its true|definitely not true
definitely not true |
# Find m if the vectors u and v are perpendicullar u=mi+3j v=(m-2)*i-j
sciencesolve | Certified Educator
You need to use the following equation that relates two perpendicular vectors, such that:
`m*(m - 2) = -((-1)*3) => m^2 - 2m = 3`
Moving the terms to one side, yields:
`m^2 - 2m - 3 = 0 => m^2 + (-3 + 1)m - 3 = 0`
`m^2 - 3m + m - 3 = 0 `
You need to group the terms, such that:
`m(m - 3) + (m - 3) = 0`
Factoring out `(m - 3)` yields:
`(m - 3)(m + 1) = 0`
Using the zero product property yields:
`{(m - 3 = 0),(m + 1 = 0):} => {(m = 3),(m = -1):}`
Hence, evaluating m, under the given conditions, yields` m = 3, m = -1.`
m = 3
giorgiana1976 | Student
We'll put the vectors u and v in the standard form:
u = xu*i + yu*j
v = xv*i + yv*j
Now, we'll write the constraint for 2 vectors to be perpendicular:
the dot product of u and v has to be zero,because the angle between u and v is 90 degrees and cos 90 = 0.
u*v = |u|*|v|*cos(u,v)
Now, we'll identify xu,xv,yu,yv from the expressions of vectors:
xu = m
xv = (m-2)
yu = 3
yv = -1
We'll calculate the product of vectors u*v:
u*v = xu*xv + yu*yv
u*v = m(m-2) + 3*(-1) (1)
But u*v = 0 (2)
We'll put (1) = (2):
m(m-2) + 3*(-1) = 0
We'll remove the brackets:
m^2 - 2m - 3 = 0
m1 = [2 + sqrt(4+12)]/2
m1 = (2 + 4)/2
m1 = 3
m2 = (2-4)/2
m2 = -1
Since it is not specified if m has to be positive or negative, both values are admissible. |
## RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3
These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3
Other Exercises
Question 1.
In the figure, compute the area of the quadrilateral.
Solution:
In the quadrilateral ABCD,
∠A = 90°, ∠CBD = 90°, AD = 9 cm, BC = 8 cm and CD = 17 cm
In right ∆BCD,
CD = BC2 + BD2 (Pythagoras Theorem)
⇒ (17)2 = (8)2 + BD2
⇒ 289 = 64 + BD2
⇒ BD2 = 289 – 64 = 225 = (15)2
∴ BD = 15 cm
Now in right ∆ABD,
BD2 = AB2 + AD2
⇒ (15)2 = AB2 + (9)2
⇒ 225 = AB2 + 81
⇒ AB2= 225 – 81 = 144 = (12)2
∴ AB = 12 cm
Question 2.
In the figure, PQRS is a square and T and U are, respectively, the mid-points of PS and QR. Find the area of ∆OTS if PQ = 8 cm.
Solution:
In square PQRS, T and U are the mid-points of the sides PS and QR
TU, QS and US are joined
PQ = 8 cm
∴ T and U are mid-points of the opposites sides PS and QR
∴ TU || PQ TO || PQ
In RQS,
T is mid-point of PS and TO || PQ
∴ O is the mid point of SQ 1 1
Question 3.
Compute the area of trapezium PQRS in the figure.
Solution:
In ∆TQR, ∠RTQ = 90°
∴ QR2 = TQ2 + RT2
⇒ (17)2 = (8)2 + RT2
⇒ 289 = 64 + RT2
⇒ RT2 = 289 – 64 = 225 = (15)2
∴ RT = 15 cm
and PQ = 8 + 8 = 16 cm
Question 4.
In the figure, ∠AOB = 90°, AC = BC, OA = 12 cm and OC = 6.5 cm. Find the area of ∆AOB.
Solution:
In ∆AOB, ∠AOB = 90°
C is a point on AB such that AC = BC Join OC
Since C is the mid-point of hypotenuse of right ∆AOB
∴ AC = CB = OC = 6.5 cm
∴ AB = 6.5 + 6.5 = 13 cm
Now in right ∆AOB
⇒ AB2 = AO2 + OB(Pythagoras Theorem)
⇒ (13)2 = (12)2 + OB2
⇒ 169 = 144 + OB2
⇒ OB2 = 169 – 144 = 25 = (5)2
∴ OB = 5 cm
Question 5.
In the figure, ABCD is a trapezium in which AB = 7 cm, AD = BC = 5 cm, DC = x cm, and distance between AB and DC is 4 cm. Find the value of x and area of trapezium ABCD.
Solution:
In the trapezium ABCD,
AB = 7 cm
AL = BM = 4 cm
AD = BC = 5 cm
Question 6.
In the figure, OCDE is a rectangle inscribed in a quadrant of a circle of radius 10 cm.If OE = 2 $$\sqrt { 5 }$$ , find the area of the rectangle.
Solution:
Radius of the quadrant of circle = 2$$\sqrt { 5 }$$ units
∴ OD diagonal of rectangle = 10 units (∵ OD = OB = OA = 10 cm)
DE = 2 $$\sqrt { 5 }$$ cm
∴ In right ∆OED,
OD2 = OE2 + DEv
(10)2 = OE2 + (2$$\sqrt { 5 }$$)2
100 = OE2 + 20
OE2 = 100 – 20 = 80
⇒ OE2 = (4$$\sqrt { 5 }$$)2
∴ OE = 4$$\sqrt { 5 }$$ cm
∴ Area of rectangle = lxb
= DE x OE
= 2$$\sqrt { 5 }$$ x 4$$\sqrt { 5 }$$
= 8 x 5 = 40 cm2
Question 7.
In the figure, ABCD is a trapezium in which AB || DC. Prove that ar( ∆AOD = ar(∆BOC).
Solution:
In trapezium ABCD, diagonals AC and BD intersect each other at O
∴ ∆ADB and ∆ACB are on the same base AB and between the same parallels
∴ ar(∆ADB = ar(∆ACD)
Subtracting, ar(AAOB) from both sides,
ar(∆ADB) – ar(∆AOB) = ar(∆ACD) – ar(∆AOB)
⇒ ar(∆AOD) = ar(∆BOC)
Question 8.
In the figure, ABCD, ABFE and CDEF are parallelograms. Prove that ar(∆ADE) = ar(∆BCF) [NCERT]
Solution:
Given : In the figure, ABCD, ABEF and CDEF are ||gms
To prove : ar(∆ADE) = ar(∆BCF)
Proof: ∴ ABCD is a ||gm
∴ AD = BC
Similarly, in ||gm ABEF
AE = BF
and in ||gm CDEF,
DE = CF
Now, in ∆ADE and ∆BCF
AD = BC (proved)
DE = CF (proved)
AE = BF (proved)
∴ ∆ADE ≅ ∆BCF
∴ ar(∆ADE) = ar(∆BCF) (∵ Congruent triangles are equal in area)
Question 9.
In the figure, ABC and ABD are two triangles on the base AB. If line segment CD is bisected by AB at O, show that ar(∆ABC) = ar(∆ABC)
Solution:
Given : In the figure, ∆ABC and ∆ABD are on the same base AB and line CD is bisected by AB at O i.e. CO = OD
To prove : ar(∆ABC) = ar(∆ABD)
Construction : Draw CL ⊥ AB and DM ⊥ AB
Question 10.
If AD is a median of a triangle ABC, then prove that triangles ADB and ADC are equal in area. If G is the mid point of median AD, prove that ar(∆BGC) = 2ar(∆AGC).
Solution:
Given : In ∆ABC, AD is its median. G is mid point of AD. BG and CF are joined
To prove :
(ii) ar(∆BGC) = 2ar(∆AGC)
Construction : Draw AL ⊥ BC
Question 11.
A point D is taken on the side BC of a AABC such that BD = 2DC. Prove that ar(∆ABD) = 2ar(∆ADC)
Solution:
Given : In ∆ABC, D is a point on BC such that
BD = 2DC
To prove : ar(∆ABD) = 2ar(∆ADC)
Construction : Draw AL ⊥ BC
Question 12.
ABCD is a parallelogram whose diagonals intersect at O. If P is any point on BO, prove that
(i) ar(∆ADO) = or (∆CDO)
(ii) ar(∆ABP) = ar(∆CBP).
Solution:
Given : In ||gm ABCD, Diagonals AC and BD intersect each other at O
P is any point on BO
AP and CP are joined
To prove :
(i) ar(∆ADO) = ar(∆CDO)
(ii) ar(∆ABP) = ar(∆CBP)
Proof:
O is the mid point of AC
∴ ar(∆ADO) = ar(∆CDO)
(ii) Since O is the mid point of AC
∴ PO is the median of ∆APC
∴ af(∆APO) = or(∆CPO) …(i)
Similarly, BO is the median of ∆ABC
∴ ar(∆ABO) = ar(∆BCO) …(ii)
Subtracting (i) from (ii),
ar(∆ABO) – ar(∆APO) = ar(∆BCO) – ar( ∆CPO)
⇒ ar(∆ABP) = ar(∆CBP)
Hence proved.
Question 13.
ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F.
(i) Prove that ar(∆ADF) = ar(∆ECF)
(ii) If the area of ∆DFB = 3 cm2, find the area of ||gm ABCD.
Solution:
Given : In ||gm ABCD, BC is produced to E such that CE = BC
AE intersects CD at F
To prove :
(i) ar(∆ADF) = ar(∆ECF)
(ii) If ar(∆DFB) = 3 cm2, find the area of (||gm ABCD)
Question 14.
ABCD is a parallelogram whose diagonals AC and BD intersect at O. A line through O intersects AB at P and DC at Q. Prove that ar(∆POA) = ar(∆QOC).
Solution:
Given : In ||gm ABCD, diagonals AC and BD intersect at O
A line through O intersects AB at P and CD at Q
To prove : ar(∆POA) = ar(∆QOC)
Proof : In ∆POA and ∆QOC,
OA = OC (O is mid-point of AC)
∠AOD = ∠COQ (Vertically opposite angles)
∠APO = ∠CQO (Alternate angles)
∴ ar(∆POA) ≅ ar(∆QOC) (AAS criterian)
∴ ar(∆POA) = ar(∆QOC)
Question 15.
In the figure, D and E are two points on BC such that BD = DE = EC. Show that ar(∆ABD) = ar(∆ADE) = ar(∆AEC). [NCERT]
Solution:
Given : D and E are two points on BC such that BD = DE = EC
AD and AE are joined
To prove : ar(∆ABD) = ar(∆ADE) = ar(∆AEC)
Construction : From A, draw AL ⊥ BC and XAY || BC
Proof: ∵ BD = DE = EC
and ∆ABD, ∆ADE and ∆AEC have equal bases and from the common vertex A
∴ ar(∆ABD) = ar(∆ADE) = ar(∆AEC)
Question 16.
Diagonals AC and BD of a quadrilateral ABCD intersect each other at P.
Show that: ar(∆APB) x ar(∆CPD) = ar(∆APD) x ar(∆BPC)
Solution:
Given : In quadrilateral ABCD, diagonal AC and BD intersect each other as P
To prove : ar(∆APB) x ar(∆CPD) = ar(APD) x ar(∆BPC)
Construction : Draw AL and CN perpendiculars on BD
Question 17.
If P is any point in the interior of a parallelogram ABCD, then prove that area of the triangle APB is less than half the area of parallelogram.
Solution:
Given : In ||gm ABCD, P is any point in the ||gm
AP and BP are joined
To prove : ar(∆APB) < $$\frac { 1 }{ 2 }$$ ar(||gm ABCD)
Construction : Draw DN ⊥AB and PM ⊥ AM
Question 18.
ABCD is a parallelogram. E is a point on BA such that BE = 2EA and F is a point on DC such that DF = 2FC. Prove that AECF is a parallelogram whose area is one third of the area of parallelogram ABCD.
Solution:
Given : In ||gm ABCD, E is a point on AB such that BE = 2EA and F is a point on CD such that DF = 2FC. AE and CE are joined
Question 19.
In a ∆ABC, P and Q are respectively, the mid-points of AB and BC and R is the mid-point of AP. Prove that
Solution:
Given : In ∆ABC,
P and Q are mid-pionts of AB and BC R is mid-point of AP, PQ, RC, RQ are joined
Question 20.
ABCD is a parallelogram, G is the point on AB such that AG = 2GB, E is a point of DC such that CE = 2DE and F is the point of BC such that BF = 2FC. Prove that:
(v) Find what portion of the area of parallelogram is the area of AEFG.
Solution:
Given : ABCD is a parallelogram and AG = 2GB, CE = 2DE and BF = 2FC
To prove :
(v) Find what portion of the area of parallelogram is the area of AFEG.
Construction : Draw EP ⊥ AB and EQ ⊥ BC
Question 21.
In the figure, CD || AE and CY || BA.
(i) Name a triangle equal in area of ACBX.
(ii) Prove that or(∆ZDE) = ar(∆CZA).
(iii) Prove that ar(∆CZY) = ar(∆EDZ).
Solution:
Given : In the figure,
CP || AE and CY || BA
To prove :
(i) Name a triangle equal in area of ∆CBX
(ii) Prove that ar(∆ZDE) = ar(∆CZA)
(iii) ar(BCZY) = ar(∆EDZ)
Proof:
(i) ∆CBX and ∆CYX are on the same base BY and between same parallels.
∴ ar(∆CBX) = ar(∆CYX)
(ii) ∆ADE and ∆ACE are on the same base AE
and between the same parallels (AE || CD)
∴ ar(∆ADE) = ar(∆ACE)
Subtracting ar(∆AZE) from both sides
⇒ ar(∆ADE) – ar(∆AZE) = ar(∆ACE) – ar(∆AZE)
⇒ ar(∆ZDE) = ar(∆ACZ)
⇒ ar∆ZDE = ar∆CZA
(iii) ∵ As ACY and BCY are on the same base CY and between the same parallels
∴ ar(∆ACY) = ar(∆BCY)
Now ar(∆ACZ) = ar(∆ZDE) (Proved)
⇒ ar(∆ACY) + ar(∆CYZ) = ar(∆EDZ)
⇒ ar(∆BCY) + ar(∆CYZ) = ar(∆EDZ)
∴ ar quad. (BCZY) = ar(EDZ)
Hence proved.
Question 22.
In the figure, PSD A is a parallelogram in which PQ = QR = RS and AP || BQ || CR. Prove that ar(∆PQE) = ar(∆CFD).
Solution:
Given : In the figure, PSDA is a ||gm in
which PQ = QR = RS and
AP || BQ || CR || DS
To prove : ar(∆PQE) = ar(∆CFD)
Construction : Join PD
Proof : ∵ PA || BQ || CR || DS
and PQ – QR = RS (Given)
∴ AB = BC = CD
∴ PQ = CD
Now in ABED, F is mid point of ED
∴ EF = FD
Similarly, EF = PE
⇒ PE = FD
In ∆PQE and ∆CFD,
∴ ∠EPQ = ∠FDC (Alternate angles)
PQ = CD
PE = FD (Proved)
∴ APQE ≅ ACFD (SAS cirterion)
∴ ar(∆PQE) = ar(∆CFD)
Question 23.
In the figure, ABCD is a trapezium in which AB || DC and DC = 40 cm and AB = 60 cm. If X and Y are, respectively the mid-points of AD and BC, prove that:
(i) XY = 50 cm
(ii) DCYX is a trapezium
(iii) ar(trap. DCYX) = $$\frac { 9 }{ 11 }$$ ar(trap. XYBA)
Solution:
Given : In the figure, ABCD is a trapezium in which AB || DC
DC = 40 cm, AB = 60 cm
X and Y are the mid-points of AD and BC respectively
To prove :
(i) XY = 50 cm
(ii) DCYX is a trapezium
(iii) ar(trap. DCYX) = $$\frac { 9 }{ 11 }$$ m(trap. XYBA)
Construction : Join DY and produce it to meet AB produced at P
Question 24.
D is the mid-point of side BC of ∆ABC and E is the mid-point of BD. If O is the mid-point of AE, prove that ar(∆BOE) = $$\frac { 1 }{ 8 }$$ ar(∆ABC).
Solution:
Given : In ∆ABC, D is mid point of BC, E is mid point BE and O is the mid point of AE. BO, AE, AD are joined.
Question 25.
In the figure, X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. Prove that ar(∆ABP) = ar(∆ACQ).
Solution:
Given : In ∆ABC, X and Y are the mid pionts of AC and AB respectively. Through A, a line parallel to BC is drawn. Join BX and CY and produce them to meet the parallel line through A, at P and Q respectively and intersect each other at O.
To prove : ar(∆ABP) = ar(∆ACQ)
Construction : Join XY and produce it to both sides
Proof : ∵ X and Y are mid points of sides AC and AB
∴ XY || BC
Similarly, XY || PQ
∆BXY and ∆CXY are on the same base XY and between the same parallels
∴ ar(∆BXY) = ar(∆CXY) …(i)
Now, trap. XYAP and trap. XYAQ are on the same base XY and between the same parallels
∴ ar(XYAP) = ar(XYAQ) …(ii)
Adding (i) and (ii),
∴ ar(∆BXY) + ar(∆YAP)
= ar(CXY) + ar(XYAQ)
⇒ ar(∆ABP) = ar(∆ACQ)
Question 26.
In the figure, ABCD and AEFD are two parallelograms. Prove that
(i) PE = FQ
(ii) ar(∆APE) : ar(∆PFA) = ar(∆QFD) : ar(∆PFD)
(iii) ar(∆PEA) = ar(∆QFD).
