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human
|
In the trapezoid shown, the ratio of the area of triangle $ABC$ to the area of triangle $ADC$ is $7:3$. If $AB + CD = 210$ cm, how long is segment $\overline{AB}$? [asy]
import olympiad; size(150); defaultpen(linewidth(0.8));
pair A = (0,0), B = (5,0), C = (3,2), D = (1,2);
draw(A--B--C--D--cycle--C);
label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,NW);
[/asy]
|
gpt
|
Let $h$ be the height of the trapezoid. The height of the trapezoid is also a height of $ABC$ and of $ADC$. Specifically, we have $[ABC] = (AB)(h)/2$ and $[ADC] = (CD)(h)/2$, so $[ABC]:[ADC] = AB:CD$. Since we are given that this area ratio equals $7:3$, we know that $AB:CD = 7:3$. Therefore, $AB = 7x$ and $CD = 3x$ for some value of $x$. Since $AB + CD = 210$ cm, we have $7x+3x=210$, so $10x=210$ and $x=21$. Therefore, $AB=7 \times 21 = \boxed{147\text{ cm}}$.
|
human
|
In the diagram, if $\angle PQR = 48^\circ$, what is the measure of $\angle PMN$? [asy]
size(6cm);
pair p = (0, 0); pair m = dir(180 - 24); pair n = dir(180 + 24); pair r = 1.3 * dir(24); pair q = 2 * 1.3 * Cos(48) * dir(-24);
label("$M$", m, N); label("$R$", r, N); label("$P$", p, 1.5 * S); label("$N$", n, S); label("$Q$", q, SE);
draw(m--q--r--n--cycle);
add(pathticks(m--p, s=4));
add(pathticks(n--p, s=4));
add(pathticks(r--p, 2, spacing=0.9, s=4));
add(pathticks(r--q, 2, spacing=0.9, s=4));
[/asy]
|
gpt
|
In $\triangle PQR$, since $PR=RQ$, then $\angle RPQ=\angle PQR = 48^\circ$.
Since $\angle MPN$ and $\angle RPQ$ are opposite angles, we have $\angle MPN = \angle RPQ=48^\circ$.
In $\triangle PMN$, $PM=PN$, so $\angle PMN = \angle PNM$.
Therefore, $$\angle PMN = \frac{1}{2}(180^\circ - \angle MPN) = \frac{1}{2}(180^\circ - 48^\circ) = \frac{1}{2}(132^\circ)=\boxed{66^\circ}.$$
|
human
|
A bottled water company has designed a new cup for its dispenser. The cup will be a right circular cone with a three-inch radius. How tall will the cup need to be to hold 93 cubic inches of water? Express your answer to the nearest whole number.
|
gpt
|
The volume of a cone with radius $r$ and height $h$ is \[\frac{1}{3} \pi r^2 h.\] Therefore, we want $h$ to satisfy \[\frac{1}{3} \pi \cdot 3^2 \cdot h \ge 93,\] or \[h \ge \frac{93}{3 \pi} \approx 9.87.\] Therefore, the height must be $\boxed{10}$ inches.
|
human
|
The truncated right circular cone has a large base radius 8 cm and a small base radius of 4 cm. The height of the truncated cone is 6 cm. How many $\text{cm}^3$ are in the volume of this solid? [asy]
import olympiad; size(150); defaultpen(linewidth(0.8)); dotfactor=4;
draw(ellipse((0,0),4,1)); draw(ellipse((0,3),2,1/2));
draw((-3.97,.1)--(-1.97,3.1)^^(3.97,.1)--(1.97,3.1));
[/asy]
|
gpt
|
[asy]
import olympiad; size(150); defaultpen(linewidth(0.8)); dotfactor=4;
draw(ellipse((0,0),4,1)); draw(ellipse((0,3),2,1/2),gray(.7));
// draw((-3.97,.1)--(-1.97,3.1)^^(3.97,.1)--(1.97,3.1));
draw((-3.97,.1)--(0,6.07)--(3.97,.1));
draw((4,0)--(0,0)--(0,6.07),linewidth(0.8));
draw((2,3)--(0,3),linewidth(0.8));
label("4",(2,3)--(0,3),S);
label("8",(4,0)--(0,0),S);
label("6",(0,0)--(0,3),W);
label("$x$",(0,2)--(0,6.07),W);
[/asy]
We "complete" the truncated cone by adding a smaller, similar cone atop the cut, forming a large cone. We don't know the height of the small cone, so call it $x$. Since the small and large cone are similar, we have $x/4=(x+6)/8$; solving yields $x=6$. Hence the small cone has radius 4, height 6, and volume $(1/3)\pi(4^2)(6)=32\pi$ and the large cone has radius 8, height 12, and volume $(1/3)\pi(8^2)(12)=256\pi$. The frustum's volume is the difference of these two volumes, or $256\pi-32\pi=\boxed{224\pi}$ cubic cm.
|
human
|
A right pyramid has a square base that measures 10 cm on each side. Its peak is 12 cm above the center of its base. What is the sum of the lengths of the pyramid's eight edges? Express your answer to the nearest whole number.
[asy]
size(150);
draw((0,0)--(3,3)--(13,3)--(10,0)--cycle,linewidth(1));
draw((0,0)--(6.5,15)--(3,3),linewidth(1));
draw((13,3)--(6.5,15)--(10,0),linewidth(1));
draw((6.5,15)--(6.5,1.5),linewidth(1));
label("12",(6.5,5),E);
draw((6.6,15)..(6.7,14.9)..(6.8,14.7)--(6.8,5.3)..(6.9,5.2)..(7,5)..(6.9,4.8)..(6.8,4.7)--(6.8,1.8)..(6.7,1.6)..(6.6,1.5),linewidth(.7));
[/asy]
|
gpt
|
To start, we can draw in a line from where the altitude meets the base to one of the bottom corners as shown:
[asy]
size(150);
draw((0,0)--(3,3)--(13,3)--(10,0)--cycle,linewidth(1));
draw((0,0)--(6.5,15)--(3,3),linewidth(1));
draw((13,3)--(6.5,15)--(10,0),linewidth(1));
draw((6.5,15)--(6.5,1.5),linewidth(1));
draw((6.5,1.5)--(10,0),linewidth(.7));
[/asy]
The length of this segment will be half the length of the diagonal of the base. The base has side $10$, so the diagonal will satisfy: $$d^2=10^2+10^2=200$$ $$d=10\sqrt{2}$$ Half of this is $5\sqrt{2}$. Now we can look at the right triangle formed by the altitude from the vertex of the pyramid, the line just drawn in, and with hypotenuse as the edge we need to find the length of. The length of this edge is: $$\sqrt{12^2+(5\sqrt{2})^2}=\sqrt{144+50}=\sqrt{194}\approx 13.928$$ The total length of all edges is: $$4(10)+4(13.928)\approx \boxed{ 96}$$
|
human
|
The lines $y=2$, $y=5$, $x=1$, and $x=a$ make a square. Find the product of the possible values for $a$.
|
gpt
|
Since the two horizontal lines are $5-2=3$ units apart, we know that the side of the square is $3$ units long. Therefore, the fourth line must be parallel to $x=1$. If $x=a$ is to the left of $x=1$, its equation is $x=1-3=-2$. If $x=a$ is to the right of $x=1$, then its equation is $x=1+3=4$. Therefore, the two values for $a$ are $4$ and $-2$, and their product is $\boxed{-8}$.
|
human
|
A right circular cone is inscribed in a right prism as shown. What is the ratio of the volume of the cone to the volume of the prism? Express your answer as a common fraction in terms of $\pi$. [asy]
import three; import graph3;
defaultpen(linewidth(0.8));
size(200);
draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--cycle);
draw((0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle);
draw((0,0,0)--(0,0,1)); draw((1,0,0)--(1,0,1)); draw((1,1,0)--(1,1,1)); draw((0,1,0)--(0,1,1));
draw(Circle((0.5,0.5,0),0.5),dashed);
draw((0.5,0.5,1)--(0.5,0,0),dashed); draw((0.5,0.5,1)--(0.5,1,0),dashed); draw((0.5,0.5,1)--(1,0.5,0),dashed); draw((0.5,0.5,1)--(0,0.5,0),dashed);
[/asy]
|
gpt
|
Since the cone is tangent to all sides of the base of the prism, the base of the prism is a square. Furthermore, if the radius of the base of the cone is $r$, then the side length of the square is $2r$.
