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10.304. Two circles are concentric, and the circumference of the smaller circle divides the larger circle into equal areas. Prove that the part of the ring enclosed between the parallel tangents to the smaller circle is equal in area to the square inscribed in the smaller circle. | ## Solution.
Let $O$ be the common center of the circles, $K$ the point of tangency of the chord $AB$ of the larger circle to the smaller circle (Fig. 10.94). Let the radius of the larger circle be $R$, and the radius of the smaller circle be $r$. Then, by the condition, $\pi r^{2}=\frac{\pi R^{2}}{2}$ and $r=\frac{R}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 48,911 |
10.305. Find the area of the circle circumscribed around a right-angled triangle, the lengths of the legs of which are the roots of the equation $a x^{2}+b x+c=0$. | ## Solution.
The radius of the circle circumscribed around a right triangle is equal to half the hypotenuse. Let \( u \) and \( v \) be the lengths of the legs, and \( w \) be the length of the hypotenuse; then \( S = \frac{\pi w^2}{4} \). Since \( u \) and \( v \) are the roots of the equation \( ax^2 + bx + c = 0 \)... | \frac{\pi(b^2-2ac)}{4a^2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,912 |
10.306. A line intersects a circle of radius $R$ at points $A$ and $B$ such that $\cup A B=45^{\circ}$, and the line perpendicular to the diameter $A M$ of the circle and passing through its center intersects it at point $K$. The line passing through point $B$ and perpendicular to $A M$ intersects it at point $C$. Find... | ## Solution.
Given that $\angle B O C=45^{\circ}$, therefore $B C=O C=\frac{R}{\sqrt{2}}, A C=R-\frac{R}{\sqrt{2}}=$ $=\frac{R(\sqrt{2}-1)}{\sqrt{2}}$ (Fig. 10.95). Since $\triangle A B C \sim \triangle A K O$, then $\frac{K O}{B C}=\frac{A O}{A C} \Rightarrow$ $\Rightarrow K O=\frac{A O \cdot B C}{A C}=\frac{R}{\sqrt... | \frac{R^{2}(3+\sqrt{2})}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,913 |
10.307. A circle is drawn through two adjacent vertices of a square such that the length of the tangent to it, drawn from the third vertex, is three times the side of the square. Find the area of the circle if the side of the square is $a$. | ## Solution.
Let $ABCD$ be the square mentioned in the problem (Fig. 10.96). The given circle passes through its vertices $A$ and $D$, $CE$ is a tangent, $AD=CD=a, CE=3a$. Extend $CD$ to intersect the circle at point $K$. Let $DK=x$. Then $CK \cdot CD = CE^2$; $(x+a) a = 9a^2$; $x = 8a$. Since $\angle ADK = 90^\circ$,... | \frac{65\pi^2}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,914 |
10.308. Given a square with side $a$. On each side of the square, outside of it, a trapezoid is constructed such that the upper bases of these trapezoids and their lateral sides form a regular dodecagon. Calculate its area. | Solution.
The desired area $S=12 S_{\triangle O A B}$, where $O A, O B$ are drawn to the adjacent vertices of a regular dodecagon (Fig. 10.97). The side of the square is $a$, so $O A=O B=\frac{a}{\sqrt{2}}$.
. Then $A B=2 h$ (since $\angle A=30^{\circ}$), $B C+A D=4 h$ (since $B C+A D=A B+C D=2 A B$). The area of the trapezoid $S=\frac{(B C+A D) h}{2}=2 h^{2}=32\left(\mathrm{~cm}^{2}\right)$, from which $h=4$ cm. Therefore, $A B=C D=8 \mathrm{~cm}, B C+... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,916 |
10.311. The lengths of the diagonals of a rhombus are in the ratio $3: 4$. How many times larger is the area of the rhombus compared to the area of the circle inscribed in it? | ## Solution.
Let point $O$ be the intersection of the diagonals and the center of the inscribed circle of radius $r$ of the rhombus $ABCD$, with $BD:AC=3:4$ (Fig. 10.100). $\frac{BO}{CO}=\frac{BD}{AC}=\frac{3}{4}$. Let $BO=3x$, then $CO=4x, BC=5x$ (from $\left.\triangle BOC\right)$. Let $P$ be the point of tangency of... | \frac{25}{6\pi} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,918 |
10.312. A regular triangle is inscribed in a circle of radius $R$, and its heights are extended to intersect the circle. These intersection points are then connected to form a new triangle. Calculate the part of the circle's area that lies outside these triangles. | ## Solution.