Solution:
Given : Two ||gm ABCD and ||gm AEFD are on the same base AD. EF is produced to meet CD at Q. Join AF and PD also
To prove :
(i) PE = FQ
(ii) ar(∆APE) : ar(∆PFA) = ar(∆QFD) : ar(∆PFD)
(iii) ar(∆PEA) = ar(∆QFD)
Proof:
(i) In ∆AEP and DFQ,
AE = DF (Opposite sides of a ||gm)
∠AEP = ∠DFQ (Corresponding angles)
∠APE = ∠DQF (Corresponding angles)
∴ ∆AEP ≅ ∆DFQ (AAS axiom)
∴ PE = QF (c.p.c.t.)
(ii) and ar(∆AEP) = ar(∆DFQ) …(i)
(iii) ∵ ∆PFA and ∆PFD are on the same base PF and between the same parallels
∴ ar(∆PFA) = ar(∆PFD) …(ii)
From (i) and (ii),
Question 27.
In the figure, ABCD is a ||gm. O is any point on AC. PQ || AB and LM || AD. Prove that ar(||gm DLOP) = ar(||gm BMOQ).
Solution:
Given : In ||gm ABCD, O is any point on diagonal AC. PQ || AB and LM || BC
To prove : ar(||gm DLOP) = ar(||gm BMOQ)
Proof : ∵ Since, a diagonal of a parallelogram divides it into two triangles of equal area.
∴ ar(∆ADC) = or(∆ABC)
⇒ ar(∆APO) + or(||gm DLOP) + ar(∆OLC)
= ar(∆AOM) + ar(||gm BMOQ) + ar( ∆OQC) …(i)
Since, AO and OC are diagonals of parallelograms AMOP and OQCL respectively,
∴ ar(∆APO) = ar(∆AMO) …(ii)
And, ar(∆OLC) = ar(∆OQC) …(Hi)
Subtracting (ii) and (iii) from (i), we get ar(||gm DLOP) = ar(||gm BMOQ)
Question 28.
In a ∆ABC, if L and M are points on AB and AC respectively such that LM || BC.
Prove that:
(i) ar(∆LCM) = ar(∆LBM)
(ii) ar(∆LBC) = ar(∆MBC)
(iii) ar(∆ABM) = ar(∆ACL)
(iv) ar(∆LOB) = ar(∆MOC).
Solution:
Given : In ∆ABC,
L and M are mid points on AB and AC
LM, LC and MB are joined
To prove :
(i) ar(∆LCM) = or(∆LBM)
(ii) ar(∆LBC) = ar(∆MBC)
(iii) ar(∆ABM) = ar(∆ACL)
(iv) ar(∆LOB) = ar(∆MOC)
Proof: ∵ L and M are the mid points of AB and AC
∴ LM || BC
(i) Now ∆LBM and ∆LCM are on the same base LM and between the same parallels
∴ar(∆LBM) = ar(∆LCM) …(i)
⇒ ar(∆LCM) = ar(∆LBM)
(ii) ∵ ∆LBC and ∆MBC are on the same base
BC and between the same parallels
∴ ar(∆LBC) = ar(∆MBC) …(ii)
(iii) a(∆LMB) = ar(∆LMC) [From (i)]
⇒ ar(∆ALM) + ar(∆LMB)
= ar(∆ALM) + ar(∆LMC) [Adding or(∆ALM) to both sides]
⇒ ar(∆ABM) = ar(∆ACL)
(iv) ∵ ar(∆LBC) = ar(∆MBC) [From (ii)]
⇒ ar(∆LBC) – ar(∆BOC) = ar(∆MBC) – ar(∆BOC)
ar(∆LBO) = ar(∆MOC)
Question 29.
In the figure, ABC and BDC are two equilateral triangles such that D is the mid-point of BC. AE intersects BC in F.
Solution:
Given : ABC and BDE are two equilateral triangles and D is mid point of BC. AE intersects BC in F
To prove :
Question 30.
In the figure, ABC is a right triangle right angled at A, BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y. Show that [NCERT]
(i) ∆MBC ≅ ∆ABD
(ii) ar(BYXD) = ar(∆MBC)
(iii) ar(BYXD) = ar(ABMN)
(iv) ∆FCB ≅ ∆ACE
(v) ar(CYXE) = 2ar(∆FCB)
(vi) ar(CYXE) = ar(∆CFG)
(vii) ar(BCED) = ar(AMBN) + ar(ACFG)
Solution:
Given : In ∆ABC, ∠A = 90°
BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively
AX ⊥ DE meeting DE at X
To prove :
(i) ∆MBC ≅ ∆ABD
(ii) ar(BYXD) = 2ar(∆MBC)
(iii) ar(BYXD) = ar(ABMN)
(iv) ∆FCB ≅ ∆ACE
(v) ar(CYXE) = 2ar(∆FCB)
(vi) ar(CYXE) = ar(ACFG)
(vii) ar(BCED) = or(AMBN) + ar(ACFG)
Construction : Join AD, AE, BF and CM
Proof:
(i) In ∆MBC and ∆ABD,
MB=AB (Sides of square)
BC = BD
∠MBC = ∠ABD (Each angle = 90° + ∠ABC)
∴ ∆MBC ≅ ∆ABD (SAS criterian)
∴ ar(∆MBC) = ar(∆ABD) …(i)
(ii) ∵ ∆ABD and rectangle BYXD are on the same base BD and between the same parallels
∴ ar(∆ABD) = $$\frac { 1 }{ 2 }$$ ar(rect. BYXD)
⇒ ar(rect. BYXD) = 2ar(∆ABD)
⇒ ar(rect. BYXD) = 2ar(∆MBC) …(ii)
(iii) Similarly, ∆MBC and square MBAN are on the same base MB and between the same parallels
∴ ar(∆MBC) = ar(sq. ABMN) …(iii)
From (ii) and (iii)
ar(sq. ∆BMN) = ar(rect. BYXD)
(iv) In AFCB and ∆ACE,
FC = AC
CB = CE (Sides of squares)
∠FCB = ∠ACE (Each = 90° + ∠ACB)
∴ ∆FCB = ∆ACE (SAS criterian)
(v) ∵ ∆FCB ≅ ∆ACE (Proved)
∴ ar(∆FCB) = ar(∆ACE)
∵∆ACE and rectangle CYXE are on the same base and between the same parallels
∴ 2ar(∆ACE) = ar(CYXC)
⇒ 2ar(∆FCB) = ar(CYXE) …(iv)
(vi) ∵ AFCB and rectangle FCAG are on the base FC and between the same parallels
∴ 2ar(∆FCB) = ar(FCAG) …(v)
From (iv) and (v)
ar(CMXE) = ar(ACFG)
(vii) In ∆ACB.
BC2 = AB2 + AC2 (By Pythagoras Theorem)
⇒ BC x BD = AB x MB + AC x FC
⇒ ar(BCED) = ar(ABMN) + ar(ACFG)
Hence proved.
Hope given RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3 are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you. |
# Grade 5 Math – Place Value- Quiz 2
Grade 5 Math - Place Value- Quiz 2
Multiple Choice Quiz
Example: What is the value of the digit 7 in the number 1,784?
Answer: 7*100= 700 is the value of the digit 7.
1 / 10
1. What is the value of the digit 0 in the number 57,013?
usechatgpt init success
usechatgpt init success
2 / 10
2. What is the value of the digit 1 in the number 5,471?
3 / 10
3. What is the value of the digit 5 in the number 57,013?
4 / 10
4. What is the value of the digit 7 in the number 29,748?
5 / 10
5. What is the value of the digit 2 in the number 32,845?
6 / 10
6. What is the value of the digit 4 in the number 2,014?
7 / 10
7. What is the value of the digit 3 in the number 5,342?
8 / 10
8. What is the value of the digit 9 in the number 6,598?
9 / 10
9. What is the value of the digit 7 in the number 8,471?
10 / 10
10. What is the value of the digit 4 in the number 9,784?
## Place value Exerice
Place value is a way to write numbers that helps us understand their value. It’s like having a secret code for numbers!
When we write a number, we use different positions for each digit to show how much it is worth. These positions are called place values.
For example, let’s look at the number 26. In this number, the digit 2 is in the tens place, and the digit 6 is in the ones place. This means that the number 26 can be read as “2 tens and 6 ones”.
When we use place value, we know that the digit in the tens place is worth 10 times more than the digit in the ones place. So in the number 26, the digit 2 in the tens place represents 2 groups of 10, or 20, and the digit 6 in the ones place represents 6 ones.
We can also use place value to write larger numbers. For example, let’s look at the number 3,581. In this number, the digit 3 is in the thousands place, the digit 5 is in the hundreds place, the digit 8 is in the tens place, and the digit 1 is in the ones place. This means that the number 3,581 can be read as “3 thousands, 5 hundreds, 8 tens, and 1 one”.
By understanding place value, we can add and subtract numbers easily, compare the value of different numbers, and even multiply and divide numbers. It’s a really important concept in math that we use all the time! |
# 3.5: Trigonometric Functions and Triangles
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In this section, we will further discuss trigonometric functions. We will examine the relationship between trigonometric functions and right triangles, and examine some more properties of trigonometric functions.
## 3.5.1 Right Triangles
In the last section, we focused on the relationships between the trigonometric functions and the unit circle. Now, we will examine the relationship of these functions and the unit circle with right triangles. In Figure $$\PageIndex{1}$$, we start by drawing a unit circle with an angle $$\theta$$ and marking the corresponding coordinates on the circle. If we drop down from these coordinates to the x axis, we can form a right triangle.
Figure $$\PageIndex{1}$$: Right triangle inside of the unit circle
This triangle has a hypotenuse of 1 because the hypotenuse is the same length as the radius of the unit circle, and side lengths of $$x=\cos{(\theta)}$$ and $$y=\sin{(\theta)}$$. If we apply the Pythagorean Theorem to this triangle, we discover an interesting identity:
\begin{align}\begin{aligned}\begin{split} a^2 + b^2 & = c^2 \\ (x)^2 + (y)^2 & = (1)^2 \\ (\cos{(\theta)})^2 + (\sin{(\theta)})^2 &= 1^2 \\ \cos^2{(\theta)} + \sin^2{(\theta)} & = 1 \end{split}\end{aligned}\end{align}
There is nothing special about the choice of $$\theta$$ shown in Figure $$\PageIndex{1}$$; this identity is true for all inputs. Notice that the input for cosine and the input for sine are the same; if the inputs are different, we cannot guarantee that the sum will be equal to 1.
This right triangle also gives us a different way of evaluating trigonometric functions, in general. With the unit circle, we saw that $$\cos{(\theta)}$$ is the x coordinate and $$\sin{(\theta)}$$ is the y coordinate, and our right triangle has a hypotenuse of 1. If we scale the triangle, the side lengths will also scale, but the size of the angles will remain the same, so the values of $$\cos{(\theta)}$$ and $$\sin{(\theta)}$$ should also remain the same. In order for this to be true, we can’t just say that cosine is the length of the adjacent side and sine is the length of the opposite; instead, we will need to divide both by the length of the hypotenuse to adjust for the scaling (see Figure $$\PageIndex{2}$$ for a visual explanation of the opposite and adjacent sides). This gives us the following identities:
Figure $$\PageIndex{2}$$: Using a right triangle to evaluate trigonometric functions
$\begin{array}{lll}{\text{1. }\sin(\theta )=\frac{\text{opposite}}{\text{hypotenuse}}}&{\qquad}&{\text{4. }\csc(\theta )=\frac{\text{hypotenuse}}{\text{opposite}}} \\ {\text{2. }\cos(\theta )=\frac{\text{adjacent}}{\text{hypotenuse}}}&{\qquad}&{\text{5. }\sec(\theta )=\frac{\text{hypotenuse}}{\text{adjacent}}} \\ {\text{3. }\tan(\theta )=\frac{\text{opposite}}{\text{adjacent}}}&{\qquad}&{\text{6. }\cot(\theta )=\frac{\text{adjacent}}{\text{opposite}}}\end{array}\nonumber$
Many people summarize the first three of these with SOH-CAH-TOA to help remember the identities. SOH-CAH-TOA stands for Sine is Opposite over Hypotenuse; Cosine is Adjacent over Hypotenuse, and Tangent is Opposite over Adjacent. The remaining three identities can then be formed from the definitions of cosecant, secant, and cotangent. Let’s take a look at how we can use right triangles to help us evaluate our trigonometric functions.
Example $$\PageIndex{1}$$: Using a Right Triangle
Suppose that $$\cos{(\theta)} = \frac{12}{13}$$. Determine all possible values of $$\sin{(\theta)}$$.
Solution
To help find the possible values of $$\sin{(\theta)}$$, we will draw a right triangle and label it using the values we already know. We know that $$\cos{(\theta)} = \frac{12}{13}$$, so we can use 12 as the length of the adjacent side and 13 as the length of the hypotenuse:
Figure $$\PageIndex{3}$$
Now, we can use the Pythagorean Theorem to find the missing side length:
\begin{align}\begin{aligned}\begin{split} a^2 + b^2 &= c^2 \\ (12)^2 + b^2 & = 13^2 \\ 144 + b^2 & = 169 \\ b^2 & = 25 \\ b & = \pm5 \end{split}\end{aligned}\end{align}
Now, we can update our drawing:
Figure $$\PageIndex{4}$$
In our drawing, we labeled all of the sides with positive values because when we measure the side of a triangle we will get a positive length. From this triangle, we get $$\sin{(\theta)} = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{5}{13}$$. However, this is not the only possible value of $$\sin{(\theta)}$$. We do not know the true value of $$\theta$$ and in our drawing assumed that it is between 0 and $$\frac{\pi}{2}$$. In reality, it could also be between $$\frac{3\pi}{2}$$ and $$2\pi$$. This would mean that $$\sin{(\theta)}$$ could also have a negative value.
The possible values of $$\sin (\theta )$$ are $$\frac{5}{13}$$ and $$-\frac{5}{13}$$
## 3.5.2 Inverse Trigonometric Functions
Often, we will have information about the side lengths of the triangle, but will want to know the value of the angle. This is where we will need the inverse trigonometric functions. Each trigonometric function has an inverse, but the inverses of sine, cosine, and tangent are the most commonly used. The notation for the functions is a bit tricky. We’ve seen that we can write $$(\sin{(\theta)})^2$$ as $$\sin^2{(\theta)}$$, however, the notation $$\sin^{-1}{(\theta)}$$ is often used to represent the inverse sine function rather than the function $$\frac{1}{\sin{(\theta)}}$$. In this book, we will instead use the notation $$\arcsin{(x)}$$ to represent the inverse sine function. This eliminates confusion over notation, but you should be aware that not all references do this. Similarly, we use $$\arccos{(x)}$$ for the inverse cosine function and $$\arctan{(x)}$$ for the inverse tangent function. These can be referred to as arcsine, arccosine, and arctangent in writing.
Notice that for each of these inverse trigonometric functions, we used $$x$$ as our input rather than $$\theta$$. This is because we are no longer inputting an angle, but rather a ratio of lengths. For these functions, our output will be an angle. Remember, when we say that two functions are inverses, we mean that there is a relationship like the following: $$\arccos{(\cos{(\theta)})}=\theta$$ and $$\cos{(\arccos{(x)})} = x$$. Another way of expressing this relationship is to say that if $$\cos{(\theta)} =x$$, then $$\arccos{(x)} = \theta$$. However, this is not exactly true here. When we look at trigonometric functions, we know that there are lots of angles that all result in the same value for sine, lots of angles that result in the same value for cosine, and lots of angles that result in the same value for tangent. Because we only want one output for each input, the inverse trigonometric functions use restricted outputs. Arccosine is restricted to output values between $$0$$ and $$\pi$$, meaning that its range is $$[0,\pi]$$. This works because every possible output of cosine shows up once for angles from 0 to $$\pi$$. Arcsine and arctangent are restricted to output values between $$-\frac{\pi}{2}$$ and $$\frac{\pi}{2}$$, meaning that they each have a range of $$[-\frac{\pi}{2}, \frac{\pi}{2}]$$. For both tangent and sine, every possible output value appears once for angles between $$-\frac{\pi}{2}$$ and $$\frac{\pi}{2}$$. By restricting the ranges, we make sure these functions are well-defined, meaning they only produce one output for each input.
In practice, we can still use the unit circle to help evaluate inverse trigonometric functions. For example, if we want to evaluate $$\arcsin{(\frac{1}{2})}$$, we will want to look at the unit circle to see where $$\sin{(\theta)}=\frac{1}{2}$$. We get two angles: $$\frac{\pi}{6}$$ and $$\frac{5\pi}{6}$$. Since the range of arcsine is restricted to $$[-\frac{\pi}{2},\frac{\pi}{2}]$$, we say that $$\arcsin{(\frac{1}{2})} = \frac{\pi}{6}$$. Similarly, we would say that $$\arccos{(\frac{1}{2})} = \frac{\pi}{3}$$, and $$\arctan{(1)} =\frac{\pi}{4}$$.
3.5: Trigonometric Functions and Triangles is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts. |
# RD Sharma Solutions for Class 8 Maths Chapter 2 - Powers Exercise 2.1
Students can refer to and download RD Sharma Solutions for Class 8 Maths Exercise 2.1 Chapter 2, Powers from the links provided below.
Our subject experts have solved the RD Sharma Solutions to ensure that the students are thorough with basic concepts and clear their doubts. Exercise 2.1 is based on the negative integral exponents. Students can download the Solutions for RD Sharma Class 8 Maths Chapter 2 and start practising them offline.