Let $h$ be the common height of the cone and the prism. Then the volume of the cone is \[\frac{1}{3} \pi r^2 h,\] and the volume of the prism is $(2r)^2 h = 4r^2 h$, so the desired ratio is \[\frac{\frac{1}{3} \pi r^2 h}{4r^2 h} = \boxed{\frac{\pi}{12}}.\]
|
human
|
The lengths of the sides of a triangle are 6 cm, 7 cm and 9 cm. In a similar triangle whose perimeter is 110 cm, what is the length of the longest side, in centimeters?
|
gpt
|
Let the ratio of side lengths between the similar triangle and the given triangle be $x$, so the lengths of the similar triangle are $6x$, $7x$, and $9x$. We are given that $6x+7x+9x=110$; solving yields $x=\frac{110}{(6+7+9)} = \frac{110}{22}=5$. The length of the longest side is thus $9x = 9 \cdot 5 = \boxed{45}$.
|
human
|
In the figure below, the smaller circle has a radius of two feet and the larger circle has a radius of four feet. What is the total area of the four shaded regions? Express your answer as a decimal to the nearest tenth.
[asy]
fill((0,0)--(12,0)--(12,-4)--(4,-4)--(4,-2)--(0,-2)--cycle,gray(0.7));
draw((0,0)--(12,0),linewidth(1));
draw((0,0)--(0,-2),linewidth(1));
draw((4,0)--(4,-4),linewidth(1));
draw((12,0)--(12,-4),linewidth(1));
draw((0.3,0)--(0.3,-0.3)--(0,-0.3));
draw((4.3,0)--(4.3,-0.3)--(4,-0.3));
draw((11.7,0)--(11.7,-0.3)--(12,-0.3));
fill(Circle((2,-2),2),white);
fill(Circle((8,-4),4),white);
draw(Circle((2,-2),2));
draw(Circle((8,-4),4));
dot((2,-2));
dot((8,-4));
[/asy]
|
gpt
|
Draw horizontal diameters of both circles to form two rectangles, both surrounding shaded regions. The height of each rectangle is a radius and the length is a diameter, so the left rectangle is 2 ft $\times$ 4 ft and the right rectangle is 4 ft $\times$ 8 ft. The shaded region is obtained by subtracting respective semicircles from each rectangle, so the total area of the shaded region in square feet is $A = [(2)(4) - \dfrac{1}{2}\pi \cdot(2)^2] +
[(4)(8) - \dfrac{1}{2}\pi \cdot(4)^2] = 40 - 10\pi \approx \boxed{8.6}$.
Equivalently, we could notice that since the right side of the figure is scaled up from the left side by a factor of 2, areas will be scaled by a factor of $2^2 = 4$, and the right shaded region will be 4 times the size of the left shaded region. Then $A = 5[(2)(4) - \dfrac{1}{2}\pi \cdot(2)^2],$ giving the same result.
|
human
|
The measure of angle $ACB$ is 40 degrees. If ray $CA$ is rotated 480 degrees about point $C$ in a clockwise direction, what will be the positive measure of the new acute angle $ACB$, in degrees?
[asy]
draw((0,0)--dir(40),linewidth(1),Arrow);
draw((0,0)--dir(0),linewidth(1),Arrow);
dot(.8dir(40));
dot(.8dir(0));
dot((0,0));
label("A",.8dir(40),NW);
label("B",.8dir(0),S);
label("C",(0,0),W);
[/asy]
|
gpt
|
A rotation by 480 degrees is the same as one by $480 - 360 = 120$. The first $40$ degrees of this rotation will be used to decrease the angle $ACB$ to $0$ degrees, leaving $80$ unused degrees, so our answer is $\boxed{80}$ degrees.
|
human
|
In the diagram, $BP$ and $BQ$ trisect $\angle ABC$. $BM$ bisects $\angle PBQ$. Find the ratio of the measure of $\angle MBQ$ to the measure of $\angle ABQ$.
|
gpt
|
Let $\angle MBQ = x$, so $\angle MBP=x$ as well. Therefore, we have $\angle PBQ = 2x$, so $\angle ABP = \angle PBQ = \angle QBC = 2x$. Finally, we have $\angle ABQ = \angle ABP + \angle PBQ = 4x$, so \[\frac{\angle MBQ}{\angle ABQ} = \frac{x}{4x} = \boxed{\frac14}.\]
|
human
|
Six congruent circles form a ring with each circle externally tangent to the two circles adjacent to it. All six circles are internally tangent to a circle $\cal C$ with radius 30. Let $K$ be the area of the region inside $\cal C$ and outside all of the six circles in the ring. Find $\lfloor K\rfloor$. (The notation $\lfloor K\rfloor$ denotes the greatest integer that is less than or equal to $K$.)
|
gpt
|
Let $r$ be the radius of each of the six congruent circles, and let $A$ and $B$ be the centers of two adjacent circles. Join the centers of adjacent circles to form a regular hexagon with side $2r$. Let $O$ be the center of $\cal C$. Draw the radii of $\cal C$ that contain $A$ and $B$. Triangle $ABO$ is equilateral, so $OA=OB=2r$. Because each of the two radii contains the point where the smaller circle is tangent to $\cal
C$, the radius of $\cal C$ is $3r$, and $K=\pi\left((3r)^2-6r^2\right)=3\pi r^2$. The radius of $\cal C$ is 30, so $r=10$, $K=300\pi$, and $\lfloor K\rfloor=\boxed{942}$.
|
human
|
In $\triangle XYZ$, we have $\angle X = 90^\circ$ and $\tan Z = 3$. What is $\cos Z$?
|
gpt
|
[asy]
pair X,Y,Z;
X = (0,0);
Y = (15,0);
Z = (0,5);
draw(X--Y--Z--X);
draw(rightanglemark(Y,X,Z,23));
label("$X$",X,SW);
label("$Y$",Y,SE);
label("$Z$",Z,N);
//label("$100$",(Y+Z)/2,NE);
label("$k$",(Z)/2,W);
label("$3k$",Y/2,S);
[/asy]
Since $\triangle XYZ$ is a right triangle with $\angle X = 90^\circ$, we have $\tan Z = \frac{XY}{XZ}$. Since $\tan Z = 3$, we have $XY = 3k$ and $XZ = k$ for some value of $k$, as shown in the diagram. Applying the Pythagorean Theorem gives $YZ^2 = (3k)^2 + k^2 = 10k^2$, so $YZ = k\sqrt{10}$.
Finally, we have $\cos Z = \frac{XZ}{YZ} = \frac{k}{k\sqrt{10}} = \frac{1}{\sqrt{10}} = \boxed{\frac{\sqrt{10}}{10}}$.
|
human
|
Semicircles of diameter 2'' are lined up as shown. What is the area, in square inches, of the shaded region in a 1-foot length of this pattern? Express your answer in terms of $\pi$.