The desired area is equal to $6 S_{1}$, where $S_{1}$ is the area of the curvilinear triangle $A P F$ (Fig. 10.101), $S_{1}$, in turn, is the difference between the sector $A O F$ and twice the area of the triangle $A O P . \angle A O F=60^{\circ}$, therefore,
$$
S_{A O F}=\frac{1}{6} \pi R^{2} . S_{\tri... | R^{2}(\pi-\sqrt{3}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,919 |
10.313. Two circles of radius $R$ intersect in such a way that each passes through the center of the other. Two other circles of the same radius have their centers at the points of intersection of the first two circles. Find the area common to all four circles. | ## Solution.
Fig. 10.102
Each of the last two circles passes through the centers of the first two (Fig. 10.102), so the length of their common chord \(O_{1} O_{2}=R\). The desired area is ... | \frac{R^{2}(2\pi-3\sqrt{3})}{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,920 |
10.314. Given a rhombus $A B C D$, the diagonals of which are 3 and 4 cm. From the vertex of the obtuse angle $B$, two heights $B E$ and $B F$ are drawn. Calculate the area of the quadrilateral $B F D E$. | ## Solution.
The area of the rhombus $S=0.5 \cdot 3 \cdot 4=6=A D \cdot B E$ (Fig. 10.103). From $\triangle A O D$, we find $A D=\sqrt{2^{2}+1.5^{2}}=2.5$ (cm) $\Rightarrow B E=6: 2.5=2.4$ (cm). From $\triangle B D E$, we have $D E=\sqrt{B D^{2}-B E^{2}}=\sqrt{3^{2}-2.4^{2}}=1.8(\mathrm{~cm})$.
We obtain $S_{B E D F}... | 4.32\mathrm{~}^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,921 |
10.315. The ratio of the magnitudes of two angles of a triangle is 2, and the difference in the lengths of the sides opposite to them is 2 cm; the length of the third side of the triangle is 5 cm. Calculate the area of the triangle. | Solution.
$\angle A=2 \angle B, B C-A C=2$ cm, $A B=5$ cm (Fig. 10.104); we need to find $S_{\triangle A B C}$. Draw the angle bisector $A D$; then $\triangle A B C \sim \triangle A D C (\angle C-$ common, $\angle B=\angle D A C) \Rightarrow \frac{A C}{B C}=\frac{C D}{A C}, A C^{2}=B C \cdot C D\left(^{*}\right)$. Sin... | \frac{15\sqrt{7}}{4}\mathrm{~}^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,922 |
10.316. In a right-angled triangle, the distance from the midpoint of the hypotenuse to one of the legs is 5 cm, and the distance from the midpoint of this leg to the hypotenuse is 4 cm. Calculate the area of the triangle. | Solution.
Let in $\triangle A B C$ (Fig. 10.105) $\angle A C B=90^{\circ}, M$ - the midpoint of $A B$, $M N \perp B C, M N=5 \text{ cm}$. Then $N$ - the midpoint of $B C$ and $A C=2 M N=10 \text{ cm}$. Drop perpendiculars $N K$ and $C P$ to $A B$. Then $N K=4 \text{ cm}, K$ - the midpoint of $B P$ and $C P=2 N K=8 \te... | \frac{200}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,923 |
10.317. In triangle $A B C$, the following are known: $B C=15 \text{ cm}, A C=14 \text{ cm}$, $A B=13$ cm. Calculate the area of the triangle enclosed between the height and the bisector drawn from vertex $B$. | ## Solution.
Let $B H$ and $B K$ be the altitude and the angle bisector of $\triangle A B C$ (Fig. 10.106). Since $15^{2}<13^{2}+14^{2}$, $\triangle A B C$ is an acute triangle and point $H$ lies on the segment $A C$. Using Heron's formula, we find: $S_{\triangle A B C}=84 \text{ cm}^2$. Then $B H=\frac{2 S_{\triangle... | 9^2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,924 |
10.318. The bases of a trapezoid are equal to $a$ and $b$. Determine the length of the segment parallel to the bases and dividing the trapezoid into two equal-area parts. | Solution.