## RD Sharma Solutions for Class 8 Maths Exercise 2.1 Chapter 2 Powers
### Access Answers to RD Sharma Solutions for Class 8 Maths Exercise 2.1 Chapter 2 Powers
1. Express each of the following as a rational number of the form p/q, where p and q are integers and q ≠ 0:
(i) 2-3
(ii) (-4)-2
(iii) 1/(3)-2
(iv) (1/2)-5
(v) (2/3)-2
Solution:
(i) 2-3 = 1/23 = 1/2×2×2 = 1/8 (we know that a-n = 1/an)
(ii) (-4)-2 = 1/-42 = 1/-4×-4 = 1/16 (we know that a-n = 1/an)
(iii) 1/(3)-2 = 32 = 3×3 = 9 (we know that 1/a-n = an)
(iv) (1/2)-5 = 25 / 15 = 2×2×2×2×2 = 32 (we know that a-n = 1/an)
(v) (2/3)-2 = 32 / 22 = 3×3 / 2×2 = 9/4 (we know that a-n = 1/an)
2. Find the values of each of the following:
(i) 3-1 + 4-1
(ii) (30 + 4-1) × 22
(iii) (3-1 + 4-1 + 5-1)0
(iv) ((1/3)-1 – (1/4)-1)-1
Solution:
(i) 3-1 + 4-1
1/3 + 1/4 (we know that a-n = 1/an)
LCM of 3 and 4 is 12
(1×4 + 1×3)/12
(4+3)/12
7/12
(ii) (30 + 4-1) × 22
(1 + 1/4) × 4 (we know that a-n = 1/an, a0 = 1)
LCM of 1 and 4 is 4
(1×4 + 1×1)/4 × 4
(4+1)/4 × 4
5/4 × 4
5
(iii) (3-1 + 4-1 + 5-1)0
(We know that a0 = 1)
(3-1 + 4-1 + 5-1)0 = 1
(iv) ((1/3)-1 – (1/4)-1)-1
(31 – 41)-1 (we know that 1/a-n = an, a-n = 1/an)
(3-4)-1
(-1)-1
1/-1 = -1
3. Find the values of each of the following:
(i) (1/2)-1 + (1/3)-1 + (1/4)-1
(ii) (1/2)-2 + (1/3)-2 + (1/4)-2
(iii) (2-1 × 4-1) ÷ 2-2
(iv) (5-1 × 2-1) ÷ 6-1
Solution:
(i) (1/2)-1 + (1/3)-1 + (1/4)-1
21 + 31 + 41 (we know that 1/a-n = an)
2+3+4 = 9
(ii) (1/2)-2 + (1/3)-2 + (1/4)-2
22 + 32 + 42 (we know that 1/a-n = an)
2×2 + 3×3 + 4×4
4+9+16 = 29
(iii) (2-1 × 4-1) ÷ 2-2
(1/21 × 1/41) / (1/22) (we know that a-n = 1/an)
(1/2 × 1/4) × 4/1 (we know that 1/a ÷ 1/b = 1/a × b/1)
1/2
(iv) (5-1 × 2-1) ÷ 6-1
(1/51 × 1/21) / (1/61) (we know that a-n = 1/an)
(1/5 × 1/2) × 6/1 (we know that 1/a ÷ 1/b = 1/a × b/1)
3/5
4. Simplify:
(i) (4-1 × 3-1)2
(ii) (5-1 ÷ 6-1)3
(iii) (2-1 + 3-1)-1
(iv) (3-1 × 4-1)-1 × 5-1
Solution:
(i) (4-1 × 3-1)2 (we know that a-n = 1/an)
(1/4 × 1/3)2
(1/12)2
(1×1 / 12×12)
1/144
(ii) (5-1 ÷ 6-1)3
((1/5) / (1/6))3 (we know that a-n = 1/an)
((1/5) × 6)3 (we know that 1/a ÷ 1/b = 1/a × b/1)
(6/5)3
6×6×6 / 5×5×5
216/125
(iii) (2-1 + 3-1)-1
(1/2 + 1/3)-1 (we know that a-n = 1/an)
LCM of 2 and 3 is 6
((1×3 + 1×2)/6)-1
(5/6)-1
6/5
(iv) (3-1 × 4-1)-1 × 5-1
(1/3 × 1/4)-1 × 1/5 (we know that a-n = 1/an)
(1/12)-1 × 1/5
12/5
5. Simplify:
(i) (32 + 22) × (1/2)3
(ii) (32 – 22) × (2/3)-3
(iii) ((1/3)-3 – (1/2)-3) ÷ (1/4)-3
(iv) (22 + 32 – 42) ÷ (3/2)2
Solution:
(i) (32 + 22) × (1/2)3
(9 + 4) × 1/8 = 13/8
(ii) (32 – 22) × (2/3)-3
(9-4) × (3/2)3
5 × (27/8)
135/8
(iii) ((1/3)-3 – (1/2)-3) ÷ (1/4)-3
(33 – 23) ÷ 43 (we know that 1/a-n = an)
(27-8) ÷ 64
19 × 1/64 (we know that 1/a ÷ 1/b = 1/a × b/1)
19/64
(iv) (22 + 32 – 42) ÷ (3/2)2
(4 + 9 – 16) ÷ (9/4)
(-3) × 4/9 (we know that 1/a ÷ 1/b = 1/a × b/1)
-4/3
6. By what number should 5-1 be multiplied so that the product may be equal to (-7)-1?
Solution:
Let us consider a number x
So, 5-1 × x = (-7)-1
1/5 × x = 1/-7
x = (-1/7) / (1/5)
= (-1/7) × (5/1)
= -5/7
7. By what number should (1/2)-1 be multiplied so that the product may be equal to (-4/7)-1?
Solution:
Let us consider a number x
So, (1/2)-1 × x = (-4/7)-1
1/(1/2) × x = 1/(-4/7)
x = (-7/4) / (2/1)
= (-7/4) × (1/2)
= -7/8
8. By what number should (-15)-1 be divided so that the quotient may be equal to (-5)-1?
Solution:
Let us consider a number x
So, (-15)-1 ÷ x = (-5)-1
1/-15 × 1/x = 1/-5
1/x = (1×-15)/-5
1/x = 3
x = 1/3
## RD Sharma Solutions for Class 8 Maths Exercise 2.1 Chapter 2 Powers
Download the free RD Sharma Solutions Chapter 2 in PDF format, which provide answers to all the questions. Class 8 Maths Chapter 2 Powers Exercise 2.1 is based on problems which include negative exponents. These solutions are prepared by experienced faculty in accordance with the CBSE syllabus for the 8th Standard. The exercise-wise solutions are explained in a simple and easily understandable language to help students excel in the annual exam. |
# How do you simplify sqrt75+sqrt108?
Oct 6, 2015
In cases like this it often helps to factorize first.
#### Explanation:
Factorizing means you write a number in the form of a multiplication of smaller numbers (primes).
So you can write:
$75 = 3 \cdot 5 \cdot 5$ and $108 = 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3$
Now isolate the squares:
$75 = 3 \cdot {5}^{2}$ and $108 = 3 \cdot {2}^{2} \cdot {3}^{2}$
You can simplify by taking the squares from under the root (where they are unsquared of course):
$\sqrt{75} + \sqrt{108} = \sqrt{3 \cdot {5}^{2}} + \sqrt{3 \cdot {2}^{2} \cdot {3}^{3}} =$
$5 \cdot \sqrt{3} + 2 \cdot 3 \cdot \sqrt{3} = 5 \sqrt{3} + 6 \sqrt{3} = 11 \sqrt{3}$ |
# Triangle & It's Important Theorems
By Sachin Awasthi|Updated : February 23rd, 2021
In continuation of the CHSL study plan, we will discuss one of the important topics from the Geometry section i.e., " Important Theorems Related to Triangle ". Questions based on these theorems are frequently asked in the exam.
### BASIC PROPORTIONALITY THEOREM
In a triangle, a line drawn parallel to one side to intersect the other side in distinct points divides the two sides in the same ratio
In continuation of the CHSL study plan, we will discuss one of the important topics from the Geometry section i.e., " Important Theorems Related to Triangle ". Questions based on these theorems are frequently asked in the exam.
### BASIC PROPORTIONALITY THEOREM
In a triangle, a line drawn parallel to one side to intersect the other side in distinct points divides the two sides in the same ratio
As per the basic proportionality theorem:
### Mid-point Theorem
Mid-point theorem is a special case of " Basic Proportionality Theorem ". In the above triangle, D and E are the midpoints of the sides AB and AC respectively.
Then, as per the midpoint theorem
DE is parallel to BC, and
DE = 1/2 BC
### Pythagoras Theorem
In right triangles, the square of the hypotenuse equals the sum of the squares of the other two sides
As per the Pythagoras theorem: AC2 = AB2 + BC2
It is important to learn the triplets that make right-angled triangles.
3, 4, 5
5, 12, 13
7, 24, 25
8, 15, 17
9, 40, 41
11, 60, 61
12, 35, 37
16, 63, 65
20, 21, 29
These triplets satisfy the condition of AC2 = AB2 + BC2
If a, b, c denote the sides of a triangle then
• If c2 < a2 + b2, triangle is acute-angled.
• If c2 = a2 + b2, triangle is right angled.
• If c2 > a2 + b2, triangle is obtuse-angled
Median: A median of a triangle is the line from a vertex to the midpoint of the opposite side. The centroid is the point at which the medians of the triangle meet. The centroid divides the medians in the ratio 2:1. The median bisects the area of the triangles.
### Theorem of Apollonius
Sum of the squares of two sides of a triangle
= 2 (median) 2 + 2 (half the third side)2In the figure AD is the median, then: AB2 + AC2 =2(AD)2 + 2(BD)2
### Angle Bisector Theorem
If AD is the bisector of angle A,
Then, AB/AC = BD/DC
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# What are the odds of rolling a straight with 6 dice?
Contents
## What are the odds of rolling a straight with 6 dice twice in a row?
The answer is 1/36. The past rolls are entirely independent of future rolls, so history doesn’t matter.
## What are the odds of rolling a straight with five dice?
Since there are 240 ways to roll a large straight in a single roll and there are 7776 rolls of five dice possible, the probability of rolling a large straight is 240/7776, which is close to 1/32 and 3.1%. Of course, it is more likely than not that the first roll is not a straight.
## What are the odds of rolling a 6 four times in a row?
The probability of rolling a 6 in 4 throws is 1−(56)4 1 − ( 5 6 ) 4 , which turns out to be just over 50%. The probability of rolling two 6’s in 24 throws of a pair of dice is 1−(3536)24 1 − ( 35 36 ) 24 , which turns out to be just under 50%.
## What are the odds of rolling at least one 6?
When you roll two dice, you have a 30.5 % chance at least one 6 will appear. This figure can also be figured out mathematically, without the use of the graphic.
## What are the odds of rolling 5 sixes with 5 dice?
Each die is independent, so the probability of getting all sixes is simply (11/36)5.
THIS IS IMPORTANT: Your question: How old do you have to be to work at a casino in Australia?
## What is a straight with dice?
When a player rolls 1,2,3,4,5,6 when rolling all 6 dice this is a Straight. When a player gets 3 sets of pairs when rolling 6 dice this is Pairs. Pairs and Straights are worth 5oo points.
## How is Yahtzee probability calculated?
The probability of rolling five of a kind of any other number is also 1/7776. Since there are a total of six different numbers on a die, we multiply the above probability by 6. This means that the probability of a Yahtzee on the first roll is 6 x 1/7776 = 1/1296 = 0.08 percent. |
# Eureka Math Grade 5 Module 6 Mid Module Assessment Answer Key
## Engage NY Eureka Math 5th Grade Module 6 Mid Module Assessment Answer Key
### Eureka Math Grade 5 Module 6 Mid Module Assessment Task Answer Key
Question 1.
Give the coordinates of each point.
A ________________
B ________________
C ________________
D ________________
E ________________
A ( 3, 4)
B (4, 2)
C ($$\frac{1}{2}$$ , $$\frac{1}{4}$$)
D ( 1, 2$$\frac{1}{2}$$)
E (1$$\frac{3}{4}$$, 4$$\frac{1}{4}$$)
Explanation :
The respective x – coordinate and y- coordinates are written of the given points .
Question 2.
Plot each point in the coordinate plane above, and label each point with F, G, or H.
F (0, 4)
G (2, 1)
H (4$$\frac{3}{4}$$, 3$$\frac{3}{4}$$)
Explanation :
The Given Points F, G and H are marked and is shown in the above graph .
Question 3.
a. Give coordinates for any three points that are on the same vertical line. Include at least one point that has a mixed number as a coordinate.
b. Give coordinates for any three points that are on the same horizontal line. Include at least one point that has a fraction as a coordinate.
a. The 3 points are P ( 3, 2 ), Q ( 3, 3 ) and R (3, 3$$\frac{1}{4}$$)
Explanation :
To form a Vertical line on the graph that mean x-coordinates will be same for all y-coordinates. then a vertical line is formed that is parallel to y-axis .
The points are Marked in the above graph .
b. The 3 points are A ( 1, 2 ), B ( 2$$\frac{1}{2}$$, 2 ) and C (3$$\frac{1}{4}$$, 2)
Explanation :
To form a Horizontal line on the graph that mean y-coordinates will be same for all x-coordinates. then a vertical line is formed that is parallel to x-axis .
The points are Marked in the above graph .
Question 4.
Garrett and Jeffrey are planning a treasure hunt. They decide to place a treasure at a point that is a distance of 5 units from the x-axis and 3 units from the y-axis. Jeffrey places a treasure at point J, and Garrett places one at point G. Who put the treasure in the right place? Explain how you know.
Jeffrey puts the treasure in the right place
Explanation :
The treasure is at at a point that is a distance of 5 units from the x-axis and 3 units from the y-axis at a point ( 5, 3)
Jeffrey marks it correct where as the Garrett marks at a point that is a distance of 3 units from the x-axis and 5 units from the y-axis at a point ( 3, 5)
Question 5.
a. Find the y-coordinates by following the rules given for each table.
Table A: Multiply by $$\frac{1}{2}$$.
x y 0 1 2 3
Table B: Multiply by $$\frac{1}{4}$$.
x y 0 1 2 3
b. Graph and label the coordinate pairs from Table A. Connect the points, and label the line a. Graph and label the coordinate pairs from Table B. Connect the points, and label the line b.
c. Describe the relationship between the y-coordinates in Table A and Table B that have the same x-coordinate.
a.
Table A: Multiply by $$\frac{1}{2}$$.
y = $$\frac{1}{2}$$(x) .
x y 0 0 1 $$\frac{1}{2}$$ 2 1 3 1$$\frac{1}{2}$$
Table B: Multiply by $$\frac{1}{4}$$.
y=$$\frac{1}{4}$$(x)
x y 0 0 1 $$\frac{1}{4}$$ 2 $$\frac{1}{2}$$ 3 $$\frac{3}{4}$$
Explanation :By following the given rule the y-coordinates are calculated and written in the above tabular columns .
b.
Explanation :
The points of tabular Column A are plotted and written as line a .
The points of tabular Column B are plotted and written as line b
All points are plotted and shown in above graph .
c.
The y coordinates in Tabular Column A are half of the y coordinates in the Tabular Column B
Question 6.
a. Use the graph to give the coordinate pairs of the points marked on the line.
x y
b. Using this rule, generate three more points that would be on this line but lie beyond the portion of the coordinate plane that is pictured.
______ _______ ________ |
# How Do You Know If A System Is Dependent Or Independent?
## What is dependent system of equations?
dependent system: A system of linear equations in which the two equations represent the.
same line; there are an infinite number of solutions to a dependent system.
inconsistent system: A system of linear equations with no common solution because they.
represent parallel lines, which have no point or line in common..
## How do you tell if a system has no solution?
A system has no solutions if the lines are parallel. When solving the system, if you get a false statement (a number equal to a different number) this means there are no solutions.
## Why do parallel lines have no solution?
Since parallel lines never cross, then there can be no intersection; that is, for a system of equations that graphs as parallel lines, there can be no solution. This is called an “inconsistent” system of equations, and it has no solution.
## Which equations have the same pair of solutions?
Systems of equations that have the same solution are called equivalent systems. Given a system of two equations, we can produce an equivalent system by replacing one equation by the sum of the two equations, or by replacing an equation by a multiple of itself.
## How do you know if an equation is equivalent?
If “x” is the same for both equations, then they are equivalent. If “x” is different (i.e., the equations have different roots), then the equations are not equivalent. For the first equation: x + 2 = 7.
## What makes an equation equivalent?
Equivalent equations are two equations that have the same solution. They are used anytime multiple equations, with the same variable, need to equal each other, just like in the story above.
## How do you tell if a system of equations has no solution or infinitely many?
A system of linear equations has no solution when the graphs are parallel. Infinite solutions. A system of linear equations has infinite solutions when the graphs are the exact same line.
## What are infinitely many solutions?
A system has infinitely many solutions when it is consistent and the number of variables is more than the number of nonzero rows in the rref of the matrix. For example if the rref is has solution set (4-3z, 5+2z, z) where z can be any real number.
## What are the 3 types of system of equations?
There are three types of systems of linear equations in two variables, and three types of solutions.An independent system has exactly one solution pair [Math Processing Error] . … An inconsistent system has no solution. … A dependent system has infinitely many solutions.
## How do you solve systems of independent equations?
The first equation tells us that y is equal to -8x + 4, so we can substitute -8x + 4 for y in the second equation. The second equation becomes 8x + 4(-8x + 4) = 0. We can then solve the equation for x, and substitute the solution for x into one of the equations to solve for y.
## What is an independent solution?
When a system is “independent,” it means that they are not lying on top of each other. There is EXACTLY ONE solution, and it is the point of intersection of the two lines. It’s as if that one point is “independent” of the others. To sum up, a dependent system has INFINITELY MANY solutions.
## How do you classify a system of equations?
A system of two equations can be classified as follows: If the slopes are the same but the y-intercepts are different, the system is inconsistent. If the slopes are different, the system is consistent and independent. If the slopes are the same and the y-intercepts are the same, the system is consistent and dependent.
## Can two different equations have the same answer?
Two different equations can never have the same answer.