[asy]import graph;
size(101);
path tophalf = Arc((0,0),1,180,0) -- Arc((2,0),1,180,0) -- Arc((4,0),1,180,0) -- Arc((6,0),1,180,0) -- Arc((8,0),1,180,90);
path bottomhalf = Arc((7,0),1,360,180) -- Arc((5,0),1,360,180) -- Arc((3,0),1,360,180) -- Arc((1,0),1,360,180) -- Arc((-1,0),1,360,270);
fill(tophalf--bottomhalf--cycle,gray(.6));
draw(tophalf,linewidth(1.5)); draw(bottomhalf,linewidth(1));
draw("$1'$",(-3,1.5)--(10,1.5),N,linewidth(1.5),Arrows(4));
draw((-3,1.25)--(-3,1.75),linewidth(1.5)); draw((10,1.25)--(10,1.75),linewidth(1.5));
dot((-5/3,0),linewidth(3)); dot((-7/3,0),linewidth(3)); dot((-3,0),linewidth(3));
dot((26/3,0),linewidth(3)); dot((28/3,0),linewidth(3)); dot((10,0),linewidth(3));[/asy]
|
gpt
|
12 inches of the pattern contains one semicircle on top and on bottom for each diameter, or $\frac{12}{2}\cdot2=12$ total semicircles. This is the equivalent of 6 full circles, so the shaded area is $6\cdot1^2\pi=\boxed{6\pi}$ square inches.
|
human
|
A spherical scoop of vanilla ice cream with radius of 2 inches is dropped onto the surface of a dish of hot chocolate sauce. As it melts, the ice cream spreads out uniformly forming a cylindrical region 8 inches in radius. Assuming the density of the ice cream remains constant, how many inches deep is the melted ice cream? Express your answer as a common fraction.
|
gpt
|
The ice cream sphere has volume $\frac{4}{3}\pi(2^3) = \frac{32\pi}{3}$ cubic inches. Let the height of the cylindrical region be $h$; then, the volume of the cylindrical region is $\pi (8^2)h=64\pi h$. Thus, we have \[\frac{32\pi}{3} = 64\pi h.\] Dividing both sides by $64\pi$ yields $h = \boxed{\frac{1}{6}}$ inches.
|
human
|
A delicious circular pie with diameter $12\text{ cm}$ is cut into three equal-sized sector-shaped pieces. Let $l$ be the number of centimeters in the length of the longest line segment that may be drawn in one of these pieces. What is $l^2$?
|
gpt
|
First of all, let us draw one of these pieces in question, labeling points of interest as necessary: [asy]
pair pA, pB, pC, pO;
pO = (0, 0);
pA = dir(150);
pB = dir(30);
pC = dir(90);
draw(pA--pO--pB);
draw(pA..pC..pB);
label("$A$", pA, W);
label("$B$", pB, E);
label("$O$", pO, S);
[/asy] We can see that the longest segment we can draw is from $A$ to $B,$ and to find $AB,$ we should create right triangles by drawing the perpendicular bisector to $AB.$ [asy]
pair pA, pB, pC, pM, pO;
pO = (0, 0);
pA = dir(150);
pB = dir(30);
pC = dir(90);
pM = 0.5 * pA + 0.5 * pB;
draw(pA--pO--pB);
draw(pA--pB);
draw(pM--pO);
draw(pA..pC..pB);
draw(rightanglemark(pO,pM,pA,2));
label("$A$", pA, W);
label("$B$", pB, E);
label("$O$", pO, S);
label("$M$", pM, N);
[/asy] Since $\angle MOB$ is half of $\angle AOB,$ which is a third of a full circle, we have $\angle MOB = 60^\circ,$ so $\triangle MOB$ is a 30-60-90 triangle. Since the diameter of the pie is $12\text{ cm},$ we see that $OB = 6\text{ cm},$ so $MO = 3\text{ cm}$ and $MB = 3\sqrt{3}\text{ cm}.$ Then, $AB = 2 \cdot MB = 6\sqrt{3}\text{ cm},$ so $l = 6\sqrt{3}.$ Finally, $l^2 = \boxed{108}.$
|
human
|
Compute $\cos 90^\circ$.
|
gpt
|
Rotating the point $(1,0)$ about the origin by $90^\circ$ counterclockwise gives us the point $(0,1)$, so $\cos 90^\circ = \boxed{0}$.
|
human
|
In right triangle $MNO$, $\tan{M}=\frac{5}{4}$, $OM=8$, and $\angle O = 90^\circ$. Find $MN$. Express your answer in simplest radical form.
|
gpt
|
Our triangle is as shown below.
[asy]
size(100);
draw((0,0)--(8,0)--(0,10)--cycle,black+linewidth(1));
draw(rightanglemark((8,0),(0,0),(0,10),20),black+linewidth(1));
label("$O$",(0,0),W);
label("$M$",(8,0),E);
label("$N$",(0,10),W);
label("8",(0,0)--(8,0),S);
[/asy]
Since $\tan{M}=\frac{5}{4}$, we have $\dfrac{NO}{OM} = \dfrac{5}{4}$, so $$NO = \frac{5}{4}OM = \frac{5}{4}\cdot 8 = 10.$$Then, from the Pythagorean Theorem, we have \begin{align*}
MN&=\sqrt{NO^2+OM^2}\\
&=\sqrt{10^2+8^2}=\sqrt{164}=\boxed{2\sqrt{41}}.\end{align*}
|
human
|
Quadrilateral $ABCD$ is a square. A circle with center $D$ has arc $AEC$. A circle with center $B$ has arc $AFC$. If $AB = 2$ cm, what is the total number of square centimeters in the football-shaped area of regions II and III combined? Express your answer as a decimal to the nearest tenth.
[asy]
path a=(7,13)..(0,0)--(20,20)..cycle;
path b=(13,7)..(0,0)--(20,20)..cycle;
draw(a);
draw(b);
dot((8.6,3.5));
label("F",(8.6,3.5),SE);
label("E",(11.4,16.5),NW);
dot((11.4,16.5));
draw((0,0)--(20,0)--(20,20)--(0,20)--cycle);
label("$A$",(0,0),SW);
label("$B$",(0,20),NW);
label("$C$",(20,20),NE);
label("$D$",(20,0),SE);
label("I",(2,19),S);
label("II",(9,13),S);
label("III",(11,7),N);
label("IV",(18,1),N);
[/asy]
|
gpt
|
Regions I, II, and III combine to form a sector of a circle whose central angle measures 90 degrees. Therefore, the area of this sector is $\frac{90}{360}\pi(\text{radius})^2=\frac{1}{4}\pi(2)^2=\pi$ square centimeters. Also, regions I and II combine to form an isosceles right triangle whose area is $\frac{1}{2}(\text{base})(\text{height})=\frac{1}{2}(AB)(BC)=\frac{1}{2}(2)(2)=2$ square centimeters. Subtracting these two areas, we find that the area of region III is $\pi-2$ square centimeters. Since region II is congruent to region III, the combined area of region II and region III is $2(\pi-2)=2\pi-4\approx \boxed{2.3}$ square centimeters.
|
human
|
A circular spinner for a game has a radius of 5 cm. The probability of winning on one spin of this spinner is $\frac{2}{5}$. What is the area, in sq cm, of the WIN sector? Express your answer in terms of $\pi$.
[asy]import graph;
draw(Circle((0,0),25),black);
draw((0,0)--(7,18),Arrow);
draw((0,0)--(0,25));
draw((0,0)--(15,-20));
label("WIN",(10,10),S);
label("LOSE",(-8,-8),N);
dot((0,0));
[/asy]
|
gpt
|
The probability of winning on one spin is equal to the ratio of the area of the WIN sector to the area of the entire circle. The area of the entire circle is $\pi \cdot 5^2 = 25\pi$. In math terms, our ratio is: $\frac{2}{5}=\frac{\text{area of the win sector}}{25\pi}$. Solving for the area of the win sector, we find it equal to $\boxed{10\pi}$ square centimeters.
|
human
|
A triangular region is enclosed by the lines with equations $y = \frac{1}{2} x + 3$, $y = -2x + 6$ and $y = 1$. What is the area of the triangular region? Express your answer as a decimal to the nearest hundredth.
|
gpt
|
The vertices of the triangle are the points where two of the lines intersect. The line $y=\frac{1}{2}x+3$ intersects $y=1$ when $$\frac{1}{2}x+3=1\Rightarrow x=-4.$$ The line $y=-2x+6$ intersects $y=1$ when $$-2x+6=1\Rightarrow x=\frac{5}{2}.$$ The line $y=\frac{1}{2}x+3$ intersects $y=-2x+6$ when $$\frac{1}{2}x+3=-2x+6\Rightarrow x=\frac{6}{5}.$$ and $$y=-2\left(\frac{6}{5}\right)+6=\frac{18}{5}$$
Thus the vertices of the triangle are $(-4,1)$, $\left(\frac{5}{2},1\right)$, and $\left(\frac{6}{5},\frac{18}{5}\right)$. We can let the base of the triangle lie along the line $y=1$. It will have length $$4+\frac{5}{2}=\frac{13}{2}.$$ The altitude from $\left(\frac{6}{5},\frac{18}{5}\right)$ to this line will have length $$\frac{18}{5}-1=\frac{13}{5}.$$ Thus the area of the triangle is $$\frac{1}{2}*\frac{13}{2}*\frac{13}{5}=\frac{169}{20}=\boxed{8.45}.$$
|
human
|
$\triangle ABC$ is similar to $\triangle DEF$ . What is the number of centimeters in the length of $\overline{EF}$ ? Express your answer as a decimal to the nearest tenth.