Let in trapezoid $ABCD$ (Fig. 10.107) $BC\|AD, EF\|BC$, the areas $S_{1}$ and $S_{2}$ of trapezoids $AEFD$ and $EBCF$ are equal, $BC=a, AD=b$. Draw $BL$ and $EN$ - the heights of trapezoids $EBCF$ and $AEFD$, respectively. Let $EF=x$, $BL=h_{1}, EN=h_{2}$. Draw $BK\|CF, EM\|CD$. Then $KF=a$, $EK=x-a, MD=x, A... | \sqrt{\frac{^{2}+b^{2}}{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,925 |
10.319. The diagonals of an isosceles trapezoid are perpendicular to each other, and its area is $a^{2}$. Determine the height of the trapezoid. | Solution.
Let in trapezoid $A B C D$ (Fig. 10.108) $A B=C D, A C \perp B D, O$ - the intersection point of $A C$ and $B D, C K$ - the height of the trapezoid. Since the trapezoid is isosceles, then $A K=\frac{A D+B C}{2}, A O=D O$, and since $\angle A O D=90^{\circ}$, then $\angle O A D=45^{\circ}$. Therefore, in the ... | a | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,926 |
10.320. The medians of one triangle are equal to the sides of another triangle. Find the ratio of the areas of these triangles. | Solution.
Extend each median of the original triangle $ABC$ by $1/3$ of its length (Fig. 10.109). The area of the resulting figure $AMBNCP$ will be $2S_{\triangle ABC} (\triangle BOC = \triangle BNC, \triangle AOB = \triangle AMB, \triangle AOC = \triangle APC)$. The lengths of the sides of each of the triangles $AOM,... | 4:3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,927 |
10.321. The medians of a triangle are 3, 4, and 5 cm. Find the area of the triangle. | ## Solution.
Let $A N, B M, C P$ be the medians of $\triangle A B C$ (Fig. 10.110), $O$-the point of their intersection, $A N=3 \text{ cm}, B M=4 \text{ cm}, C P=5 \text{ cm}$. On the extension of segment $B M$ beyond point $M$, lay off segment $D M=O M$. Then $D O=B O=$ $=\frac{2}{3} B M=\frac{8}{3}$ cm. Drop a perpe... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,928 |
10.322. In a circle with center $O$, a chord $A B$ is drawn, intersecting the diameter at point $M$ and forming an angle of $60^{\circ}$ with the diameter. Find $O M$, if $A M=10 \, \text{cm}$, and $B M=4 \, \text{cm}$. | ## Solution.
Draw $O P \perp A B$ (Fig. 10.111). Then $A P = B P = 7$ cm and $M P = 3$ cm. Since $\angle P M O = 60^{\circ}$, then $\angle M O P = 30^{\circ}$ and $O M = 2 M P = 6$ cm.
Answer: 6 cm.
 $AB=CD, AC=10$ cm, $CE$ is the height. Let $AE=x$ cm, $CE=y$ cm. Since $AE=\frac{AD+BC}{2}$, then by the condition $xy=48$ cm. From $\triangle AEC\left(\angle AEC=90^{\circ}\right): AE^{2}+CE^{2}=AC^{2} ; x^{2}+y^{2}=100$. We solve the system of equations: $\left\{\begin{arr... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,930 |
10.324. A circle is inscribed in a triangle. The lines connecting the center of the circle with the vertices divide the area of the triangle into parts with areas 4, 13, and $15 \mathrm{~cm}^{2}$. Find the sides of the triangle. | ## Solution.
Let the sides of the triangle be denoted by $a, b, c$. Then the areas of the parts of the triangle are $\frac{1}{2} a r, \frac{1}{2} b r, \frac{1}{2} c r$, i.e., $a r=8, b r=26, c r=30$, from which we get $a=\frac{8}{r}, b=\frac{26}{r}, c=\frac{30}{r}$. Using Heron's formula, we find
$S=\sqrt{\frac{32}{r}... | \frac{8}{\sqrt{3}},\frac{26}{\sqrt{3}},\frac{30}{\sqrt{3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,931 |
10.325. The base of the triangle is 20 cm, the medians of the lateral sides are 18 and 24 cm. Find the area of the triangle. | ## Solution.