## Can a system of two linear equations have exactly two solutions?
Most linear systems you will encounter will have exactly one solution. However, it is possible that there are no solutions, or infinitely many. (It is not possible that there are exactly two solutions.) The word unique in this context means there is a solution, and it’s the only one. |
# MODULE II
Topics
1.SIMULTANEOUS LINEAR EQUATIONS WITH TWO UNKNOWNS 2. SIMULTANEOUS LINEAR EQUATIOS WITH THREE UNKNOWNS 3. QUADRATIC EQUATION AND SOLUTION 4. NATURE OF ROOTS IN QUADRATIC EQUATION 5. INEQUALITY WITH GRAPHS 6. INDICES
7.SURDS
8. LOGARITHM
Linear Equation
A linear equation with two unknowns is given by ax + by + c = 0. Where a,b,c are real numbers
Simultaneous linear equations with two unknowns Pair of Linear Equations: a1 x b1 y c1 0 and a2 x b2 y c2 0 Ex: 2x+4y+6=0 And x+5y+8=0 .
Geometrical Forms Pair of Equations will: • Intersect • Parallel No Solution Unique Solution • Coincide Infinitely Many Solutions .
Methods to Solve Linear Equation Algebraically • Substitution Method • Elimination Method • Cross Multiplication Method .
Solve 7x – 15y = 2 x + 2y = 3 Step(i) We pick either of the equation and write one variable in terms of other.1) Substitution method: Eg:. Let us consider 2nd equation x + 2y = 3 therefore x = 3 – 2y .
7(3 – 2y) – 15y =2 21 – 14y -15 y = 2 29y =19 y = 19/29.Step (ii) Substitute the value of x in 1st equation. x = 49/29 . Step (iii) Substitute y = 19/29 in any of the equations we get.
Elimination Method This is done by making the coefficient of either x or y same. 9x-4y=2000 7x-3y=2000 (Ans: x=2000.2. y=4000) .
3) Cross multiplication method: If the two equations are : a1 x b1 y c1 0 a2 x b2 y c2 0 b1c2 b2 c1 Then. x a1b2 a2b1 c1a2 c2 a1 and y a1b2 a2b1 .
y=10) .Q) Solve by cross multiplication: 2x+3y-46=0 3x+5y-74=0 Ans:(x=8.
y=5.Questions(Substitution method) • Solve: (i) x + y = 14 and x – y = 4(Ans: x=9.y=4) (ii) s – t =3 and s/3 + t/2 = 6 (Ans: t=6.m=-1) .s=9) (iii) 3x/2 – 5y/3 = -2 and x/3 + y/2 =13/6 (x=2.y=3) (iv) Solve 2x + 3y = 11 and 2x – 4y = -24 and hence find the value of m for which y = mx + 3.(x=-2.
-1/7) (iii)3x – 5y – 4 = 0 and 9x = 2y + 7 (iv)x/2 + 2y/3 = -1 and x – y/3 = 3 .Questions(Elimination method) • Solve: (i)x + y = 5 and 2x – 3y = 4(Ans:19/5.6/5) (ii)3x + 4y = 2 and 2x – 2y =2 (Ans: 6/7.
Questions(Cross multiplication method) • Solve: (i)x – 3y – 3 = 0 and 3x – 9y – 2 = 0 (ii)2x + y = 5 and 3x + 2y = 8 (iii)3x – 5y = 20 and 6x – 10y = 40 (iv)x – 3y – 7 = 0 and 3x – 3y = 15 .
Find them. (Ans: 13.39) .Word Problems • Q) The difference between two numbers is 26 and one number is 3 times the other.
Find the cost of each ball and each bat. (Ans: Rs. 3800. 50) . 1750.• Q) The coach of a cricket team buys 7 bats and 6 balls for Rs. 500 and Rs. Later he buys 3 bats and 5 balls for Rs.
Q) A fraction becomes 9/11. If 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction. (Ans: 7/9) . if 2 is added to both the numerator and denominator.
Five years ago.Q) Five years hence. Jacob’s age was 7 times that of his son. the age of Jacob will be three times that of his son. What are their present ages? (Ans: 40 years and 10 years) .
x represents a variable and a. the linear coefficient and the constant term. c are called respectively the quadratic coefficient. b. b.Quadratic Equation • In mathematics a quadratic equation is a polynomial equation of the second degree. 2 . • The constants a. c represents constants with a≠0.The general form is ax bx c 0 where.
• The solution of the quadratic equation are called as ROOTS of the equation.Solution of Quadratic Equation • There are always two solutions for a quadratic equation. .
Methods • Two methods to solve a quadratic Equation: • By factoring (works sometimes) • By Quadratic Formula (always works) .
For example: Solve 4 x 2 25 We can write the aboveeq n as (2 x) 2 (5) 2 0 (2 x 5) ( 2 x 5) 0 either ( 2 x 5) 0 or ( 2 x 5) 0 5 x 2 or 5 x 2 .Factoring • This method is used where quadratic equation can be easily resolved into linear factors.
Factoring Example 2) Solve: x2 6x 8 0 We can solvethe abovequadratic eq by splitting the middleterm. x2 4x 2x 8 0 x( x 4) 2( x 4) 0 ( x 2) ( x 4) 0 x 2 or x 4 n .
Solve: (Ans: x=3/2.Questions 1.x=-3/2) 9x 2 4x2 7 6x 1 2 3 4x 3 2 . Solve: x b a b b x b a (Ans: x=a.x=b²/a) 3.Solve: x 3x 4 2 2.
2.b.c ax bx c 0 2 The solution is given by: b b 2 4ac b b 2 4ac x and 2a 2a . By Quadratic Formula This method works all the time Put the equation in standard form: Determine the values of a.
Questions • Solve: x²+9x=-8 (Ans: x=-1 and x=-8) • Solve: 2x²-10x+5=0 (Ans: x=(5+√15)/2 and (5-√15)/2) .
QUESTIONS Solve the following equations: 1) 6 1 x7 4x 2 3 3 5 3 5 8 2) x2 x6 x3 x 1 2( x 1) 3) 4 x x 3 3 5 ax 2a x 3a x 4) a 2a 3a .
Solve the following equations: 1) 25 x 2 16 x x3 2) x2 5( x 11) 2( 45 2 x 2 ) 3( x 2 9) 3) 7 x2 9 x2 3 4) 3 x 2 14 x 11 0 5) x 2 ( p q ) x pq 0 x 3 10 3 x 3 x a a b 7) a x b a 8) x 2 2 3 x 1 0 6) 9) x 2 ( 3 3) x ( 3 2) 0 .
x=480 ..x=-2 4. x 1 x 1 x x 21 6 a 1 a 1 Ans: x=4/13 and x=9/13 2...Special Cases 1. 6 6 6 . x 12 a x x 12 a x Ans: x=3a.x=-4a 3. Ans: x=3. 2 x 1 3x 4 7 Ans: x=4.
Solve the following equations: 1) 1 5 x 1 3x 2 2) 3x 10 9 x 7 9 .
Roots of Quadratic Equations The general quadratic equation is : ax 2 bx c 0 The two roots of the quadratic equation are givenby : b b 2 4ac b b 2 4ac x and 2a 2a where. So. the roots of the quadratic equation are : b D b D x and 2a 2a . the exp ression b 2 4ac is called discri min ant of the equation and is denoted by D.
then both the roots are complex and unequal. If D=0. If D<0. 2. b2 4ac 3. . If D>0. both the roots are unequal. both the roots are real and equal.Nature of the Roots Since the roots of the quadratic equation are given by: b b 2 4ac b b 2 4ac x and 2a 2a The nature of the roots depends on the numerical value of We obtain the following results: 1.
m=-2) .Questions • Find the nature of the roots: • i) x²+2x+3=0 • ii) Find what values of m will the equation (m+1)x²+2(m+3)x+(2m+3)=0 will have equal roots. (Ans: m=3.
Let and be the roots of the quadratic equation ax 2 bx c 0. Sum of the roots are : b coefficien of x t sum of the roots a coefficien of x 2 t c cons tan t term product of the roots a coefficien of x 2 t .Sum and Product of the roots of quadratic equations: Let.
1 1 . formthe quadratic equation whose roots are : (i ) 3. find the valuesof (i ) 2 2 (ii) 3 3 (iii) 4 4 2) If .3 2 (iii) . are the roots of x 2 2 x 3 0. 3 (ii) 2 3 . are the roots of 2 x 2 3 x 7 0. and 1 1 (iv) .Questions 1) If .
Questions on nature of roots 1) Find k . (ii) one of the roots of equation x 2 4 x k 0 is 2(1 3 ) (iii) one of the roots of equation x 2 6 x k 0 is 3 i 2 (iv) one root of the equation x 2 6 x k 0 is double the other. if (i ) the roots of 2 x 2 3 x k 0 are equal. Ans:K=8 . |
# Shopping & Dining with Proportions
## Presentation on theme: "Shopping & Dining with Proportions"— Presentation transcript:
Shopping & Dining with Proportions
Cornell Notes with Summary Essential Question: How are percents used practically when solving problems dealing with discounts, sales tax, and tips? NOTE TO TEACHERS: This slide includes the title for the notes, and the instructions regarding note-taking style. On subsequent pages, the slide title indicates the topic for the left-hand side of their Cornell Notes.
Discount An amount by which the cost of an item is reduced. The total price of an item is the cost minus discount. At a local retail store, you find a pair of jeans that are 30% off. Originally, they cost \$ How much money is the sale price of the jeans? Set up a percent proportion. % = is = x Of \$59.99 Use CMAD to find the discount amount (cross multiply and divide) . 30/100 = x/\$ cross multiply \$59.99 by 30, then divide the product by 100. Your quotient is \$18.00 after you round to the nearest cent. Subtract the discount from the jeans’ original price. \$ \$18.00 Find the final cost of the jeans: \$41.99
You try it! Find the total cost of a dinning room set that is \$1,200 when it is discounted 25%. Set up a percent proportion. % = is = x Of \$1,200 Use CMAD to find the discount amount (cross multiply and divide) . 25/100 = x/\$1,200 cross multiply \$1,200 by 25, then divide the product by 100. Your quotient is \$ Subtract the discount from the dining room set’s original price. \$1,200 - \$300 Find the final cost of the jeans: \$900
Sales Tax An additional charge added to the cost of an item. The total price of an item is the cost plus tax. At a local retail store, you purchase a t-shirt for \$ How much money will you need to purchase the shirt if sales tax is 7.5%? Set up a percent proportion. % = is = x Of \$14.99 Use CMAD to find the sales tax (cross multiply and divide). 7.5/100 = x/\$ cross multiply \$14.99 by 7.5, then divide the product by 100. Your quotient is \$1.12 after you round to the nearest cent. Add the tax to the t-shirt’s original price. \$ \$1.12 Final cost of the t-shirt: \$16.11
You try it! Find the total cost of a TV that is \$298 when sales tax is 6.5%. Set up a percent proportion. % = is = x Of \$298 Use CMAD to find the sales tax (cross multiply and divide) . 6.5/100 = x/\$298 cross multiply \$298 by 6.5, then divide the product by 100. Your quotient is \$19.37. Add sales tax to the TV’s original price. \$298 + \$19.37 Find the final cost of the TV: \$317.37
Tip or Gratuity An additional charge added to the cost of a service. The total price is the cost of the service plus tip. At a local restaurant, you purchase a drink and some dinner. The restaurant bill comes to \$17.45 including tax. The waitress did her job well, so you decide to leave a 17% tip. How much money will you spend on dinner? Set up a percent proportion. % = is = x Of \$17.45 Use CMAD to find the tip amount (cross multiply and divide) 17/100 = x/\$ cross multiply \$17.45 by 17, then divide the product by Your quotient is \$2.97 after you round to the nearest cent. Add the tip to the restaurant bill. \$ \$2.97 Final cost of dinner: \$20.42 *Would you leave a \$2.97 tip, or would you round it up to \$3.00?
You try it! Find the cost lunch after leaving a 16% tip if the bill comes to \$12.57. Set up a percent proportion. % = is = x Of \$12.57 Use CMAD to find the tip amount (cross multiply and divide) 16/100 = x/\$ cross multiply \$12.57 by 16, then divide the product by 100. Your quotient is \$2.01 after you round to the nearest cent. Add the tip to the restaurant bill. \$ \$2.01 Final cost of dinner: \$14.58 *Would you leave a \$2.01 tip, or would you round it to \$2.00?
Putting it all together
You have \$150 to spend shopping at your favorite retail store. They’re having a sale on a huge selection of clothing, and you’re excited to shop. Most items are 30-50% off and you hope you can buy all the items on your list. However, you’re having dinner with your friends afterwards at your favorite restaurant, so you know you’ll spend at least \$10.98 before tax and tip on dinner. You must tip the waitress 15-20%, and tax in your area is 6.5%. NOTE: for this activity, adjust the TAX PERCENT to match your local sales tax
Putting it all together
Shop wisely, you only have \$150, and you must buy dinner. Using the original prices listed below, pick at least 2 shirts, one pair of pants, and a pair of shoes you can afford – their discounts are also listed. Remember, tax is 6.5% at both the retail store and the restaurant. Dinner will cost \$10.98 before tax & a 15-20% tip. NOTE: for this activity, adjust the TAX PERCENT to match your local sales tax Sale prices (before tax): khakis \$32.50; t-shirt \$13.65; dress shirt - \$19.25; dress shoes - \$34.30; jeans - \$34.80; graphic t-shirt - \$17.49; Hawaiian print shirt - \$11.50; tennis shoes - \$40.30 \$65.00 – Khakis – 50% off \$58.00 – Jeans – 40% off \$21.00 – T-Shirt – 35% off \$24.99 – Graphic T-Shirt – 30% off \$35.00 – Dress Shirt – 45% off \$23.00 – Hawaiian Print Shirt – 50% off \$49.00 – Dress Shoes – 30% off \$62.00 – Tennis Shoes – 35% off
Summary Answer the Essential Question in two or more complete sentences. EQ: How are percents used practically when solving problems dealing with discounts, sales tax, and tips?
Homework Using the order calculated during class today, describe in at least one paragraph the decisions that were made in selecting menu items, as well as tip percentage. Include an explanation of how the totals were calculated. |
## Two Divisibility Problems
2016 May 9
Here are two interesting problems regarding divisibility that I found interesting.
Prove that $$(n-1)^2 \mid n^k-1$$ if and only if $$(n-1) \mid k$$.
Prove that $$F_{kn} \mid F_n$$ for all integer $$n$$ and $$k$$, where $$F_i$$ denotes the ith Fibonacci number.
The first is from Math Stackexchange. By breaking this problem down into a few lemmas, it can be done without too much difficulty.
Lemma #1: The sum
$a^0+a^1+...+a^b$
is equal to
$\frac{a^{b+1}-1}{a-1}$
This lemma can be proven easily, since when the two are set equal to each other, we recieve a truth statement:
$\frac{a^{b+1}-1}{a-1}=a^0+a^1+...+a^b$
$a^{b+1}-1=(a-1)(a^0+a^1+...+a^b)$
$a^{b+1}-1=-a^0+a^1-a^1+a^2-a^2+...+a^b-a^b+a^{b+1}$
$a^{b+1}-1=a^{b+1}-a^0$
$a^{b+1}-1=a^{b+1}-1$
And there is our truth statement, so Lemma #1 is proven.
Lemma #2: The quotient
$\frac{a^b}{a-1}$
is equal to
$a^{b-1}+a^{b-2}+...+a^2+a^1+a^0+\frac{1}{a-1}$
By Lemma #1, the following is true:
$\frac{a^b-1}{a-1}=a^0+a^1+...+a^{b-1}$
From this we can algebraically derive Lemma #2.
$\frac{a^{b}-1}{a-1}=a^0+a^1+...+a^{b-1}$
$\frac{a^{b}}{a-1}-\frac{1}{a-1}=a^0+a^1+...+a^{b-1}$
$\frac{a^{b}}{a-1}=a^0+a^1+...+a^{b-1}+\frac{1}{a-1}$
$\frac{a^{b}}{a-1}=a^0+a^1+...+a^{b-1}+\frac{1}{a-1}$
Which is equivalent to the statement made by Lemma #2, so Lemma #2 is thus proven.
Lemma #3: The quotient
$\frac{a^b+a^{b-1}+...+a^1+a^0}{a-1}$
is equal to
$a^{b-1}+2a^{b-2}+...+(b-1)a^1+ba^0+\frac{b+1}{a-1}$
This can be proven by splitting up the quotient into the sum of many quotients:
$\frac{a^b}{a-1}+\frac{a^{b-1}}{a-1}+...+\frac{a^1}{a-1}+\frac{a^0}{a-1}$
By Lemma #2, we can expand each of these quotients to form
$(a^{b-1}+...+a^0+\frac{1}{a-1})+(a^{b-2}+...+a^0+\frac{1}{a-1})+...+(a^1+a^0+\frac{1}{a-1})+(a^0+\frac{1}{a-1})$
Notice now that when we combine like terms, there will be $$1$$ term in the form $$a^{b-1}$$, $$2$$ of the form $$a^{b-2}$$, and so on, and $$b-1$$ of the form $$a^1$$, $$b$$ of the form $$a^0$$, and $$b+1$$ of the form $$\frac{1}{a-1}$$, so the whole thing condenses to
$a^{b-1}+2a^{b-2}+...+(b-1)a^1+ba^0+\frac{b+1}{a-1}$
Proving Lemma #3.