[asy]
draw((0,0)--(8,-2)--(5,4)--cycle);
label("8cm",(2.5,2),NW);
label("5cm",(6.1,1),NE);
draw((12,0)--(18,-1.5)--(15.7,2.5)--cycle);
label("$A$",(8,-2),SE);
label("3cm",(16.9,0.5),NE);
label("$B$",(5,4),N);
label("$C$",(0,0),SW);
label("$D$",(18,-1.5),SE);
label("$E$",(15.7,2.5),N);
label("$F$",(12,0),N);
[/asy]
|
gpt
|
Because $\triangle ABC \sim \triangle DEF,$ we know that: \begin{align*}
\frac{EF}{ED} &= \frac{BC}{BA} \\
\frac{EF}{3\text{ cm}} &= \frac{8\text{ cm}}{5\text{ cm}} \\
EF &= \frac{8\text{ cm}\cdot3\text{ cm}}{5\text{ cm}} = \boxed{4.8}\text{ cm}.
\end{align*}
|
human
|
Three coplanar squares with sides of lengths two, four and six units, respectively, are arranged side-by-side, as shown so that one side of each square lies on line $AB$ and a segment connects the bottom left corner of the smallest square to the upper right corner of the largest square. What is the area of the shaded quadrilateral? [asy] size(150); defaultpen(linewidth(0.9)+fontsize(10));
fill((2,0)--(6,0)--(6,3)--(2,1)--cycle,gray(0.8));
draw(scale(2)*unitsquare);
draw(shift(2,0)*scale(4)*unitsquare);
draw(shift(6,0)*scale(6)*unitsquare);
draw((0,0)--(12,6));
real d = 1.2; pair d2 = (0.9,0);
pair A = (-d,0), B = (12+d,0); dot(A,linewidth(3)); dot(B,linewidth(3)); label("A",A,(0,-1.5)); label("B",B,(0,-1.5)); draw(A-d2--B+d2,Arrows(4));
label("2",(1,2.7)); label("4",(4,4.7)); label("6",(9,6.7));
[/asy]
|
gpt
|
[asy]size(150); defaultpen(linewidth(0.9)+fontsize(10));
fill((2,0)--(6,0)--(6,3)--(2,1)--cycle,gray(0.8));
draw(scale(2)*unitsquare);
draw(shift(2,0)*scale(4)*unitsquare);
draw(shift(6,0)*scale(6)*unitsquare);
draw((0,0)--(12,6));
real d = 1.2; pair d2 = (0.9,0);
pair A = (-d,0), B = (12+d,0); dot(A,linewidth(3)); dot(B,linewidth(3)); label("A",A,(0,-1.5)); label("B",B,(0,-1.5)); draw(A-d2--B+d2,Arrows(4));
label("2",(1,2.7)); label("4",(4,4.7)); label("6",(9,6.7)); label("6",(12.7,3)); label("3",(6.7,1.5)); label("1",(2.5,0.5)); label("$2$",(1,-0.7)); label("$4$",(4,-0.7)); label("$6$",(9,-0.7));
[/asy] Consider the three right triangles $T_1, T_2, T_3$ formed by the line $AB$, the segment connecting the bottom left corner of the smallest square to the upper right corner of the largest square, and a side of the smallest, medium, and largest squares, respectively. Since all three triangles share an angle, it follows that they must be similar. Notice that the base of $T_3$ is equal to $2+4+6 = 12$, and its height is equal to $6$. This, the height-to-base ratio of each of $T_1$ and $T_2$ is equal to $6/12 = 1/2$. Since the base of $T_1$ is $2$ and the base of $T_2$ is $2+4 = 6$, it follows that their heights are, respectively, $2 \cdot (1/2) = 1$ and $6 \cdot (1/2) = 3$. The shaded region is a trapezoid with bases $1$ and $3$ and altitude $4$, and area $\frac{4(1+3)}{2} = \boxed{8}$.
|
human
|
We have a triangle $\triangle ABC$ and a point $K$ on $BC$ such that $AK$ is an altitude of $\triangle ABC$. If $AC = 10,$ $BK = 7$, and $BC = 13,$ then what is the area of $\triangle ABC$?
|
gpt
|
First, we sketch! [asy]
pair A, B, C, K;
A = (0, 8);
B = (-7, 0);
C = (6, 0);
K = (0, 0);
draw(A--B--C--cycle);
draw(A--K);
label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$K$", K, NE);
label("10", C--A, NE);
label("7", B--K, N);
label("13", B--C, S);
draw(rightanglemark(A,K,B,10));
[/asy] We now see that $CK = BC - BK = 6.$ That means $\triangle AKC$ is a $3:4:5$ right triangle, so $AK = 8.$ At this point, we can see that the area of $\triangle ABC$ is $\frac{1}{2} \cdot AK \cdot BC = \frac{1}{2} \cdot 8 \cdot 13 = \boxed{52}.$
|
human
|
An ice cream cone has radius 1 inch and height 4 inches, What is the number of inches in the radius of a sphere of ice cream which has the same volume as the cone?
|
gpt
|
A cone with radius $r$ and height $h$ has volume $\frac{1}{3}\pi r^2 h$; here, our cone has volume $\frac{1}{3}\pi (1^2)(4)=\frac{4}{3}\pi$. A sphere with radius $r$ has volume $\frac{4}{3}\pi r^3$, so we set up the equation \[\frac{4}{3}\pi r^3=\frac{4}{3}\pi.\] Solving for $r$ yields $r^3=1$, so $r = 1$. The sphere's radius is $\boxed{1}$ inch.
|
human
|
In the two concentric circles shown, the radius of the outer circle is twice the radius of the inner circle. What is the area of the gray region, in square feet, if the width of the gray region is 2 feet? Express your answer in terms of $\pi$.
[asy]
filldraw(circle((0,0),4),gray);
filldraw(circle((0,0),2),white);
draw((2,0)--(4,0),linewidth(1));
label("$2^{\prime}$",(3,0),N);
[/asy]
|
gpt
|
The radius of the inner circle must be 2 feet. The area of the gray region is the area of the outer circle minus the area of the inner circle, or just $\pi\cdot 4^2 - \pi\cdot 2^2 = \boxed{12\pi}$.
|
human
|
Compute $\sin 300^\circ$.
|
gpt
|
Let $P$ be the point on the unit circle that is $300^\circ$ counterclockwise from $(1,0)$, and let $D$ be the foot of the altitude from $P$ to the $x$-axis, as shown below.
[asy]
pair A,C,P,O,D;
draw((0,-1.2)--(0,1.2),p=black+1.2bp,Arrows(0.15cm));
draw((-1.2,0)--(1.2,0),p=black+1.2bp,Arrows(0.15cm));
A = (1,0);
O= (0,0);
label("$x$",(1.2,0),SE);
label("$y$",(0,1.2),NE);
P = rotate(300)*A;
D = foot(P,A,-A);
draw(O--P--D);
draw(rightanglemark(O,D,P,2));
draw(Circle(O,1));
label("$O$",O,NW);
label("$P$",P,SE);
//label("$A$",A,SE);
label("$D$",D,N);
[/asy]
Triangle $POD$ is a 30-60-90 triangle, so $DO = \frac{1}{2}$ and $DP = \frac{\sqrt{3}}{2}$. Therefore, the coordinates of $P$ are $\left(\frac12,-\frac{\sqrt{3}}{2}\right)$, so $\sin300^\circ = \boxed{-\frac{\sqrt{3}}{2}}$.
|
human
|
In the diagram, points $U$, $V$, $W$, $X$, $Y$, and $Z$ lie on a straight line with $UV=VW=WX=XY=YZ=5$. Semicircles with diameters $UZ$, $UV$, $VW$, $WX$, $XY$, and $YZ$ create the shape shown. What is the area of the shaded region?