Let $A F$ and $C Q$ be the medians of $\triangle A B C$ (Fig. 10.113), $M$ the point of their intersection, $A C=20 \mathrm{~cm}, A F=18 \mathrm{~cm}, C Q=24 \mathrm{~cm}$. Then $A M=\frac{2}{3} A F=12 \mathrm{~cm}$, $C M=\frac{2}{3} C Q=16 \mathrm{~cm}$. Since $A M^{2}+M C^{2}=A C^{2}$, $\triangle A M C$... | 288\mathrm{~}^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,932 |
10.326. The medians of a triangle are 5, 6, and 5 m. Find the area of the triangle. | Solution.
The area of $\triangle ABC$ is 3 times the area of $\triangle AOC$ (where $O$ is the point of intersection of the medians; see Fig. 10.114). We have $S_{\triangle AOC} = \frac{1}{2} AC \cdot OK = KC \cdot OK$. But $OK = \frac{1}{2} BK = 2M$, $KC = \sqrt{OC^2 - OK^2}$, where $OC = \frac{2}{3} MC = \frac{10}{3... | 16M^2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,933 |
10.327. Determine the area of a triangle if two of its sides are 1 and $\sqrt{15}$ cm, and the median to the third side is 2 cm. | Solution.
Let $B D$ be the median of $\triangle A B C$ (Fig. 10.115), $A B=1$ cm, $B C=\sqrt{15}$ cm, $B D=2$ cm. On the extension of segment $B D$ beyond point $D$, we lay off segment $D E=B D$. $A B C E$ is a parallelogram, so $A C^{2}+B E^{2}=2(A B+B C)^{2}$; $A C^{2}=2(1+15)-16 ; A C=4$ cm.
Then $A C^{2}=A B^{2}+... | \frac{\sqrt{15}}{2}\mathrm{~}^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,934 |
10.328. The sides of a triangle are 3, 4, and 5 cm. Determine the areas of the triangles into which the given triangle is divided by the height and median drawn to the largest side. | ## Solution.
Let in $\triangle ABC$ (Fig. 10.116) $BC=3$ cm, $AC=4$ cm, $AB=5$ cm, $CK$ - median, $CM$ - altitude. Since $AB^2=BC^2+AC^2$, $\triangle ABC$ is a right triangle. Then $S_{\triangle ABC}=\frac{1}{2} BC \cdot AC=6 \text{ cm}^2$; $BM=\frac{BC^2}{AB}=\frac{9}{5} \text{ cm}$; $CM=\frac{BC \cdot AC}{AB}=\frac{... | 3^2;0.84^2;2.16^2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,935 |
10.330. The lengths of the legs of a certain right triangle are the roots of the equation $a x^{2}+b x+c=0$. Find the radius of the circle inscribed in this triangle. | ## Solution.
Without loss of generality, we can assume that $a>0$ (the case $a<0$ is considered analogously). Then its hypotenuse is $\sqrt{m^{2}+n^{2}}$, and the radius of the inscribed circle is $r=\frac{m+n-\sqrt{m^{2}+n^{2}}}{2}$. By the condition,
$$
\begin{aligned}
& m+n=-\frac{b}{a} ; m n=\frac{c}{a} \cdot r=\... | -\frac{b+\sqrt{b^{2}-2}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,936 |
10.331. In a rectangle with sides $a$ and $b$, the angle bisectors of all angles are drawn until they intersect each other. Find the area of the quadrilateral formed by the angle bisectors. | Solution.
$K L M N$ is a quadrilateral formed by the mutual intersection of the angle bisectors of rectangle $A B C D$ (Fig. 10.118), $A B=a$, $A D=b$. Without loss of generality, assume that $b>a$. Since $A B C D$ is not a rhombus, it follows from the properties of the angle bisectors of a parallelogram that $K L M N... | \frac{(b-)^{2}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,937 |
10.332. Determine the sides of a right-angled triangle, given that its perimeter is $2 p$, and the area is $m^{2}$. | ## Solution.