Now we can prove that $$(n-1)^2 \mid n^k-1$$ if and only if $$(n-1) \mid k$$ using our three Lemmas. By Lemma #1, the quotient of $$n^k-1$$ and $$n-1$$ is given by the sum
$n^{k-1}+n^{k-2}+...+n^1+n^0$
meaning that $$(n-1)^2 \mid n^k-1$$ if and only if
$(n-1) \mid n^{k-1}+n^{k-2}+...+n^1+n^0$
If that is true, then their quotient will be an integer. By Lemma #3, the quotient is given by the sum
$n^{k-2}+2n^{k-3}+...+(k-2)n^1+(k-1)n^0+\frac{k}{n-1}$
All terms of this sum must be integers except for the last term,
$\frac{k}{n-1}$
The entire sum will be an integer if and only if that last term is an integer, and the last term is an integer if and only if $$(n-1) \mid k$$, proving our initial statement.
The second problem is from a pattern that I noticed whilst playing around with the Fibonacci Sequence. My conjecture was that, if we let $$F_n$$ denote the nth Fibonacci number (starting with $$F_1=F_2=1$$), then $$F_n \mid F_{kn}$$ for positive integer $$k$$ and $$n$$. The proof of this statement reuqires only one Lemma before it can be proven using mathematical induction.
Lemma #4: $$F_a=F_bF_{a-b+1}+F_{b-1}F_{a-b}$$
By the definition of the Fibonacci Sequence,
$F_a=F_{a-1}+F_{a-2}$
and by substitution,
$F_a=2F_{a-2}+F_{a-3}$
and
$F_a=3F_{a-3}+2F_{a-4}$
This pattern necessarily continues on forever, because whenever another substitution happens, one coefficient is shifted and the coefficient of the new term is the sum of those of the coefficients of the previous two. Now Lemma #4 is proven.
Now we are ready to prove the theorem. First we show that it holds for $$k=2$$. By Lemma #4,
$F_{2n}=F_nF_{n+1}+F_{n-1}F_n$
$F_{2n}=F_n(F_{n+1}+F_{n-1})$
This shows that $$F_n \mid F_{2n}$$.
Now suppose that, for some $$k$$, my conjecture is true, so that $$F_n \mid F_{kn}$$. Then, by Lemma #4, by letting $$a=(k+1)n$$ and $$b=kn$$,
$F_{(k+1)n}=F_{kn}F_{n+1}+F_{kn-1}F_n$
Now, since we assumed that $$F_n \mid F_{kn}$$, we can say that $$F_{kn}=c*F_n$$ for some integer $$c$$, and by substitution,
$F_{(k+1)n}=c*F_nF_{n+1}+F_{kn-1}F_n$
We can now factor out $$F_n$$:
$F_{(k+1)n}=F_n(cF_{n+1}+F_{kn-1})$
Showing that if $$F_{kn}$$ is divisible by $$F_n$$, so is $$F_{(k+1)n}$$. However, since we have already proven that $$F_n \mid F_{2n}$$, we know the statement must be true for all positive integral $$k$$, thus proving that $$F_n \mid F_{kn}$$ for all positive integral $$k$$ and $$n$$. |
Congruent Shapes
Start Practice
## What Are Congruent Shapes?
Congruent shapes have the exact same shape and size.
Of the shapes below, only the triangles are congruent.
If you put two congruent shapes on top of each other, the top one completely covers the bottom one.
Tip: congruent shapes can have different colors.
To test if two shapes are congruent, they need to have:
✅ the same shape
✅ the same size
### Are These Shapes Congruent?
Let's check.
(1) Are they the same shape?
✅ Yes, they're the same shape, a circle.
(2) Are they the same size?
One circle is big. The other one is smaller than it.
❎ No, they're not the same size.
👉 These shapes are not congruent.
### Example 2
Are these congruent?
Let's check!
(1) Are they the same shape?
One shape has 5 sides. The other shape has 8 sides.
❎ No, they're not the same shape.
(2) Are they the same size?
❎ No, they're not the same size.
This pair does not check out on the two tests.
👉 These shapes are not congruent.
### Example 3
(1) Are these triangles the same shape?
✅ Yes, they're the same shape, a triangle.
(2) Are they the same size?
✅ Yes, they're the same size.
👉 These two shapes are congruent! 😃
Their colors are different.
Are they still congruent?
Yes, because color doesn't matter when looking for congruence.
The triangles are congruent because they have the same shape and size.
### Example 4
Are these congruent? 🤔
It might seem like they're not.
But let's check.
(1) Are they the same shape?
✅ Yes, they're both triangles. They are the same kind of triangle.
Their sides and angles are the same.
(2) Are they the same size?
It's tough to tell.
What if we rotate it to the right?
Like this:
Do their sizes look the same now?
✅ Yes, they have the same size!
This means that these triangles are congruent.
Tip: Congruent shapes match even if you flip, slide, or turn them.
Great job learning about congruent shapes!
Now, ace the practice. 💪 It'll help you remember what congruent shapes are for longer.
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# How do you find the partial quotient?
## How do you find the partial quotient?
How To Divide Using The Partial Quotients Method?
1. Step 1: Subtract from the dividend an easy multiple (for example 100×, 10×, 5× 2×, etc.)
2. Step 2: Repeat the subtraction until the large number has been reduced to zero or the remainder is less than the divisor.
## What is a partial quotient?
A partial quotient refers to a method used in solving large division mathematical problems. The method uses simple logic by allowing the student to see the problem in a less abstract form.
What is known as quotient method in accounts?
The quotient is a number obtained by dividing the dividend by divisor which is a simple division of numerator and denominator. A dividend is a numerator and a divisor is a denominator and the result obtained is the quotient. Refer the below example: Here 5 is the dividend, 25 is the divisor and 5 is the quotient.
What is a quotient in math?
: the number obtained by dividing one number by another Dividing 10 by 5 gives a quotient of 2. quotient.
### Why is partial quotient important?
The “partial quotients” strategy uses place value and allows students to build on multiplication facts with friendly numbers. The students can multiply 4 x 20 over and over again or use higher multiples of ten efficiently; they all reach the same solution. The “partial quotient” way will work with any division problem.
### How does partial quotient work?
The Partial Quotients method is one of these strategies. It is a mental math based approach that will enhance number sense understanding. Students solve the equation by subtracting multiples until they get down to 0, or as close to 0 as possible.
What is standard algorithm in division?
Standard algorithm: One of the conventional algorithms used in the United States based on place value and properties of operations for addition, subtraction, multiplication, and division.
What is quotient method?
The quotient can be calculated by dividing dividend with divisor. Quotient = Dividend ÷ Divisor. This is the most common method used to solve problems on division.
#### What is the quotient in dividing 6 by 3?
2
For example, if we divide the number 6 by 3, the result so obtained is 2, which is the quotient. It is the answer from the division process. |
# How do you write the equation of the parabola in vertex form given points (3,1) the vertex(5,9)?
Jul 23, 2017
$y = - 2 {\left(x - 5\right)}^{2} + 9$
#### Explanation:
Given -
$\left(3 , 1\right)$ [point on the parabola]
Vertex $\left(5 , 9\right)$
Formula for the parabola in vertex form
$y = a {\left(x - h\right)}^{2} + k$
Here $\left(h , k\right)$ coordinates of the vertex.
Let us substitute $\left(5 , 9\right)$ in the formula.
$y = a {\left(x - 5\right)}^{2} + 9$
To find the value of $a$, substitute $\left(3 , 1\right)$
$1 = a {\left(3 - 5\right)}^{2} + 9$
$1 = a 4 + 9$
Solve for $a$
$4 a + 9 = 1$
$4 a = 1 - 9 = - 8$
$a = - \frac{8}{4}$
$a = - 2$
The equation is -
$y = - 2 {\left(x - 5\right)}^{2} + 9$ |
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# Inverse Property of Addition in Decimal Equations
## Solve one - step equations with decimals by using subtraction.
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Practice Inverse Property of Addition in Decimal Equations
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Inverse Property of Addition in Decimal Equations
Have you ever ran a half marathon? Well, Kelly ran one and then ran a 5k too. Take a look.
Kelly ran a half marathon in 2.01 hours. She ran a 5k on the next day. Her total time for both races was 2.29.
What was Kelly's time for the 5k?
Can you write an equation with a variable that shows these values?
Then can you use the inverse property of addition to solve it?
Use this Concept to learn all about the inverse property of addition in decimal situations. Then you will be able to accomplish this task.
### Guidance
The word “ inverse ” is one that is used in mathematics.
In fact, as you get into higher levels of math, you will see the word “inverse” used more often.
What do we mean when we use the word inverse?
The word inverse means opposite. Sometimes numbers are opposites of each other, and sometimes operations are opposites of each other too. Subtraction is called the inverse , or opposite, of addition. If you’ve ever used addition to check subtraction calculations, you already have a certain idea of the relationship between addition and subtraction.
The inverse property of addition states that for every number $x, x + (-x) = 0$ . In other words, when you add $x$ to its opposite, $-x$ (also known as the additive inverse ), the result is zero.
Adding a negative number is the same as subtracting that number.
Let’s put whole numbers in place of $x$ to make the property clear: $3 + (-3) = 0$ .
The inverse property of addition may seem obvious, but it has important implications for our ability to solve variable equations which aren’t easily solved using mental math—such as variable equations involving decimals.
Remember back to our work with equations and expressions in an earlier Concept?
Think about an equation. An equation is a number sentence where one expression is equal to another expression. The inverse property of addition lets us add or subtract the same number to both sides of the equation without changing the solution to the equation. This technique is called an inverse operation and it lets us get the variable $x$ alone on one side of the equation so that we can find its value. Take a look at how it’s done.
$x + 5 & = 15\\x + 5 - 5 & = 15 - 5 \rightarrow \text{inverse operation means we subtract} \ 5 \ \text{from both sides}\\x & = 10$
Notice how, on the left side of the equation, $+ 5 - 5 = 0$ , and cancels itself out leaving the $x$ alone on the left side.
Our answer is that $x$ is equal to 10.
Now let’s look at one where the two expressions in the equation are decimals.
$x + 39.517 = 50.281$
First, we need to isolate the variable. To do this, we can use the inverse operation. This is an addition problem, so we use subtraction to get the variable by itself. We subtract 39.517 from both sides of the equation.
$x + 39.517 & = 50.281\\x + 39.517 - 39.517 & = 50.281 - 39.517 \rightarrow \text{inverse operations: subtract} \ 39.517 \ \text{from both sides}\\x & = 10.764$
Notice how we subtracted 39.517 from both sides of the equation. Remember, both expressions on either side of an equation must be equal at all times . Whatever operation you perform to one side of the equation, you must perform to the other side.
In this instance, on the left side of the equation, $+ 39.517 - 39.517 = 0$ , and cancels itself out leaving the $x$ alone on the left side. On the right side of the equation, 50.281 - 39.517 gives us the value of $x$ .
Our answer is that $x$ is equal to 10.764.
$x - 43.27 = 182.205$
First, we need to isolate the variable. To do this, we can use the inverse operation. Notice that this is a subtraction problem. To isolate the variable we are going to use the inverse of subtraction which is addition.
$x - 43.27 & = 182.205\\x - 43.27 + 43.27 & = 182.205 + 43.27 \rightarrow \text{inverse operations: add} \ 43.27 \ \text{to both sides}\\x & = 225.475$
Notice how we added 43.27 to both sides of the equation . Remember, both expressions on either side of an equation must be equal at all times . Whatever operation you perform to one side of the equation, you must perform to the other side. In this instance on the left side of the equation $-43.27 + 43.27 = 0$ , and cancels itself out leaving the $x$ alone on the left side. On the right side of the equation, 182.205 + 43.27 gives us the value of $x$ .
Our answer is that $x$ is equal to 225.475.
Use the inverse property of addition to solve each equation.
#### Example A
$x+5.678=12.765$
Solution: $7.087$
#### Example B
$x-4.32=19.87$
Solution: $24.19$
#### Example C
$x+123.578=469.333$
Solution: $345.755$
Here is the original problem once again. Use what you have learned to solve it.
Kelly ran a half marathon in 2.01 hours. She ran a 5k on the next day. Her total time for both races was 2.29.
What was Kelly's time for the 5k?
Can you write an equation with a variable that shows these values?
Then can you use the inverse property of addition to solve it?
First, let's write an equation using the given information. We don't know the time for the 5k, so this is our variable.
$x + 2.01 = 2.29$
Now we can use the inverse operation for addition to solve.
$x = 2.29 - 2.01$
$x = .28$
Kelly ran the 5k in 28 minutes.
### Vocabulary
Inverse
means opposite
means that when you add two opposite values together that the answer is 0.
using the opposite operation to solve for an unknown variable in an equation.
### Guided Practice
Here is one for you to try on your own.
At Saturday’s track meet, Liz ran 1.96 kilometers less than Sonya. Liz ran 1.258 kilometers. How many kilometers did Sonya run?
Anytime you see the key words “less than,” you know you’re going to write a subtraction equation.
In this problem, we are trying to find out the number of kilometers Sonya ran. We’ll give that value the variable $x$ . We know how many kilometers Liz ran, and we know the subtraction relationship between the Liz’s distance and Sonya’s distance, so we can write the equation $x - 1.96 = 1.258$ . Now we can solve for the value of $x$ using inverse operations.
$x - 1.96 & = 1.258\\x - 1.96 + 1.96 & = 1.258 + 1.96 \rightarrow \text{inverse operation, add} \ 1.96 \ \text{to both sides}\\x & = 3.218$
### Practice
Directions : Solve each equation for the missing variable using the inverse operation.
1. $x + 2.39 = 7.01$
2. $x + 5.64 = 17.22$
3. $x + 8.07 = 18.l2$
4. $x + 14.39 = 7.342$
5. $x + 21.3 = 87.12$
6. $x + 31.9 = 7.22$
7. $x + 18.77 = 97.12$
8. $x + 21.31 = 27.09$
9. $x + 18.11 = 87.22$
10. $818.703 = 614.208 + x$
11. $x + 55.27 = 100.95$
Directions: Use equations to solve each word problem. Each answer should have an equation and a value for the variable.
12. Jamal’s leek and potato soup calls for 2.45 kg more potatoes than leeks. Jamal uses 4.05 kg of potatoes. How many kilograms of leeks does he use? Write an equation and solve.
13. He distance from Waterville to Longford is 118.816 kilometers less than the distance from Transtown to Longford. The distance from Waterville to Longford is 67.729 kilometers. What is the distance from Transtown to Longford? Write an equation and solve.
14. Sabrina spent $25.62 at the book fair. When she left the fair, she had$6.87. How much did money did she take to the fair? Write an equation and solve.
15. Mr. Bodin has 11.09 liters of a cleaning solution, which is a combination of soap and water. If there are 2.75 liters of soap in the solution, how many liters of water are in the solution? Write an equation and solve.
### Vocabulary Language: English
The additive inverse or opposite of a number x is -1(x). A number and its additive inverse always sum to zero.
Inverse
Inverse
Inverse operations are operations that "undo" each other. Multiplication is the inverse operation of division. Addition is the inverse operation of subtraction.
The inverse property of addition states that the sum of any real number and its additive inverse (opposite) is zero. If $a$ is a real number, then $a+(-a)=0$. |
# Calculating Parts of a Circle
## Arc Lengths
It’s like working out the amount of crust on your slice of pizza!
To work out the length of an arc we need to remember that it is simply a fraction of the circle.
The fraction of the arc can be calculated by dividing the angle of the two radii between the arc by 360.
This formula can be used to work out the length of an arc:
= Angle/360 x circumference
= Angle/360 x π x diameter
Eg. If I have an arc that is enclosed by an angle of 20 degrees and my diameter is 4 cm the arc length
= 20/360 x π x 4
= 0.698 cm (to 3 dp)
## Area of sectors
To work out the area of a sector I can use a similar method. As I want a fraction of the area of the circle.
=Angle/360 x area of a circle
=Angle/360 x π x r x r
For example I want to work out the area of the pink sector of this circle that has a radius of 5 cm I simply use the formula:
= Angle/360 x π x r x r
= 300/360 x π x 5 x 5
= 65.45 cm2 (to 2 dp)
## Angles of sectors
We can also reverse the above formulas to work out the angles of sectors.
For example, if I have a sector with an area of 5 cm2 and a radius of 3 cm I can work out the angle of the sector.
Start by using the original formula and inputting the information that you know:
5 = Angle/360 x π x 3 x 3
5 = Angle/360 x π x 9
5/9π = Angle/360
5/9π x 360 = 64.29
Angle = 64.29 (to 2 dp) |
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# Matrix Multiplication
## Multiply rows by columns in different matrices
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Multiplying Two Matrices
Mr. Hwan writes the following matrices on the board and asks his students to multiply them.
[3210][41]
When they are done multiplying, Mr. Hwan asks, "What element is in the first row, second column of your answer?"
Wanda says that because the two matrices do not have the same dimensions they can't be multiplied. Therefore, there is no answer.
Xavier says that the product is a 2 x 1 matrix so there is no element in the first row, second column.
Zach says that the element requested is 8.
Who is correct?
### Guidance
To multiply matrices together we will be multiplying each element in each row of the first matrix by each element in each column in the second matrix. Each of these products will be added together to get the result for a particular row and column as shown below.
[acbd][egfh]=[ae+bgce+dgaf+bhcf+dh]
#### Example A
Multiply the matrices:
[2135][6401]
Solution:
By following the rule given above, we get:
[2(6)+3(4)1(6)+5(4)2(0)+3(1)1(0)+5(1)]=[12+126+200+30+5]=[02635]
#### Example B
Multiply the matrices:
174302[5182]
Solution:
In order to multiply these two matrices we need to extend the pattern given in the guidance to apply to a 3×2\begin{align*}3\times2\end{align*} matrix and a 2×2\begin{align*}2\times2\end{align*} matrix. It is important to note that matrices do not need to have the same dimensions in order to multiply them together. However, there are limitations that will be explored in the next topic. All matrix multiplication problems in this topic are possible.