[asy]
size(5cm); defaultpen(fontsize(9));
pair one = (1, 0);
pair u = (0, 0); pair v = u + one; pair w = v + one; pair x = w + one; pair y = x + one; pair z = y + one;
path region = u{up}..{down}z..{up}y..{down}x..{up}w..{down}v..{up}u--cycle;
filldraw(region, gray(0.75), linewidth(0.75));
draw(u--z, dashed + linewidth(0.75));
// labels
label("$U$", u, W); label("$Z$", z, E);
label("$V$", v, 0.8 * SE); label("$X$", x, 0.8 * SE);
label("$W$", w, 0.8 * SW); label("$Y$", y, 0.8 * SW);
[/asy]
|
gpt
|
The area of a semi-circle with radius $r$ is $\frac{1}{2}\pi r^2$ so the area of a semi-circle with diameter $d$ is $\frac{1}{2}\pi \left( \frac{1}{2}d \right)^2 = \frac{1}{8}\pi d^2$.
The semicircles with diameters $UV$, $VW$, $WX$, $XY$, and $YZ$ each have equal diameter and thus equal area. The area of each of these semicircles is $\frac{1}{8}\pi(5^2)=\frac{25}{8}\pi$.
The large semicircle has diameter $UZ = 5(5)=25$, so has area $\frac{1}{8}\pi (25^2)=\frac{625}{8}\pi$.
The shaded area equals the area of the large semicircle, minus the area of two small semicircles, plus the area of three small semicircles, which equals the area of the large semicircle plus the area of one small semicircle. Therefore, the shaded area equals $$\frac{625}{8}\pi + \frac{25}{8}\pi = \frac{650}{8}\pi = \boxed{\frac{325}{4}\pi}.$$
|
human
|
A section is cut out of a circular piece of paper having radius four inches, as shown. Points A and B are then glued together to form a right circular cone. What is the circumference of the base of the resulting cone? Express your answer in terms of $\pi$. (The $270^\circ$ sector forms the cone.)
[asy]import graph;
draw(Circle((0,0),42.4),linewidth(1));
draw((0,0)--(30,30),linewidth(2));
draw((0,0)--(30,-30),linewidth(2));
label("$A$",(30,30),E);
label("$B$",(30,-30),E);
label("4''",(15,15),NW);
draw((3,3)--(6,0));
draw((6,0)--(3,-3));
[/asy]
|
gpt
|
The circumference of the whole circle is $2 \pi \cdot 4 = 8 \pi$. Then the circumference of the base of the cone is \[\frac{270^\circ}{360^\circ} \cdot 8 \pi = \boxed{6 \pi}.\]
|
human
|
What is the volume in cubic inches of a right, rectangular prism with side, front and bottom faces having an area 15 square inches, 10 square inches and 6 square inches, respectively?
|
gpt
|
Let the sides of the prism have lengths $x$, $y$, and $z$. We have the equations $xy=15$, $yz=10$ and $xz=6$. Multiplying these equations together, we have $xy\cdot yz \cdot xz = 15\cdot10\cdot6 \Rightarrow x^2y^2z^2=900$. Since the volume of the prism is equal to $xyz$, we take the square root of both sides to get the volume as $\sqrt{900}=\boxed{30}$ cubic inches.
|
human
|
In right $\Delta ABC$, $\angle CAB$ is a right angle. Point $M$ is the midpoint of $\overline{BC}$. What is the number of centimeters in the length of median $\overline{AM}$? Express your answer as a decimal to the nearest tenth. [asy] pair A,B,C,M;
A = (0,0); B = (4,0); C = (0,3); M = (B+C)/2;
draw(M--A--B--C--A);
label("$A$",A,W); label("$B$",B,E);
label("$C$",C,W); label("$M$",M,NE);
label("3 cm",A--C,W); label("4 cm",A--B,S);
[/asy]
|
gpt
|
The length of the median to the hypotenuse of a right triangle is half the hypotenuse. The hypotenuse of $\triangle ABC$ is $\sqrt{3^2+4^2} = 5$, so $AM = BC/2 = \boxed{2.5}$.
|
human
|
Given that $m \angle A= 60^\circ$, $BC=12$ units, $\overline{BD} \perp \overline{AC}$, $\overline{CE} \perp \overline{AB}$ and $m \angle DBC = 3m \angle ECB$, the length of segment $EC$ can be expressed in the form $a(\sqrt{b}+\sqrt{c})$ units where $b$ and $c$ have no perfect-square factors. What is the value of $a+b+c$?
[asy]
draw((0,0)--(8,.7)--(2.5,5)--cycle);
draw((0,0)--(4.2,3.7));
draw((8,.7)--(1.64,3.2));
label("$B$",(0,0),W);
label("$C$",(8,.7),E);
label("$D$",(4.2,3.7),NE);
label("$E$",(1.64,3.2),NW);
label("$A$",(2.5,5),N);
[/asy]
|
gpt
|
The diagram the problem gives is drawn very out of scale so we redraw the diagram, this time with $\overline{AC}$ as the base:
[asy]
draw((0,0)--(1+sqrt(3),0)--(1,sqrt(3))--cycle);
label("$A$",(0,0),SW); label("$C$",(1+sqrt(3),0),SE); label("$B$",(1,sqrt(3)),N);
draw((1,0)--(1,sqrt(3)));
label("$D$",(1,0),S);
draw((1+sqrt(3),0)--(.75,1.3));
label("$E$",(.75,1.3),W);
label("$y$",(2.2,.4),NW);
label("$3y$",(.95,1.55),SE); label("$60^\circ$",(.1,0),NE);
[/asy] All angles are given in degrees.
Let $\angle ECB = y$, so $\angle DBC=3y$. From $\triangle AEC$ we have $\angle ACE = 180^\circ-60^\circ-90^\circ= 30^\circ$.
Now let $EC$ and $BD$ intersect at $F$. $\angle BFE=\angle DFC$ by vertical angles and $\angle BEF=\angle CDF=90^\circ$, so $\angle FBE=\angle FCD$, which is equal to 30 degrees. Now summing the angles in $\triangle ABC$, we have $60^\circ+30^\circ+3y+y+30^\circ=180$, solving yields $4y=60$ so $y=15$ and we see $\triangle BDC$ is a 45-45-90 triangle. Also, $\triangle ABD$ is a 30-60-90 triangle.
Let $ AD = x$, so $AB = 2x$ and $DB = DC = x\sqrt{3}$. $BC = x\sqrt{3}\sqrt{2} = x\sqrt{6}$. We are given that this equals 12, so we find $x = 12/\sqrt{6} = 2\sqrt{6}$. It follows that the area of $\triangle ABC$ can be found via \[(1/2)(AC)(BD)=(1/2)(x+x\sqrt{3})(x\sqrt{3})=12\sqrt{3}+36.\] To find $EC$, notice that the area of $\triangle ABC$ can also be written as $(1/2)(AB)(EC)$. Thus, \[(1/2)(4\sqrt{6})(EC)=12\sqrt{3}+36 \Rightarrow EC = 3(\sqrt{2}+\sqrt{6}).\] Hence $a=3$, $b=2$, and $c=6$, so $a+b+c=\boxed{11}$.
|
human
|
Parallelogram $ABCD$ has vertices $A(3,3)$, $B(-3,-3)$, $C(-9,-3)$, and $D(-3,3)$. If a point is selected at random from the region determined by the parallelogram, what is the probability that the point is not above the $x$-axis? Express your answer as a common fraction.