Now, let’s multiply each of the rows in the first matrix by each of the columns in the second matrix to get:
174302[5182]=1(5)+3(1)7(5)+0(1)4(5)+2(1)1(8)+3(2)7(8)+0(2)4(8)+2(2)=5335+020+28+656+0324=8352225628
#### Example C
Multiply the matrices:
[327815]209
Solution:
Sometimes, not only are the matrices different dimensions, but the result has dimensions other than either of the original matrices as is the case in this example. Multiply the matrices row by column to get:
[327815]209=[3(2)+7(0)+1(9)2(2)+8(0)+5(9)]=[6+0+94+045]=[1541]
Intro Problem Revisit When the two matrixes are multiplied, the resulting matrix is:
[78]
The matrices can be multiplied, but there is no first row, second column in the resulting matrix. The element 8 is in the second row, first column. Therefore Xavier is right.
### Guided Practice
Multiply the matrices together.
1.
[4513][21]
2.
325[214]
3.
[52][1347]
1.
[4513][21]=[4(2)+1(1)5(2)+3(1)]=[8110+3]=[913]
2.
325[214]=3(2)2(2)5(2)3(1)2(1)5(1)3(4)2(4)5(4)=641032512820
3.
[52][1347]=[5(1)+2(3)5(4)+2(7)]=[5+620+14]=[1134]
### Explore More
Multiply the matrices together.
1. .
[2013][4235]
1. .
[27][1543]
1. .
351246[1432]
1. .
[8315][47]
1. .
148[236]
1. .
[51][923106]
1. .
[4513][26]
1. .
[247]315
1. .
155232411264321510
1. .
[124]351112267
1. .
[61112][21116]
1. .
419526738121
1. .
[213]458
1. .
[10][15]
1. .
2174212[53] |
# McGraw Hill My Math Grade 4 Chapter 10 Lesson 4 Answer Key Model Decimals and Fractions
All the solutions provided in McGraw Hill My Math Grade 4 Answer Key PDF Chapter 10 Lesson 4 Model Decimals and Fractions will give you a clear idea of the concepts.
## McGraw-Hill My Math Grade 4 Answer Key Chapter 10 Lesson 4 Model Decimals and Fractions
You can write a decimal as a fraction. Both a decimal and a fraction show part of a whole.
Draw It
Write 0.4 as a fraction with a denominator of 10.
1. Write the decimal in the place-value chart.
2. Shade the grid to represent four tenths.
Shade four of the ten equal parts.
3. Write a fraction.
Think of 0.4 as four tenths as $$\frac{4}{10}$$.
Check:
The number line shows that 0.4 is 4 out of 10 or $$\frac{4}{10}$$.
1)
2)
3)
Try It
Write 0.45 as a fraction with a denominator of 100.
Use a number line.
So, 0.45 can be written as .
0.45 can be written as 45 divided by 100.
Question 1.
Shade the models below to show how 0.7 and 0.70 are equivalent.
0.7 is indicated with an yellow color, where 7 parts are filled out of 10 total parts.
So, 7 divided by 10 gives 0.7 in decimal.
0.70 is indicated with an yellow color, where 70 parts are filled out of 100 total parts.
So, 70 divided by 100 gives 0.70 in decimal.
Question 2.
Write two fractions that represent the models in Exercise 1.
Answer: similar to Question1, 0.2 and 0.20 are equivalent.
0.2 is indicated with an pink color, where 2 parts are filled out of 10 total parts.
So, 2 divided by 10 gives 0.2 in decimal.
0.20 is indicated with an pink color, where 20 parts are filled out of 100 total parts.
So, 20 divided by 100 gives 0.20 in decimal.
Question 3.
Mathematical PRACTICE Draw a Conclusion Explain how you could write $$\frac{9}{10}$$ as a fraction with a denominator of 100 without using models.
Answer: $$\frac{9}{10}$$ is equivalent to $$\frac{90}{100}$$ with a denominator of 100.
Since 1 tenth is equal to 10 hundredths, 9 tenths is equal to 90 hundredths. You could also find the equivalent fraction by multiplying the numerator and denominator by 10.
Question 4.
Explain how you could write 0.78 as a fraction with a denominator of 100 without using models.
Answer: $$\frac{78}{100}$$ is the fractional form of 0.78 decimal.
To convert decimal into the fraction by multiplying the numerator and denominator by 100.
Practice It
Write each decimal as a fraction with a denominator of 10. Shade the grid.
Question 5.
0.2 = ______________
Answer: $$\frac{2}{10}$$ is the fractional form of 0.2 decimal.
To convert decimal into the fraction by multiplying the numerator and denominator by 10.
The yellow color shaded parts in the above represents the given number.
Question 6.
0.7 = ______________
Answer: $$\frac{7}{10}$$ is the fractional form of 0.7 decimal.
To convert decimal into the fraction by multiplying the numerator and denominator by 10.
The yellow color shaded parts in the above represents the given number.
Question 7.
0.3 = ______________
Answer: $$\frac{3}{10}$$ is the fractional form of 0.3 decimal.
To convert decimal into the fraction by multiplying the numerator and denominator by 10.
The yellow color shaded parts in the above represents the given number.
Question 8.
0.5 = ______________
Answer: $$\frac{5}{10}$$ is the fractional form of 0.5 decimal.
To convert decimal into the fraction by multiplying the numerator and denominator by 10.
The yellow color shaded parts in the above represents the given number.
Question 9.
0.6 = ______________
Answer: $$\frac{6}{10}$$ is the fractional form of 0.6 decimal.
To convert decimal into the fraction by multiplying the numerator and denominator by 10.
The yellow color shaded parts in the above represents the given number.
Question 10.
0.1 = ______________
Answer: $$\frac{1}{10}$$ is the fractional form of 0.1 decimal.
To convert decimal into the fraction by multiplying the numerator and denominator by 10.
The yellow color shaded parts in the above represents the given number.
Write each decimal as a fraction with a denominator of 100. Shade the grid.
Question 11.
0.39 = ______________
Answer: $$\frac{39}{100}$$ is the fractional form of 0.39 decimal.
To convert decimal into the fraction by multiplying the numerator and denominator by 100.
The pink color shaded parts in the above represents the given number.
Question 12.
0.71 = ______________
Answer: $$\frac{71}{100}$$ is the fractional form of 0.71 decimal.
To convert decimal into the fraction by multiplying the numerator and denominator by 100.
The pink color shaded parts in the above represents the given number.
Question 13.
0.12 = ______________
Answer: $$\frac{12}{100}$$ is the fractional form of 0.12 decimal.
To convert decimal into the fraction by multiplying the numerator and denominator by 100.
The pink color shaded parts in the above represents the given number.
Question 14.
0.09 = ______________
Answer: $$\frac{9}{100}$$ is the fractional form of 0.09 decimal.
To convert decimal into the fraction by multiplying the numerator and denominator by 100.
The pink color shaded parts in the above represents the given number.
Question 15.
0.23 = ______________
Answer: $$\frac{23}{100}$$ is the fractional form of 0.23 decimal.
To convert decimal into the fraction by multiplying the numerator and denominator by 100.
The pink color shaded parts in the above represents the given number.
Question 16.
0.02 = ______________
Answer: $$\frac{2}{100}$$ is the fractional form of 0.02 decimal.
To convert decimal into the fraction by multiplying the numerator and denominator by 100.
The pink color shaded parts in the above represents the given number.
Apply It
Question 17.
There are 10 cars on the race track. If 0.3 of the cars are red, what fraction of the cars are not red? Write as a fraction with a denominator of 10.
Answer: 0.3 which is 3 out of 10 cars are in red color. So, remaining cars which are not in red color are 7. Therefore, if you divide 7 by 10, then the fraction is $$\frac{7}{10}$$
Question 18.
On Monday, it snowed 0.6 of an inch. Write 0.6 as a fraction with a denominator of 10.
Answer: $$\frac{6}{10}$$ is the fractional form of 0.6 with the denominator 10.
Question 19.
There are 100 students at the gymnastics competition. 57 hundredths, or 0.57 of the students are nine years old. Write 0.57 as a fraction with a denominator of 100.
Answer: $$\frac{57}{100}$$ is the fractional form of 0.57 with the denominator 100.
Question 20.
The fourth grade class is donating 100 clothing items to charity. 63 hundredths of the items are sweaters and 0.20 of the items are jeans. What part of the clothing items are neither sweaters nor jeans’ Write the number as a fraction with a denominator of 100.
Total number of sweaters out of 100= 63
Total number of jeans out of 100= 20
The items which are neither sweaters nor jeans are 100-63-20 = 17 out of 100
If you divide 17 by 100, $$\frac{17}{100}$$ is the fractional part.
Question 21.
Mathematical PRACTICE Explain to a Friend Explain to a classmate why 0.8 and 0.80 name the same number.
Answer: 0.8 is equivalent to 0.80.
Since 1 tenth is equal to 10 hundredths, 8 tenths is equal to 80 hundredths. You could also find the equivalent fraction by multiplying the numerator and denominator by 10.
Question 22.
How can I use models to relate fractions and decimals?
Answer: Both decimals and fractions are represented using models. The coloured parts out of the total number of parts are used to illustrate the example of both fractions and decimals.
For instance, Suppose there are 10 components, 4 of which are blue. Therefore, $$\frac{4}{10}$$ is in the fractional component and 0.4 is in the decimal part when you divide 4 by 10.
### McGraw Hill My Math Grade 4 Chapter 10 Lesson 4 My Homework Answer Key
Practice
Write each decimal as a fraction with a denominator of 10. Shade the grid.
Question 1.
0.1 = ________________
Answer: $$\frac{1}{10}$$ is the fractional form of 0.1 decimal.
To convert decimal into the fraction by multiplying the numerator and denominator by 10.
The yellow color shaded parts in the above represents the given number.
Question 2.
0.3 = ________________
Answer: $$\frac{3}{10}$$ is the fractional form of 0.3 decimal.
To convert decimal into the fraction by multiplying the numerator and denominator by 10.
The yellow color shaded parts in the above represents the given number.
Question 3.
0.9 = ________________
Answer: $$\frac{9}{10}$$ is the fractional form of 0.9 decimal.
To convert decimal into the fraction by multiplying the numerator and denominator by 10.
The yellow color shaded parts in the above represents the given number.
Write each decimal as a fraction with a denominator of 100. Shade the grid.
Question 4.
0.17 = ________________
Answer: $$\frac{17}{100}$$ is the fractional form of 0.17 decimal.
To convert decimal into the fraction by multiplying the numerator and denominator by 100.
The pink color shaded parts in the above represents the given number.
Question 5.
0.24 = ________________
Answer: $$\frac{24}{100}$$ is the fractional form of 0.24 decimal.
To convert decimal into the fraction by multiplying the numerator and denominator by 100.
The pink color shaded parts in the above represents the given number.
Question 6.
0.88 = ________________
Answer: $$\frac{88}{100}$$ is the fractional form of 0.88 decimal.
To convert decimal into the fraction by multiplying the numerator and denominator by 100.
The pink color shaded parts in the above represents the given number.
Problem Solving
Mathematical PRACTICE Model Math For Exercises 7-9, refer to the table that shows the part of pets that Josiah’s family has for each type of pet.
Question 7.
Write the part of pets that are cats as a fraction with a denominator of 10.
Answer: The total number of cats are 0.2
0.2 in decimal is converted into fraction by dividing 2 by 10, which is $$\frac{2}{10}$$
Question 8.
Write the part of pets that are fish as a fraction with a denominator of 10.
Answer: The total number of cats are 0.4
0.4 in decimal is converted into fraction by dividing 4 by 10, which is $$\frac{4}{10}$$
Question 9.
Write the part of pets that are fish as a fraction with a denominator of 100.
Answer: The total number of cats are 0.4
0.4 in decimal is converted into fraction with a denominator of 100 by dividing 4 by 100, which is $$\frac{4}{100}$$
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# Difference between revisions of "2006 AMC 10A Problems/Problem 16"
## Problem
A circle of radius 1 is tangent to a circle of radius 2. The sides of $\triangle ABC$ are tangent to the circles as shown, and the sides $\overline{AB}$ and $\overline{AC}$ are congruent. What is the area of $\triangle ABC$? $[asy] size(200); pathpen = linewidth(0.7); pointpen = black; real t=2^0.5; D((0,0)--(4*t,0)--(2*t,8)--cycle); D(CR((2*t,2),2)); D(CR((2*t,5),1)); D('B', (0,0),SW); D('C',(4*t,0), SE); D('A', (2*t, 8), N); D((2*t,2)--(2*t,4)); D((2*t,5)--(2*t,6)); MP('2', (2*t,3), W); MP('1',(2*t, 5.5), W);[/asy]$
$\mathrm{(A) \ } \frac{35}{2}\qquad\mathrm{(B) \ } 15\sqrt{2}\qquad\mathrm{(C) \ } \frac{64}{3}\qquad\mathrm{(D) \ } 16\sqrt{2}\qquad\mathrm{(E) \ } 24\qquad$
## Solution
Let the centers of the smaller and larger circles be $O_1$ and $O_2$ , respectively. Let their tangent points to $\triangle ABC$ be $D$ and $E$, respectively. We can then draw the following diagram:
$[asy] size(200); pathpen = linewidth(0.7); pointpen = black; pointfontpen = fontsize(10); real t=2^0.5; D((0,0)--(4*t,0)--(2*t,8)--cycle); D(CR(D((2*t,2)),2)); D(CR(D((2*t,5)),1)); D('B', (0,0),SW); D('C',(4*t,0), SE); D('A', (2*t, 8), N); pair A = foot((2*t,2),(2*t,8),(4*t,0)), B = foot((2*t,5),(2*t,8),(4*t,0)); D((2*t,0)--(2*t,8)); D((2*t,2)--D(A)); D((2*t,5)--D(B)); D(rightanglemark((2*t,2),A,(4*t,0))); D(rightanglemark((2*t,5),B,(4*t,0))); MP('2', (2*t,3), W); MP('1',(2*t, 5.5), W); MP("F",(2*t,0)); MP("O_1",(2*t,5),W); MP("O_2",(2*t,2),W); MP("D",B,NE); MP("E",A,NE); [/asy]$
We see that $\triangle ADO_1 \sim \triangle AEO_2 \sim \triangle AFC$. Using the first pair of similar triangles, we write the proportion:
$\frac{AO_1}{AO_2} = \frac{DO_1}{EO_2} \Longrightarrow \frac{AO_1}{AO_1 + 3} = \frac{1}{2} \Longrightarrow AO_1 = 3$
By the Pythagorean Theorem, we have $AD = \sqrt{3^2-1^2} = \sqrt{8}$.
Now using $\triangle ADO_1 \sim \triangle AFC$,
$\frac{AD}{AF} = \frac{DO_1}{FC} \Longrightarrow \frac{2\sqrt{2}}{8} = \frac{1}{FC} \Longrightarrow FC = 2\sqrt{2}$
Hence, the area of the triangle is $\frac{1}{2}\cdot AF \cdot BC = \frac{1}{2}\cdot AF \cdot (2\cdot CF) = AF \cdot CF = 8\left(2\sqrt{2}\right) = \boxed{16\sqrt{2}\ \mathrm{(D)}}$ |
# Spherical trigonometry
Missing image
RechtwKugeldreieck.png
Right spherical triangle
Spherical trigonometry is a part of spherical geometry that deals with polygons (especially triangles) on the sphere and explains how to find relations between the involved angles. This is of great importance for calculations in astronomy and navigation.
On the surface of a sphere, the closest analogue to straight lines are great circles, i.e. circles whose center coincide with the center of the sphere. (For examples, meridians and the equator are great circles on the Earth.) As lines on a plane, great circles on a sphere are the closest connection of two points (if you constrain yourself to lines on the sphere). (cf. geodesic)
An area on the sphere which is bounded by arcs of great circles is called a spherical polygon. Note that, unlike the case on a plane, spherical "biangles" (two-sided analogs to triangle) are possible (think about peeling an orange).
The sides of these polygons are most conveniently specified not by their length but by the angle under which its endpoints appear when looked at from the sphere's center. Note that this arc angle, measured in radians, and multiplied by the sphere's radius, is the arc length.
Hence, a spherical triangle is specified as usual by its corner angles and its sides, but the sides are not given by their length, but by their arc angle.
Remarkably, the sum of the corner angle is not 180°, as in a planar triangle, but always larger. This surplus is called the spherical excess E: E = α + β + γ − 180°. It allows calculation of the surface area A surrounded by the triangle, which is simply given by A = R2 · E. Here, R is the radius of the sphere; if R = 1, then A = E. In other words: E is the solid angle, as measured in steradians, spanned up by the triangle. This formula is an application of the Gauss-Bonnet theorem.
To solve a geometric problem on the sphere, one dissects the relevant figure into right spherical triangles (i.e.: one of the triangle's corner angles is 90°) because one can then use Neper's pentagon:
Missing image
Neper's_Circle.png
Neper's Circle shows the relations of parts of a right spherical triangle
Neper's pentagon (also known as Neper's circle) is a mnemonic aid aid to easily find all relations between the angles in a right spherical triangle:
Write the six angles of the triangle in the form of a circle, sticking to the order as they appear in the triangle (i.e.: start with a corner angle, write the arc angle of an attached side next to it, proceed with the next corner angle, etc. and close the circle). Then cross out the 90° corner angle and replace the arc angles adjacent to it by their complement to 90° (i.e. replace, say, a by 90° - a). The five numbers that you now have on your paper form Neper's Pentagon (or Neper's Circle). For them, it holds that the cosine of each angle is equal to
• the product of the cotangents of the angles written next to it, and is also equal to
• the product of the sines of the two angles written opposed to it.
See also the Haversine formula, which relates the lengths of sides and angles in spherical triangles in a numerically stable form for navigation.
• Art and Cultures
• Countries of the World (http://www.academickids.com/encyclopedia/index.php/Countries)
• Space and Astronomy |
# 22 Surface Area of Common Solids
You may use a calculator throughout this module.
We will now turn our attention from two-dimensional figures to three-dimensional figures, which we often call solids, even if they are hollow inside.
In this module, we will look the surface areas of some common solids. (We will look at volume in a later module.) Surface area is what it sounds like: it’s the sum of the areas of all of the outer surfaces of the solid. When you are struggling to wrap a present because your sheet of wrapping paper isn’t quite big enough, you are dealing with surface area.
There are two different kinds of surface area that are important: the lateral surface area (LSA) and total surface area (TSA).
To visualize the difference between LSA and TSA, consider a can of food. The lateral surface area would be used to measure the size of the paper label around the can. The total surface area would be used to measure the amount of sheet metal needed to make the can. In other words, the total surface area includes the top and bottom, whereas the lateral surface area does not.
# Surface Area: Rectangular Solids
A rectangular solid looks like a rectangular box. It has three pairs of equally sized rectangles on the front and back, on the left and right, and on the top and bottom.
A cube is a special rectangular solid with equally-sized squares for all six faces.
The lateral surface area is the combined total area of the four vertical faces of the solid, but not the top and bottom. If you were painting the four walls of a room, you would be thinking about the lateral surface area.
The total surface area is the combined total area of all six faces of the solid. If you were painting the four walls, the floor, and the ceiling of a room, you would be thinking about the total surface area.
For a rectangular solid with length , width , and height
For a cube with side length
Note: These dimensions are sometimes called base, depth, and height.
Exercises
1. Find the lateral surface area of this rectangular solid.
2. Find the total surface area of this rectangular solid.
# Surface Area: Cylinders
As mentioned earlier in this module, the lateral surface area of a soup can is the paper label, which is a rectangle. Therefore, the lateral surface area of a cylinder is a rectangle; its width is equal to the circumference of the circle, , and its height is the height of the cylinder.
Since a cylinder has equal-sized circles at the top and bottom, its total surface area is equal to the lateral surface area plus twice the area of one of the circles.
For a cylinder with radius and height
Be aware that if you are given the diameter of the cylinder, you will need to cut it in half before using these formulas.
Exercises
A cylinder has a diameter of and a height of .
1. Find the lateral surface area.
2. Find the total surface area.
# Surface Area: Spheres
The final solid of this module is the sphere, which can be thought of as a circle in three dimensions: every point on the surface of a sphere is the same distance from the center. Because of this, a sphere has only one important measurement: its radius. Of course, its diameter could be important also, but the idea is that a sphere doesn’t have different dimensions such as length, width, and height. A sphere has the same radius (or diameter) in every direction.
We would need to use calculus to derive the formula for the surface area of a sphere, so we’ll just assume it’s true and get on with the business at hand. Notice that, because a sphere doesn’t have top or bottom faces, we don’t need to worry about finding the lateral surface area. The only surface area is the total surface area.
For a sphere with radius or diameter
or
Coincidentally, the surface area of a sphere is times the area of the cross-sectional circle at the sphere’s widest part. You may find it interesting to try to visualize this, or head to the kitchen for a demonstration: if you cut an orange into four quarters, the peel on one of those quarter oranges has the same area as the circle formed by the first cut.
Exercises
1. Find the surface area of this sphere.
1. Find the surface area of this sphere. |
# Age Problems II – Problems on Age for SSC Exams
In this post, we will discuss Problems on Age that are frequently asked in SSC Exams and solve them in minimum time.
Problems on age are the word problems that are solved framing equations. Making tables, framing equations are all conventional methods that take a lot of time in SSC Exams. Since SSC Exams have a time constraint; one should know the smart approach towards these problems.
In this post, we will discuss problems on age that are frequently asked in SSC Exams. These problems on age are solved using the smart method which will help you save time. Before moving further let's have a quick review of the basics of Age Problems.
Problems on Age: Example
Question: Ranjan’s and Anurag’s ages are in the ratio 4:5. Four years hence, their age ratio will become 5:6. What is Anurag’s present age?
1) 20 years 2) 16 years 3) 24 years 4) Data Insufficient 5) None of these
Solution:
Regular approach towards Problems on Age
Step 1:
Let us assume Ranjan’s present age to be as R.
Let us assume Anurag’s present age to be as A.
Step 2:
As mentioned in the question, The ratio of Ranjan’s age to Anurag’s age is
R: A = 4: 5 => R/A = 4/5 => R = 4x ; A = 5x (i)
Step 3:
As mentioned in the question, after 4 years
(R + 4)/ (A + 4) = 5/6 (ii)
Step 4:
By substituting equation (i) in equation (ii), we get
(4x + 4) / (5x + 4) = 5/6
Step 5:
By cross multiplication, we get
6(4x + 4) = 5(5 + 4)
24x + 24 = 25x + 20
x = 4
Step 6:
By substituting the x value in equation (i) we get,
Anurag’s present age to be 5 x 4 = 20
Therefore, the correct answer is option 1; 20 years.
Smart Approach towards Problems on Age
Step 1:
Eliminate all the options that are not multiples of 5 because the options are on the basis of Anurag's age
Option 2 and 3 are not multiples of 5 so we eliminate both the options.
Option 4; Data insufficient is also eliminated as the data is sufficient to solve the problem.
Therefore the only options left are- options 1 and 5.
Step 2:
Option 1; 20 years – To verify whether this option is correct or not, we should know that the option should be a multiple of 5 and after 4 years his age should be a multiple of 6 [20 + 4 = 24]. Hence, this option satisfies both the conditions.
Therefore, the correct answer is option 1; 20 years.
Problems on Age: Example
Question: The ages of Samina and Suhana are in the ratio of 7:3. After 6 years, the ratio of their ages will be 5:3. What is the difference in their ages?
1) 6 years 2)8 years 3)10 years 4) 12 years 5) None of these
Solution:
Regular approach towards Problems on Age
Step 1:
Let us assume the present age of Samina be Sa
Let us assume the present age of Suhana be Su.
Step 2:
As mentioned in the question, the ratio of the ages of Samina and Suhana is
Sa: Su = 7: 3 => R/A = 7/3 =>Sa = 7x Su = 3x (i)
Step 3:
As mentioned in the question after 6 years, their ratio will be 5:3.
This can be denoted as
(7x + 6)/( 3x + 6 ) = 5/3 (ii)
By cross multiplication we get,
21x + 18 = 15x + 30
6x = 12
x = 2
Step 4:
By substituting the value of x in the equation (ii) we get,
7x – 3x = (7 x 2) – (3 x 2) = 14 – 6 = 8
Therefore, the correct option is option 2; 8 years.
Smart Approach towards Problems on Age
Step 1:
Let us assume the present age of Samina be Sa
Let us assume the present age of Suhana be Su
Step 2:
As mentioned in the question, the ratio of the ages of Samina and Suhana is
Sa: Su = 7: 3 => R/A = 7/3 => Sa = 7x Su = 3x
Then the difference between their ages will be
7x - 3x = 4x
Step 3:
By using the elimination method, eliminate options which don't satisfy the condition.
Option 1: years
If 6 years is the difference between Samina and Suhana then,
4x = 6
x = 1.5
Since in age problems the ages are considered to be an integer, we eliminate this option.
Option 2: 8 years
If 8 years is the difference between Samina and Suhana then,
4x = 8
x = 2
Then Samina age will be 7 x 2 = 14 and Suhana age will be 3 x 2 = 6,
As 14 - 6 = 8 years
Therefore, this option satisfies our condition.
Hence the answer is option 2: 8 years.
Problems on Age: Example
Question: After 5 years, the average age of a daughter and her mother will become 29.5 years. If today, the ratio of their ages is 2:5, what is the present age of daughter?
1) Cannot be determined 2) 25 years 3) 21 years 4)14 years 5) None of these
Solution:
Regular approach towards Problems on Age
Step 1:
Let us assume the present age of daughter to be as ‘D.’
Let us assume the present age of Mother to be as ‘M.’
Let us assume the present average to be as ‘A.’
The ratio of D: M = 2: 5
Step 2:
After 5 years,
The daughters age = D + 5
The mothers age = M + 5
The average of their age = A + 5 = 29.5 (As the numbers increase by 5, the average also increases by 5.
Step 3:
To find the present average we need to
(D + M) / 2 = 29.5 – 5
(D + M) / 2 = 24.5
D + M = 24.5 x 2
D + M = 49
Step 4:
To find the present age of the daughter we need to multiply 2/7 (ratio of the daughter) with the present average.
D = 2/ 7 x 49
D = 14 years
Therefore, the correct answer is option 4: 14 years
Smart approach towards Problems on Age
This question can be easily solved using the smart method.
In the question, the present ratio between the ages of the daughter and mother is 2: 5.
The daughter's present age should be a multiple of 2.
Option 1: Can not be determined - The question contains all the necessary data to find the ages. Hence, this option is eliminated
Option 2: 25 years - Because 25 is not a multiple of 2, this option is eliminated.
Option 3: 21 years - Because 21 is not a multiple of 2, this option is eliminated.
Option 4: 14 years - 14 is a multiple of 2, therefore this option is taken into consideration.
By taking the second condition this option can be verified.
Therefore the correct answer is option 4: 14 years.
Watch our expert faculty explain the smart method.
Practice Problems on Age for SSC Exams
Question: The ratio of ages of a couple is 4:3. After 4 years, this ratio will be 9:7. If at the time of marriage, the ratio was 5:3, then how many years ago were they married?
1) 10 years 2)12 years 3) 7years 4) 9 years 5) None of these
Question: The ratio between the present ages of Rajesh and Uma is 5:3 respectively. The ratio between Rajesh’s age 4 years ago and Uma’s age 4 years hence is 1:1. What is the ratio between Rajesh’s age 4 years hence and Uma’s age 4 years ago?
1) 2:1 2) 3:1 3) 5:2 4) 4:3 5) None of these |
# Question #80df0
May 23, 2017
$\left(1 , 3\right)$
#### Explanation:
$\textcolor{red}{- 2 x} + 5 y = 13 \to \left(1\right)$
$\textcolor{red}{16 x} + 3 y = 25 \to \left(2\right)$
$\textcolor{b l u e}{\text{solving by elimination}}$ means attempting to eliminate one of the variables x or y from an equation leaving the equation with only one variable which we can solve.
$\text{note that multiplying " color(red)(-2x)" in (1) by 8 gives}$
$\textcolor{red}{- 16 x} \text{ and adding it to " color(red)(16x)" in (2) will }$
$\textcolor{b l u e}{\text{eliminate"" the x-term}}$
$\text{multiply ALL terms in " (1)" by 8}$
$\Rightarrow \textcolor{red}{- 16 x} + 40 y = 104 \to \left(3\right)$
$\text{adding equations " (2)" and " (3)" term by term gives}$
$\left(16 x - 16 x\right) + \left(3 y + 40 y\right) = \left(25 + 104\right)$
$\Rightarrow 43 y = 129 \leftarrow \textcolor{red}{\text{ variable x is eliminated}}$
$\text{divide both sides by 43}$
$\frac{\cancel{43} y}{\cancel{43}} = \frac{129}{43}$
$\Rightarrow y = 3$
$\text{substitute this value in either " (1)" or } \left(2\right)$
$16 x + 9 = 25 \leftarrow \text{ substituting in } \left(2\right)$
$\text{subtract 9 from both sides}$
$16 x \cancel{+ 9} \cancel{- 9} = 25 - 9$
$\Rightarrow 16 x = 16 \Rightarrow x = 1$
$\left(1 , 3\right) \text{ is the point of intersection of the 2 linear equations}$ graph{(y-2/5x-13/5)(y+16/3x-25/3)=0 [-10, 10, -5, 5]} |
Math
# Are all Integers and Fractions Rational Numbers? [Explained]
Rational numbers can be written in the form p/q, where p and q are integers and q ≠ 0. Integers are numbers that can be positive, negative, or zero but cannot be a fraction. This implies that all the natural numbers, 0 (zero), and the negative of all the natural numbers are sets of integers. Integers can be denoted by Z and can be expressed as Z = {-5,-4,-3, -2, -1, 0,1,2,3,4,5…}.
Rational numbers can be written as p/q, where p and q are integers and q ≠ 0. If the signs of the numerator and denominator are positive or negative, the rational number is known as a Positive Rational Number (For example, -2/-1, -4/-2, 6/3). If the signs of the numerator and denominator are opposite, the rational number is a Negative Rational Number (For example, -2/1, 9/-3, etc.)
Thus, the set of rational numbers contains all integers, i.e., all integers are rational numbers. For example, -2 and 4 can be written in the form p/q, 2/1 =2, and 4/1 = 4, so they are rational numbers.
## Explanation
A number can be written as p/q, q ≠ 0 where p and q are whole numbers known as fractions. A fraction is part of a whole number. Kindly note that all counting numbers including 0 (zero) form the set of whole numbers. Its set is represented by R. Therefore, R = {0, 1, 2, 3, 4,5, 6,7,8,9,10…} is the set of the whole numbers.
A fraction has two parts, namely, the numerator and denominator. The number on the top is called the numerator, and the number on the bottom is called the denominator. For example 2/3, 7/8, 9/4, etc. There are types of fractions: proper, improper, mixed, and unit.
A fraction is always positive, whereas a rational number can be positive or negative. Thus, the rational numbers contain all fractions, i.e., all fractions are integers.
In conclusion, integers and fractions are a subset of rational numbers, i.e., fractions and integers are rational numbers.
### Bolarinwa Olajire
A tutor with a demonstrated history of working in the education industry. Skilled in analytical skills. Strong education professional with a M. SC focused in condensed matter. You can follow me on Twitter by clicking on the icon below to ask questions. |
## Section5.6Circular Motion Bootcamp
### Subsection5.6.6Miscellaneous
Figure 5.6.30 shows a pebble P stuck at the outside of a tire. As the tire rolls, the pebble's position in space changes. Suppose tire does not slide but rolls smoothly. The tire has radius $R$ and is rolling at a constant speed $v_0 \text{.}$ At $t=0\text{,}$ the pebble was at the origin.
Find the expressions for (a) position, (b) velocity, and (c) acceleration vectors (i.e., magnitudes and directions) at an arbitraru instant $t$ with respect to the coordinate system given in the figure.
Hint
(a) Introduce another coordinate system with origin at the center of the tire at all times and $x$ and $y$ axes parallel to the one given. Find position in that coordinate system first. Then translate that to the given coordinate system. (b) and (c): use derivatives.
(a) $\vec r = (-R\sin\theta + v_0 t)\; \hat i + (-R\cos\theta)\; \hat j\text{,}$ (b) $((1 -\cos\theta)\;v_0 , (-\sin\theta)v_0)\text{,}$ (c) $\text{.}$
Solution 1 (a)
(a) Let us introduce a coordinate system as shown in Figure 5.6.31. For convenience, I will define $\theta$ that measures angle from negative $y$ axis as shown.
Let $(x,y)$ and $(x',y')$ be the coordinates of P at this arbitrary instant $t$ with respect to the two coordinate systems. It is immediately clear from the figure that
\begin{align*} \amp y = y',\ \ x = x' + v_0 t.\\ \amp x' = -R\sin\theta,\ \ y'=-R\cos\theta. \end{align*}
Therefore, position $\vec r$ in original coordinate system will be
\begin{equation*} \vec r = (-R\sin\theta + v_0 t)\; \hat i + (-R\cos\theta)\; \hat j. \end{equation*}
It magnitude is
\begin{align*} r \amp = \sqrt{ x^2 + y^2 }\\ \amp = \sqrt{(-R\sin\theta + v_0 t)^2 + (-R\cos\theta)^2 }\\ \amp = \sqrt{R^2 - (2Rv_0\sin\theta) t + v_0^2t^2 }. \end{align*}
We can indicate direction in space by angle with respect to $x$ axis by using arctangent.
\begin{equation*} \theta_x = \tan^{-1}(y/x) =\tan^{-1}\left( \frac{-R\cos\theta}{-R\sin\theta + v_0 t} \right). \end{equation*}
Solution 2 (b)
(b) Now, that we know the components of position, we can obtain components of velocity by taking derivatives. Thus,
\begin{align*} \amp v_x = \frac{dx}{dt} = (-R\cos\theta)\frac{d\theta}{dt} + v_0 \\ \amp v_y = \frac{dy}{dt} = (-R\sin\theta)\frac{d\theta}{dt} \end{align*}
Since the tire is not slipping, the rate at which angle $\theta$ changes is related to the speed with which center of the tire is moving. The tire has rotated by $R\Delta\theta\text{,}$ its center has moved $v_0\Delta t\text{.}$ Equating them we get
\begin{equation*} R\Delta\theta = v_0\Delta t. \end{equation*}
This gives
\begin{equation*} \frac{d\theta}{dt} = \frac{v_0}{R}. \end{equation*}
Using this, we rewrite the components of velocity as
\begin{align*} \amp v_x = (1 -\cos\theta)\;v_0 \\ \amp v_y = (-\sin\theta)v_0. \end{align*}
Therefore, magnitude of velocity, i.e., speed is
\begin{equation*} v = \sqrt{v_x^2 + v_y^2} = v_0\sqrt{ 2 - 2 \cos\theta} = 2 v_0 \left| \sin(\theta/2) \right|. \end{equation*}
Note: this tells us that the speed of pebble when touching the ground is zero (since $\theta=0$) and when at the very top, it is $(2\;v_0)$ (since $\theta=\pi$). That is, speed of the pebble in $Oxy$ coordinate system changes as it goes around.