|
gpt
|
Let us first call the point where the $x$-axis intersects side $\overline{AB}$ point $E$ and where it intersects $\overline{CD}$ point $F$. [asy]
draw((-12,0)--(6,0),Arrows);
draw((0,-6)--(0,6),Arrows);
for(int i = -11; i < 6; ++i)
{
draw((i,.5)--(i,-.5));
}
for(int i = -5; i < 6; ++i)
{
draw((.5,i)--(-.5,i));
}
dot((3,3));
dot((-3,-3));
dot((-9,-3));
dot((-3,3));
dot((0,0));
dot((-6,0));
draw((3,3)--(-3,-3)--(-9,-3)--(-3,3)--cycle, linewidth(.65));
draw((0,0)--(-6,0), linewidth(.65));
label("A",(3,3),NE);
label("B",(-3,-3),SE);
label("C",(-9,-3),SW);
label("D",(-3,3),NW);
label("F",(-6,0),NW);
label("E",(0,0),NW);
[/asy] Now, since the $x$-axis is parallel to bases $\overline{AD}$ and $\overline{BC}$ of the parallelogram, $\overline{EF}$ is parallel to the two bases and splits parallelogram $ABCD$ into two smaller parallelograms $AEFD$ and $EBCF$. Since the height of each of these parallelograms is $3$ and the length of their bases equals $AD=BC=6$, both parallelograms must have the same area. Half of parallelogram $ABCD$'s area is above the $x$-axis and half is below, so there is a $\boxed{\frac{1}{2}}$ probability that the point selected is not above the $x$-axis.
|
human
|
How many square units are in the area of the parallelogram with vertices at (0, 0), (6, 0), (2, 8) and (8, 8)?
|
gpt
|
The measure of the base of the parallelogram is 6 units and the height of the parallelogram is 8 units. Therefore, the area of the parallel is $(6)(8)=\boxed{48}$ square units. [asy]
size(4cm);
defaultpen(linewidth(0.6));
draw((-1,0)--(10,0),EndArrow(6));
draw((0,-1)--(0,10),EndArrow(6));
draw((0,0)--(6,0)--(8,8)--(2,8)--cycle);
dot((0,0));dot((6,0));dot((8,8));dot((2,8));[/asy]
|
human
|
Let $\triangle ABC$ be a right triangle such that $B$ is a right angle. A circle with diameter of $BC$ meets side $AC$ at $D.$ If the area of $\triangle ABC$ is $150$ and $AC = 25,$ then what is $BD$?
|
gpt
|
We might try sketching a diagram: [asy]
pair pA, pB, pC, pO, pD;
pA = (-15, 0);
pB = (0, 0);
pC = (0, 20);
pO = (0, 10);
pD = (-9.6, 7.2);
draw(pA--pB--pC--pA);
draw(pD--pB);
draw(circle(pO, 10));
label("$A$", pA, SW);
label("$B$", pB, S);
label("$C$", pC, N);
label("$D$", pD, W);
[/asy] Since $BC$ is a diameter of the circle, that makes $\angle BDC$ a right angle. That means that $BD$ is an altitude of $\triangle ABC.$ Then, we use the area formula to find $150 = \frac{1}{2} \cdot AC \cdot BD,$ where $AC = 25.$ Solving, we have $BD = \boxed{12}.$
|
human
|
In convex quadrilateral $ABCD$, $AB=BC=13$, $CD=DA=24$, and $\angle D=60^\circ$. Points $X$ and $Y$ are the midpoints of $\overline{BC}$ and $\overline{DA}$ respectively. Compute $XY^2$ (the square of the length of $XY$).
|
gpt
|
We begin by drawing a diagram: [asy]
pair A,B,C,D,X,Y,H;
A=(-12,12*sqrt(3)); D=(0,0); C=(12,12*sqrt(3)); B=(0,5+12*sqrt(3)); X=(B+C)/2; Y=(A+D)/2; H=(A+C)/2;
draw(A--B--C--D--cycle); draw(X--Y);
label("$A$",A,W); label("$B$",B,N); label("$C$",C,E); label("$D$",D,S); label("$X$",X,NE); label("$Y$",Y,SW);
label("$24$",D--C,SE); label("$13$",A--B,NW); label("$60^\circ$",(0,4)); draw(B--D,heavycyan); draw(A--C,heavycyan); label("$H$",H,NW);
[/asy] We draw diagonals $\overline{AC}$ and $\overline{BD}$ and let the intersection point be $H$. Since $\angle ADC=60^\circ$ and $AD=CD$, $\triangle ACD$ is equilateral, so $AC=24$. Since $ABCD$ has two pairs of equal sides, it is a kite, and so its diagonals are perpendicular and $\overline{BD}$ bisects $\overline{AC}$. Thus, \[AH=HC=24/2=12.\]Applying the Pythagorean Theorem on $\triangle BHC$ and $\triangle CHD$ gives \[BH=\sqrt{BC^2-HC^2}=\sqrt{13^2-12^2}=5\]and \[HD=\sqrt{CD^2-HC^2}=\sqrt{24^2-12^2}=12\sqrt{3}.\][asy]
size(180);
pair A,B,C,D,X,Y,H;
A=(-12,12*sqrt(3)); D=(0,0); C=(12,12*sqrt(3)); B=(0,5+12*sqrt(3)); X=(B+C)/2; Y=(A+D)/2; H=(A+C)/2;
draw(A--B--C--D--cycle); draw(X--Y);
label("$A$",A,W); label("$B$",B,N); label("$C$",C,E); label("$D$",D,S); label("$X$",X,NE); label("$Y$",Y,SW);
draw(B--D,heavycyan); draw(A--C,heavycyan); label("$H$",H,NW);
pair W; W = (C+D)/2; draw(X--W--Y,dashed); label("$Y'$",W,SE);
draw(rightanglemark(B,H,C,20),heavycyan);
[/asy]
Let $Y'$ be the midpoint of $\overline{CD}$. We look at triangle $BCD$. Since segment $\overline{XY'}$ connects midpoints $X$ and $Y'$, it is parallel to $\overline{BD}$ and has half the length of $\overline{BD}$. Thus, \[XY' = \frac{1}{2}(BH+HD)=\frac{1}{2}(5+12\sqrt{3}).\]Now, we look at triangle $ACD$. Similarly, since $Y$ and $Y'$ are midpoints, $\overline{YY'}$ is parallel to $\overline{AC}$ and has half the length of $\overline{AC}$, so \[YY' = 24/2=12.\]Since $\overline{BD} \perp \overline{AC}$, we have $\overline{XY'}\perp \overline{YY'}$, so $\angle XY'Y=90^\circ$. Finally, we use the Pythagorean theorem on $\triangle XY'Y$ to compute \begin{align*}
XY^2=YY'^2+XY'^2&=12^2+\left(\frac{1}{2}(5+12\sqrt{3})\right)^2\\
&=144+\frac{1}{4}(25+120\sqrt{3}+144\cdot 3) \\
&= \boxed{\frac{1033}{4}+30\sqrt{3}}. \end{align*}
|
human
|
The surface area of a sphere with radius $r$ is $4\pi r^2$. Including the area of its circular base, what is the total surface area of a hemisphere with radius 6 cm? Express your answer in terms of $\pi$.
[asy]
import markers;
size(150);
import geometry;
draw((0,-7)--(0,-1),Arrow);
draw((10,10)--(5,5),Arrow);
label("half of sphere",(10,10),N);
label("circular base",(0,-7),S);
draw(scale(1,.2)*arc((0,0),10,0,180),dashed);
draw(scale(1,.2)*arc((0,0),10,180,360));
draw(Arc((0,0),10,0,180));
[/asy]
|
gpt
|
The base of the hemisphere is a circle with radius 6 and area $6^2\pi=36\pi$. The curved top of the hemisphere has half the surface area of a full sphere, which has surface area $4\pi(6^2)=144\pi$, so the curved top of the hemisphere has $144\pi/2=72\pi$. The total surface area of the hemisphere is $36\pi+72\pi=\boxed{108\pi}$.
|
human
|
A fly is on the edge of a ceiling of a circular room with a radius of 58 feet. The fly walks straight across the ceiling to the opposite edge, passing through the center of the circle. It then walks straight to another point on the edge of the circle but not back through the center. The third part of the journey is straight back to the original starting point. If the third part of the journey was 80 feet long, how many total feet did the fly travel over the course of all three parts?
|
gpt
|
The fly's journey traces out the three sides of triangle. Because one side of this triangle is a diameter of the circular ceiling and the other two sides are chords of this circle, the triangle is a right triangle. If the radius of the circular room is 58 feet, then the diameter is $2 \times 58 = 116$ feet. This is the hypotenuse of the right triangle. One of the legs is 80 feet, so the other leg must be equal to $\sqrt{116^2 - 80^2} = \sqrt{(13{,}456 - 6400)} = \sqrt{7056} = 84$ feet. The total distance traveled by the fly is $116 + 84 + 80 = \boxed{280}$ feet.
|
human
|
What is the ratio of the volume of a cube with edge length six inches to the volume of a cube with edge length one foot? Express your answer as a common fraction.
|
gpt
|
Each side of the smaller cube is half that of the larger cube, so the ratio of the volumes is $\left( \frac{1}{2} \right) ^3 = \boxed{\frac{1}{8}}.$
|
human
|
Line $l_1$ has equation $3x - 2y = 1$ and goes through $A = (-1, -2)$. Line $l_2$ has equation $y = 1$ and meets line $l_1$ at point $B$. Line $l_3$ has positive slope, goes through point $A$, and meets $l_2$ at point $C$. The area of $\triangle ABC$ is $3$. What is the slope of $l_3$?
|
gpt
|
We find the coordinates of point $B$ by solving $3x-2y = 1$ and $y = 1$ simultaneously. With $y=1,$ we get $3x-2=1,$ and so $x=1.$ Thus, $B=(1,1).$ The distance from $A$ to line $l_2$ is $1 - (-2) = 3,$ so we have \[\tfrac{1}{2} \cdot BC \cdot 3 = [\triangle ABC] = 3,\]and thus $BC = 2.$ Therefore, either $C = (3, 1)$ or $C = (-1, 1).$ If $C = (3, 1),$ then the slope of $l_3$ is $\tfrac{1-(-2)}{3-(-1)} = \tfrac{3}{4},$ and if $C=(-1,1)$, then $l_3$ is a vertical line, so its slope is undefined. Therefore, the answer is $\boxed{\tfrac34}.$
[asy]
size(6cm);
pair A=(-1,-2),B=(1,1),C=(3,1),C2=(-1,-1);
filldraw(A--B--C--cycle,gray);
draw((-4,0)--(5,0), EndArrow); label("$x$",(5,0),E);
draw((0,-4)--(0,3),EndArrow); label("$y$",(0,3),N);
real l1( real x) { return (3*x-1)/2; }
real l2 (real x) { return 1; }
real l3 (real x) { return 3/4*x-5/4; }
draw(graph(l1, -2, 2),Arrows); draw(graph(l2, -2, 4.5),Arrows); draw(graph(l3, -3, 4),Arrows);
dot("$A$",A,NW);
dot("$B$",B,NNW);
dot("$C$",C,NNW);
label("$l_1$",(2,2.5),N);
label("$l_2$",(-2,1),NW);
label("$l_3$",(3.5,1.5),N);
[/asy]
|
human
|
In $\triangle XYZ$, we have $\angle X = 90^\circ$ and $\tan Z = 7$. If $YZ = 100$, then what is $XY$?
|
gpt
|
[asy]
pair X,Y,Z;
X = (0,0);
Y = (14,0);
Z = (0,2);
draw(X--Y--Z--X);
draw(rightanglemark(Y,X,Z,23));
label("$X$",X,SW);
label("$Y$",Y,SE);
label("$Z$",Z,N);
label("$100$",(Y+Z)/2,NE);
label("$k$",(Z)/2,W);
label("$7k$",Y/2,S);
[/asy]
Since $\triangle XYZ$ is a right triangle with $\angle X = 90^\circ$, we have $\tan Z = \frac{XY}{XZ}$. Since $\tan Z = 7$, we have $XY = 7k$ and $XZ = k$ for some value of $k$, as shown in the diagram. Applying the Pythagorean Theorem gives $(7k)^2 + k^2 = 100^2$, so $50k^2 = 100^2$, which gives $k^2 = 100^2/50 = 200$. Since $k$ must be positive, we have $k = \sqrt{200} = 10\sqrt{2}$, so $XY = 7k = \boxed{70\sqrt{2}}$.
|
human
|
Triangle $ABC$ has a perimeter of 2007 units. The sides have lengths that are all integer values with $AB< BC \leq AC$. What is the smallest possible value of $BC - AB$?
|
gpt
|
Since $AB$ and $BC$ are positive integers and $AB < BC$, $BC - AB$ must be at least 1.
The triangle with side lengths $AB = 650$, $BC = 651$, and $AC = 706$ satisfies the given conditions, and for this triangle $BC - AB = 1$.
Therefore, the smallest possible value of $BC - AB$ is $\boxed{1}$.
|
human
|
Circle $T$ has its center at point $T(-2,6)$. Circle $T$ is reflected across the $y$-axis and then translated 8 units down. What are the coordinates of the image of the center of circle $T$?
|
gpt
|
Since the image is reflected across the $y$-axis first, we will just change the sign of the $x$-coordinate, which will give us $(2, 6)$. Next the image is shifted down 8 units so we will subtract 8 from the $y$-coordinate, giving our image a final center of $\boxed{(2, -2)}$.
|
human
|
In triangle $ABC$, $AB = 9$, $BC = 12$, $AC = 15$, and $CD$ is the angle bisector. Find the length of $CD$.
|
gpt
|
$\triangle ABC$ is a right triangle, since $9^2 + 12^2 = 15^2$. Thus, $\angle ABC = 90^\circ$.
[asy]
unitsize(0.3 cm);
pair A, B, C, D;
A = (0,9);
B = (0,0);
C = (12,0);
D = (0,4);
draw(A--B--C--cycle);
draw(C--D);
label("$A$", A, NW);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, W);
[/asy]
By the angle bisector theorem, $BD/AD = BC/AC$, so \[BD = \frac{BC}{BC + AC} \cdot AB = \frac{4}{9} \cdot 9 = 4.\] Then by use of the Pythagorean Theorem on right triangle $BCD$, $CD = \sqrt{BC^2 + BD^2} = \sqrt{12^2 + 4^2} = \sqrt{160} = \boxed{4 \sqrt{10}}$.
|
human
|
A circle has a radius of three inches. The distance from the center of the circle to chord $CD$ is two inches. How many inches long is chord $CD$? Express your answer in simplest radical form.
|
gpt
|
Call the center of the circle O, and call the point where the radius of the circle bisects the chord E. Thus, the line segment from the center of the circle to point E has length 2, and we have $\triangle ODE$ with a leg of 2 and a hypotenuse of 3. Thus, the other leg, DE has length $\sqrt{5}$, and since DE is $\frac{CD}{2}$, we have $CD = \boxed{2\sqrt{5}}$.
|
human
|
Circles $A,B$, and $C$ are externally tangent to each other and internally tangent to circle $D$. Circles $B$ and $C$ are congruent. Circle $A$ has radius 1 and passes through the center of $D$. What is the radius of circle $B$?
[asy]unitsize(1cm);
pair A,B,C,D;
A=(-1,0);
B=(0.66,0.88);
C=(0.66,-0.88);
D=(0,0);
draw(Circle(A,1),linewidth(0.7));
draw(Circle(B,0.88),linewidth(0.7));
draw(Circle(C,0.88),linewidth(0.7));
draw(Circle(D,2),linewidth(0.7));
label("$A$",A,A);
label("$B$",B,B);
label("$C$",C,C);
label("$D$",(-1.2,1.6),NW);
[/asy]
|
gpt
|
Let $E,H$, and $F$ be the centers of circles $A,B$, and $D$, respectively, and let $G$ be the point of tangency of circles $B$ and $C$. Let $x=FG$ and $y=GH$. Since the center of circle $D$ lies on circle $A$ and the circles have a common point of tangency, the radius of circle $D$ is $2$, which is the diameter of circle $A$. Applying the Pythagorean Theorem to right triangles $EGH$ and $FGH$ gives \[
(1+y)^{2}= (1+x)^{2} + y^{2} \quad\text{and}\quad (2-y)^{2}= x^{2} + y^{2},
\] from which it follows that \[
y= x + \frac{x^2}{2} \quad\text{and}\quad y= 1 - \frac{x^2}{4}.