The direction of velocity can again be given with respect to the $x$ axis by using arctangent.
\begin{equation*} \theta_x = \tan^{-1}(v_y/v_x) = \tan^{-1}\left( \frac{-\sin\theta}{1 -\cos\theta} \right). \end{equation*}
You can further simplify this expression by using double angle formulas. Try that. You should be able to express this without arctangent operation.
Solution 3 (b)
(c) Now, that we know the components of velocity, we can obtain components of acceleration by taking derivatives. Thus,
\begin{align*} \amp a_x = \frac{dv_x}{dt} = (-v_0 \sin\theta)\frac{d\theta}{dt} \\ \amp a_y = \frac{dv_y}{dt} = (-v_0\cos\theta)\frac{d\theta}{dt} \end{align*}
Replacing $d\theta/dt$ by $v_0/R$ as found in (b), we get
\begin{equation*} a_x = -\frac{v_0^2}{R}\;\sin\theta;\ \ a_y = -\frac{v_0^2}{R}\;\cos\theta. \end{equation*}
Therefore, magnitude of acceleration is
\begin{equation*} a = \sqrt{a_x^2 + a_y^2} = \frac{v_0^2}{R}. \end{equation*}
The direction with respect to $x$ axis will again be obtained by using arctangent.
\begin{equation*} \theta_x = \tan^{-1}(a_y/a_x) = \tan^{-1}(\cot(\theta)). \end{equation*}
Of course, we can write $cot(\theta)$ as $\tan(\pi/2 - \theta)\text{.}$ Hence
\begin{equation*} \theta_x = \frac{\pi}{2} -\theta. \end{equation*}
This is actually direction towards the moving center of the tire. Take values of $\theta=0, \pi/2, \pi, 3\pi/2, 2\pi$ and draw them on the corresponding $Oxy$ plane. You will see that the direction is always towards the center of the tire as shown in Figure 5.6.32. |
### #ezw_tco-2 .ez-toc-title{ font-size: 120%; ; ; } #ezw_tco-2 .ez-toc-widget-container ul.ez-toc-list li.active{ background-color: #ededed; } chapter outline
‘Quad’ means square, and the variable is squared (order 2). Thus, a quadratic equation is an algebraic equation of the second degree written in the form:
Labels
ax2 + bx + c, here ‘a’ and ‘b’ are the coefficients, ‘x’ is the variable, ‘c’ is a constant
ax2 + bx + c, here ‘a’ and ‘b’ are the coefficients, ‘x’ is the variable, ‘c’ is a constant
This is the standard form of the quadratic equation. For an equation to be quadratic, the coefficient of x2 will be a non-zero term (a ≠0)
Some examples of quadratic equations are:
x2 + 2x – 15 = 0, here a = 1, b = 2, and c =-15
x2 – 49x = 0, here a = 1, b = -49, and c = 0
Sometimes the quadratic equations are outside the standard form and are disguised. In such cases, they were arranged and brought into the standard form. Some examples of such instances are shown below:
2x2 -36 = x
Rearranging this equation, we get
2x2 – x – 36 = 0, this equation is now in the standard form
Similarly, for the equation
(a – 4)2 – 9 = 0
=> a2 – 8a + 16 – 9 = 0
=> a2 – 8a + 7 = 0
## How to Solve Quadratic Equations
The solutions to the quadratic equations are its two roots, also called zeros. The simplest way to find the two roots is by using the quadratic formula:
x = ${x=\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}}$
• ${x=\dfrac{-b+\sqrt{b^{2}-4ac}}{2a} }$ |
# SAT Math : Plane Geometry
## Example Questions
### Example Question #1 : How To Find The Area Of A Right Triangle
Figure NOT drawn to scale.
is a right triangle with altitude . What percent of is shaded in?
Explanation:
The altitude of a right triangle from the vertex of its right angle - which, here, is - divides the triangle into two triangles similar to each other and to the large triangle. From the Pythagorean Theorem, the hypotenuse of has length
.
The similarity ratio of to is the ratio of the lengths of the hypotenuses:
The ratio of the areas of two similar triangles is the square of their similarity ratio, which here is
Therefore, the area of is
the overall area of . This makes the closest response.
### Example Question #11 : Triangles
The perimeter of a right triangle is 40 units. If the lengths of the sides are , , and units, then what is the area of the triangle?
Explanation:
Because the perimeter is equal to the sum of the lengths of the three sides of a triangle, we can add the three expressions for the lengths and set them equal to 40.
Perimeter:
Simplify the x terms.
Simplify the constants.
Subtract 8 from both sides.
Divide by 4
One side is 8.
The second side is
.
The third side is
.
Thus, the sides of the triangle are 8, 15, and 17.
The question asks us for the area of the triangle, which is given by the formula (1/2)bh. We are told it is a right triangle, so we can use one of the legs as the base, and the other leg as the height, since the legs will intersect at right angles. The legs of the right triangle must be the smallest sides (the longest must be the hypotenuse), which in this case are 8 and 15. So, let's assume that 8 is the base and 15 is the height.
The area of a triangle is (1/2)bh. We can substitute 8 and 15 for b and h.
.
The answer is 60 units squared.
### Example Question #2 : How To Find The Area Of A Right Triangle
The vertices of a right triangle on the coordinate axes are at the origin, , and . Give the area of the triangle.
Explanation:
The triangle in question can be drawn as the following:
The lengths of the legs of the triangle are 12, the distance from the origin to , and 8, the distance from the origin to . The area of a right triangle is equal to half the product of the lengths of the legs, so set in the formula:
### Example Question #123 : Plane Geometry
Three points in the xy-coordinate system form a triangle.
The points are .
What is the perimeter of the triangle?
Explanation:
Drawing points gives sides of a right triangle of 4, 5, and an unknown hypotenuse.
Using the pythagorean theorem we find that the hypotenuse is .
### Example Question #124 : Plane Geometry
Based on the information given above, what is the perimeter of triangle ABC?
Explanation:
Consult the diagram above while reading the solution. Because of what we know about supplementary angles, we can fill in the inner values of the triangle. Angles A and B can be found by the following reductions:
A + 120 = 180; A = 60
B + 150 = 180; B = 30
Since we know A + B + C = 180 and have the values of A and B, we know:
60 + 30 + C = 180; C = 90
This gives us a 30:60:90 triangle. Now, since 17.5 is across from the 30° angle, we know that the other two sides will have to be √3 and 2 times 17.5; therefore, our perimeter will be as follows:
### Example Question #11 : Triangles
Give the perimeter of the provided triangle.
Explanation:
The figure shows a right triangle. The acute angles of a right triangle have measures whose sum is , so
Substituting for :
This makes a 45-45-90 triangle.
By the 45-45-90 Triangle Theorem, legs and are of the same length, so
.
Also by the 45-45-90 Triangle Theorem, the length of hypotenuse is equal to that of leg multiplied by . Therefore,
.
The perimeter of the triangle is
### Example Question #2 : How To Find The Perimeter Of A Right Triangle
What is the perimeter of the triangle above?
Explanation:
The figure shows a right triangle. The acute angles of a right triangle have measures whose sum is , so
Substituting for :
This makes a 45-45-90 triangle. By the 45-45-90 Triangle Theorem, the length of leg is equal to that of hypotenuse , the length of which is 12, divided by . Therefore,
Rationalize the denominator by multiplying both halves of the fraction by :
By the same reasoning, .
The perimeter of the triangle is
### Example Question #1 : Apply The Pythagorean Theorem To Find The Distance Between Two Points In A Coordinate System: Ccss.Math.Content.8.G.B.8
If and , how long is side ?
Not enough information to solve
Explanation:
This problem is solved using the Pythagorean theorem . In this formula and are the legs of the right triangle while is the hypotenuse.
Using the labels of our triangle we have:
### Example Question #1 : How To Find The Length Of The Hypotenuse Of A Right Triangle : Pythagorean Theorem
If one of the short sides of a 45-45-90 triangle equals 5, how long is the hypotenuse?
√15
5
π
√10
5√2
5√2
Explanation:
Using the Pythagorean theorem, x2 + y2 = h2. And since it is a 45-45-90 triangle the two short sides are equal. Therefore 52 + 52 = h2 . Multiplied out 25 + 25 = h2.
Therefore h2 = 50, so h = √50 = √2 * √25 or 5√2.
### Example Question #1 : How To Find The Length Of The Hypotenuse Of A Right Triangle : Pythagorean Theorem
The height of a right circular cylinder is 10 inches and the diameter of its base is 6 inches. What is the distance from a point on the edge of the base to the center of the entire cylinder?
√(34)
4π/5
√(43)/2
3π/4
√(34)
Explanation:
The best thing to do here is to draw diagram and draw the appropiate triangle for what is being asked. It does not matter where you place your point on the base because any point will produce the same result. We know that the center of the base of the cylinder is 3 inches away from the base (6/2). We also know that the center of the cylinder is 5 inches from the base of the cylinder (10/2). So we have a right triangle with a height of 5 inches and a base of 3 inches. So using the Pythagorean Theorem 3+ 5= c2. 34 = c2, c = √(34). |
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# 3.3E: Graphs of Polynomial Functions (Exercises)
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$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
Find the C and t intercepts of each function. $1. C\left(t\right)=2\left(t-4\right)\left(t+1\right)(t-6) 2. C\left(t\right)=3\left(t+2\right)\left(t-3\right)(t+5)$ $3. C\left(t\right)=4t\left(t-2\right)^{2} (t+1) 4. C\left(t\right)=2t\left(t-3\right)\left(t+1\right)^{2}$ $5. C\left(t\right)=2t^{4} -8t^{3} +6t^{2} 6. C\left(t\right)=4t^{4} +12t^{3} -40t^{2}$
Use your calculator or other graphing technology to solve graphically for the zeros of the function. $7. f\left(x\right)=x^{3} -7x^{2} +4x+30 8. g\left(x\right)=x^{3} -6x^{2} +x+28$
Find the long run behavior of each function as $$t\to \infty$$ and $$t\to -\infty$$ $9. h\left(t\right)=3\left(t-5\right)^{3} \left(t-3\right)^{3} (t-2) 10. k\left(t\right)=2\left(t-3\right)^{2} \left(t+1\right)^{3} (t+2)$ $11. p\left(t\right)=-2t\left(t-1\right)\left(3-t\right)^{2} 12. q\left(t\right)=-4t\left(2-t\right)\left(t+1\right)^{3}$
Sketch a graph of each equation. $13. f\left(x\right)=\left(x+3\right)^{2} (x-2) 14. g\left(x\right)=\left(x+4\right)\left(x-1\right)^{2}$ $15. h\left(x\right)=\left(x-1\right)^{3} \left(x+3\right)^{2} 16. k\left(x\right)=\left(x-3\right)^{3} \left(x-2\right)^{2}$ $17. m\left(x\right)=-2x\left(x-1\right)(x+3) 18. n\left(x\right)=-3x\left(x+2\right)(x-4)$
Solve each inequality. $19. \left(x-3\right)\left(x-2\right)^{2} >0 20. \left(x-5\right)\left(x+1\right)^{2} >0$ $21. \left(x-1\right)\left(x+2\right)\left(x-3\right)<0 22. \left(x-4\right)\left(x+3\right)\left(x+6\right)<0$
Find the domain of each function. $23. f\left(x\right)=\sqrt{-42+19x-2x^{2} } 24. g\left(x\right)=\sqrt{28-17x-3x^{2} }$ $25. h\left(x\right)=\sqrt{4-5x+x^{2} } 26. k\left(x\right)=\sqrt{2+7x+3x^{2} }$ $27. n\left(x\right)=\sqrt{\left(x-3\right)\left(x+2\right)^{2} } 28. m\left(x\right)=\sqrt{\left(x-1\right)^{2} (x+3)}$ $29. p\left(t\right)=\frac{1}{t^{2} +2t-8} 30. q\left(t\right)=\frac{4}{x^{2} -4x-5}$ Write an equation for a polynomial the given features.
1. Degree 3. Zeros at x = -2, x = 1, and x = 3. Vertical intercept at (0, -4)
2. Degree 3. Zeros at x = -5, x = -2, and x = 1. Vertical intercept at (0, 6)
3. Degree 5. Roots of multiplicity 2 at x = 3 and x = 1, and a root of multiplicity 1 at x = -3. Vertical intercept at (0, 9)
4. Degree 4. Root of multiplicity 2 at x = 4, and a roots of multiplicity 1 at x = 1 and x = -2. Vertical intercept at (0, -3)
5. Degree 5. Double zero at x = 1, and triple zero at x = 3. Passes through the point (2, 15)
6. Degree 5. Single zero at x = -2 and x = 3, and triple zero at x = 1. Passes through the point (2, 4)
Write a formula for each polynomial function graphed.
37. 38. 39.
40. 41. 42.
43. 44.
Write a formula for each polynomial function graphed.
45. 46.
47. 48.
49. 50.
1. A rectangle is inscribed with its base on the x axis and its upper corners on the parabola $$y=5-x^{2}$$. What are the dimensions of such a rectangle that has the greatest possible area?
A rectangle is inscribed with its base on the x axis and its upper corners on the curve $$y=16-x^{4}$$. What are the dimensions of such a rectangle that has the greatest possible area?203
3.4 Factor Theorem and Remainder Theorem |
5.6 Transformations of the Sine and Cosine Graphs Wed Nov 12 Do Now Use the sine and cosine values of 0, pi/2, pi, 3pi/2 and 2pi to sketch the graph of.
Presentation on theme: "5.6 Transformations of the Sine and Cosine Graphs Wed Nov 12 Do Now Use the sine and cosine values of 0, pi/2, pi, 3pi/2 and 2pi to sketch the graph of."— Presentation transcript:
5.6 Transformations of the Sine and Cosine Graphs Wed Nov 12 Do Now Use the sine and cosine values of 0, pi/2, pi, 3pi/2 and 2pi to sketch the graph of f(x) = sin x and g(x) = cos x
Review of graphs Y = sin x Period of 2pi Amplitude of 1 Goes through (0, 0)
Review of Graphs Y = cos x Period of 2pi Amplitude of 1 Goes through (0, 1)
Transformations We are interested in the graphs of functions in the form
The Constant A Recall that coefficients of functions result in a vertical shift / shrink The constant A affects the amplitude of sine and cosine. The amplitude = A If A is negative, the graph is upside down
Ex Graph the following 1) 2) 3)
The Constant B Recall that coefficients of X result in a horizontal stretch / shrink The constant B affects the period The period of these graphs is
Ex Sketch a graph of the following 1) 2)
The Constant C The constant C, like in previous functions, results in a horizontal shift C units right / left This is also known as a phase shift Ex: sin (x – 4) is a shift to the right 4 units Cos (x + pi) is a shift to the left pi units
The Constant D The constant D results in a vertical shift D units up / down Ex: y = sin x + 1 shifts up 1 unit Ex: y = cos x – 4 shifts down 4 units Notice no parenthesis
Combined Transformations When working with multiple transformations, we want to rewrite the functions This helps you see the phase shift
How to graph 1) determine the period, amplitude, and shifts 2) graph and shift the period, and split it into 4 regions 3) plot a point in between each region, including the amplitude and shifts in your calculations 4) connect the points in the correct sine or cosine wave
Ex Sketch a graph of
Ex Sketch a graph of
Closure Graph HW: p.523 #1-25 odds
5.6 Transformations of Sine and Cosine cont’d Thurs Nov 13 Do Now Graph the following 1) y = sin(1/2 x)] 2) y = - 2cos( 2x )
HW Review: p.523 #1-25 odds
Review of Sine and Cosine Recall the transformations A affects the amplitude B affects the period C/B affects the phase shift D affects the vertical shift
Ex Graph
Matching On p.522
Closure What kind of transformations can affect the sine and cosine graphs? How do we determine what transformations occur? HW: p.523 #27-43 odds
5.6 Addition and Multiplication of Ordinates Fri Nov 14 Do Now Graph
HW Review: p.523 #27-43
Graphs of Sums: Additions of Ordinates When graphing a sum of 2 trigonometric functions, we use a strategy called addition of ordinates
Properties of sums The period of a sum will be the least common multiple of every period Graph each important point by adding the y- values of each trig function
ex Graph y = 2sin x + sin 2x
Damped Oscillation: Multiplication of Ordinates We’ll just graph these
Finding zeros (review) To find zeros of a function, 1) Graph function 2) 2 nd -> calc -> zeros 3) Left bound – pick a point slightly left of the zero you want 4) Right bound – pick a point slight right of the zero you want 5) Guess – hit enter
ex Solvethe zeros ofon the interval [-12,12]
closure What is addition of ordinates? How do we graph these functions? HW: p.524 #45-73 odds
5.6 Other Trig Transformations Mon Nov 17 Do Now Graph y = csc x and y = tan x on your calculator
HW Review: p.524 #45-73 odds
Review: f(x) = tan x and cot x The period of tangent and cotangent is pi Each period is separated by vertical asymptotes Amplitude does not affect the graph drastically
Basic graphs Y = tan xy = cot x
Review: f(x) = csc x and sec x The period of csc x and sec x is 2pi Vertical asymptotes occur every half period The amplitude represents how close to the center each curve gets
Basic Graphs Y = csc x y = sec x
Transformations Transformations affect these 4 graphs the same way
Ex Sketch the graph of
Ex Sketch the graph of
Closure Graph HW: p. 525 #89-97 odds CH 5 Test soon
5.6 Review Tues Nov 18 Do Now Sketch the graph of
HW Review p.525 #89-97
Transformations Review Basic graphs Transformations Period, Amplitude, Phase shift, Vertical shift
Closure What are some identifying properties of trigonometric functions and their graphs? HW: p.529 #1-83 odds skip 53 55 due Thursday SGO Assessment Wed Nov 19 Ch 5 Test Fri Nov 21
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