\] The solutions of this system are $(x,y)=(2/3, 8/9)$ and $(x,y)=(-2, 0)$. The radius of circle $B$ is the positive solution for $y$, which is $\boxed{\frac{8}{9}}$.
[asy]unitsize(2.2cm);
pair A,B,C,D;
A=(-1,0);
B=(0.66,0.88);
C=(0.66,-0.88);
D=(0,0);
draw(Circle(A,1),linewidth(0.7));
draw(Circle(B,0.88),linewidth(0.7));
draw(Circle(C,0.88),linewidth(0.7));
draw(Circle(D,2),linewidth(0.7));
label("$E$",A,W);
label("$H$",B,N);
label("$y$",(1,1.2),S);
label("$y$",(0.66,0.44),E);
label("$G$",(0.66,0),S);
label("$y$",(0.2,0.6),N);
label("$x$",(0.45,-0.1),S);
draw((0,0)--(1.2,1.6),linewidth(0.7));
label(scale(0.7)*rotate(55)*"$2-y$",(0.33,0.44),E);
label("1",(-0.8,0.2),N);
label("1",(-0.7,0),S);
draw((-1,0)--(0.66,0.88)--(0.66,0)--cycle,linewidth(0.7));
[/asy]
|
human
|
Square $BCFE$ is inscribed in right triangle $AGD$, as shown below. If $AB = 28$ units and $CD = 58$ units, what is the area of square $BCFE$?
[asy]
draw((9,15.6)--(10.7,14.6)--(11.7,16.3));
draw((0,0)--(10,17.3)--(40,0)--cycle);
draw((7,0)--(19,0)--(19,12.1)--(7,12.1)--cycle);
label("$A$",(0,0),SW);
label("$B$",(7,0),S);
label("$C$",(19,0),S);
label("$D$",(40,0),SE);
label("$E$",(7,12.1),NW);
label("$F$",(19,12.1),NE);
label("$G$",(10,17.3),N);
[/asy]
|
gpt
|
$\angle CDF = \angle AEB$ and $\angle BAE = \angle CFD$, so we know that $\bigtriangleup AEB \sim \bigtriangleup FDC$. Thus, denoting the side length of $BEFC$ as $x$, we can create the ratios: $\frac{28}{x} = \frac{x}{58} \Rightarrow x^2 = 1624$, and since $x^2$ is the area of square $BCFE$, $\boxed{1624}$ is the answer.
|
human
|
Points $A$, $B$, $C$, and $T$ are in space such that each of $\overline{TA}$, $\overline{TB}$, and $\overline{TC}$ is perpendicular to the other two. If $TA = TB = 10$ and $TC = 9$, then what is the volume of pyramid $TABC$?
|
gpt
|
[asy]
import three;
triple A = (4,8,0);
triple B= (4,0,0);
triple C = (0,0,0);
triple D = (0,8,0);
triple P = (4,8,6);
draw(B--P--D--A--B);
draw(A--P);
draw(B--D,dashed);
label("$T$",A,S);
label("$B$",B,W);
label("$C$",D,E);
label("$A$",P,N);
[/asy]
We can think of $TAB$ as the base of the pyramid, and $\overline{CT}$ as the height from apex $C$ to the base, since $\overline{CT}$ is perpendicular to face $ABT$. The area of right triangle $ABT$ is $(10)(10)/2 = 50$ square units, so the volume of the pyramid is $\frac13([ABT])(CT) = \frac13(50)(9) = \boxed{150}$ cubic units.
|
human
|
A semicircle of diameter 1 sits at the top of a semicircle of diameter 2, as shown. The shaded area inside the smaller semicircle and outside the larger semicircle is called a $\textit{lune}$. Determine the area of this lune. Express your answer in terms of $\pi$ and in simplest radical form.
[asy]
fill((0,2.73)..(1,1.73)--(-1,1.73)..cycle,gray(0.7));
draw((0,2.73)..(1,1.73)--(-1,1.73)..cycle,linewidth(0.7));
fill((0,2)..(2,0)--(-2,0)..cycle,white);
draw((0,2)..(2,0)--(-2,0)..cycle,linewidth(0.7));
draw((-1,1.73)--(1,1.73),dashed);
label("2",(0,0),S);
label("1",(0,1.73),S);
[/asy]
|
gpt
|
First note that the area of the region determined by the triangle topped by the semicircle of diameter 1 is \[
\frac{1}{2}\cdot\frac{\sqrt{3}}{2} + \frac{1}{2}\pi\displaystyle\left(\frac{1}{2}\displaystyle\right)^2 =
\frac{\sqrt{3}}{4} + \frac{1}{8}\pi.
\] The area of the lune results from subtracting from this the area of the sector of the larger semicircle, \[
\frac{1}{6}\pi(1)^2 = \frac{1}{6}\pi.
\] So the area of the lune is \[
\frac{\sqrt{3}}{4} + \frac{1}{8}\pi -\frac{1}{6}\pi=\boxed{\frac{\sqrt{3}}{4} - \frac{1}{24}\pi}.
\]
[asy]
fill((0,2.73)..(1,1.73)--(-1,1.73)..cycle,gray(0.5));
draw((0,2.73)..(1,1.73)--(-1,1.73)..cycle,linewidth(0.7));
fill((0,2)..(2,0)--(-2,0)..cycle,white);
fill((0,2)..(1,1.73)--(-1,1.73)..cycle,gray(0.7));
fill((0,0)--(1,1.73)--(-1,1.73)--cycle,gray(0.9));
draw((0,2)..(2,0)--(-2,0)..cycle,linewidth(0.7));
draw((-1,1.73)--(1,1.73),dashed);
label("2",(0,0),S);
label("1",(0,1.73),SW);
draw((0,0)--(0,1.73),dashed);
label("1",(-0.5,0.87),SW);
label("1",(0.5,0.87),SE);
label("$\frac{\sqrt{3}}{2}$",(0,0.87),E);
[/asy] Note that the answer does not depend on the position of the lune on the semicircle.
|
human
|
A pentagon is formed by cutting a triangular corner from a rectangular piece of paper. The five sides of the pentagon have lengths $13,$ $19,$ $20,$ $25$ and $31,$ in some order. Find the area of the pentagon.
|
gpt
|
Let the sides of the pentagon be $a,$ $b,$ $c,$ $d$ and $e,$ and let $r$ and $s$ be the legs of the triangular region cut off, as shown.[asy]
size(6cm);
pair A=(0,0),B=(0,5),C=(8,5),D=(8,0),E=(8,2),F=(5.5,5);
draw(A--B--C--D--A^^E--F);
label("$c$",A--B,W);
label("$d$",B--F,N);
label("$e$",E--F,SW);
label("$a$",E--D,dir(0));
label("$b$",D--A,S);
label("$r$",F--C,N);
label("$s$",C--E,dir(0));
[/asy] By the Pythagorean theorem, $r^2+s^2=e^2.$ Furthermore, we have $r=b-d$ and $s=c-a,$ which are integers because $a,b,c,$ and $d$ are integers. Thus, $e$ must be the hypotenuse of some Pythagorean triple. The possibilities for that triple are $$\{5,12,13\},\quad\{12,16,20\},\quad\{15,20,25\},\quad\{7,24,25\}.$$Also, the leg lengths $r=b-d$ and $s=c-a$ must be among the pairwise differences of the given numbers. Since $16,$ $15$ and $24$ do not appear among any of the pairwise differences of $\{13,19,20,25,31\},$ the only possible triple is $\{5,12,13\}.$ Then we may take $r=b-d=5$ and $s=c-a=12,$ and this forces $a=19,$ $b=25,$ $c=31,$ $d=20$ and $e=13.$ Hence, the area of the pentagon is $$bc - \frac12 rs = 31 \cdot 25 -\frac 12(12\cdot 5)= 775-30=\boxed{745}.$$
|
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