Let $a$ and $b$ be the legs, and $c$ be the hypotenuse of the given triangle. Using the properties of a right triangle, we can set up the following system:
$$
\left\{\begin{array}{l}
a+b+c=2 p \\
\frac{a b}{2}=m^{2}, \\
a^{2}+b^{2}=c^{2}
\end{array} \quad \text { We have: } a+b=2 p-c, a^{2}+2 a b+b^{2}=(... | \frac{p^{2}-^{2}}{p};\frac{p^{2}+^{2}\\sqrt{(p^{2}+^{2})^{2}-8p^{2}^{2}}}{2p} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,938 |
10.333. Parallelogram $A B C D$, where $A B=153 \mathrm{~cm}, A D=180$ cm, $B E=135$ cm ( $B E$ - height), is divided into three equal-area figures by lines perpendicular to $A D$. At what distance from point $A$ are the points of intersection of these perpendiculars with $A D$? | ## Solution.
Let $K M$ and $L H$ be the lines dividing the parallelogram $A B C D$ (Fig. 10.119) into three equal-area figures, and $D F$ be the height of the parallelogram. Then $B K=E M=H D=L F$. Let $B K=x$ cm. From $\triangle A E B\left(\angle A E B=90^{\circ}\right)$: $A E=\sqrt{A B^{2}-B E^{2}}=72$ cm. Then $A M... | 96 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,939 |
10.334. Inside a square with side $a$, a semicircle is constructed on each side as a diameter. Find the area of the rosette bounded by the arcs of the semicircles. | ## Solution.
Chord $O A$ subtends an arc of $90^{\circ}$ (Fig. 10.120); therefore, the area of half a petal is $\frac{\pi a^{2}}{16}-\frac{a^{2}}{8}=\frac{a^{2}(\pi-2)}{16}$.
From this, the desired area $S=8 \cdot \frac{a^{2}(\pi-2)}{16}=\frac{a^{2}(\pi-2)}{2}$.
Answer: $\frac{a^{2}(\pi-2)}{2}$. | \frac{^{2}(\pi-2)}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,940 |
10.335. The perimeter of a sector is 28 cm, and its area is $49 \mathrm{~cm}^{2}$. Determine the length of the arc of the sector. | ## Solution.
The perimeter of the sector $p=2 r+l$, where $r$ is the radius of the sector, $l$ is the length of the arc of the sector. The area of the sector $S=\frac{r l}{2}$.
Then $\left\{\begin{array}{l}2 r+l=28, \\ r l=98,\end{array}\left\{\begin{array}{l}r=\frac{28-l}{2} \\ r l=98\end{array}, l(28-l)=196 ; l^{2}... | 14 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,941 |
10.336. An equilateral triangle $ABC$ with side $a=$ $=2$ cm has a circle inscribed in it; point $A$ is the center of a second circle with radius 1 cm. Find the area of the intersection of these circles. | ## Solution.
The desired area is the sum of the areas of segment $D m E$ of a circle with radius $R=\sqrt{3} / 3$ cm centered at $O$, inscribed in $A B C$, and segment of a circle with radius $r=1$ cm (Fig. 10.121),
. Hence, $2 x \sqrt{3}+2 x=a, x=\frac{a}{2(\sqrt{3}+1)}=\frac{a(\sqrt{3}-1)}{4}$. The area of each of the circles is $\pi x^{2}=\frac{\pi a^{2}(2-\sqrt{3})}{8}$; therefore, the desired area $S=\frac{a^... | \frac{^{2}(2\sqrt{3}-6\pi+3\pi\sqrt{3})}{8} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,943 |
10.338. A curvilinear triangle is formed by three equal mutually tangent arcs of circles of radius $R$. Find the area of this triangle. | Solution.
The desired area is the difference between the area of the equilateral $\triangle ABC$ with side $2R$ and the sum of the areas of the sectors bounded by the given arcs (Fig. 10.123): $S=S_{\triangle ABC}-3 S_{\text{sect}}=\frac{4 R^{2} \sqrt{3}}{4}-3 \cdot \frac{\pi R^{2}}{6}=\frac{R^{2}(2 \sqrt{3}-\pi)}{2}$... | \frac{R^{2}(2\sqrt{3}-\pi)}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,944 |
10.339. The center of an equilateral triangle with a side length of 6 cm coincides with the center of a circle with a radius of 2 cm. Determine the area of the part of the triangle that lies outside this circle. | Solution.
Let $O$ be the center of the equilateral $\triangle ABC$ (Fig. 10.124), $AB=6$ cm. The radius $r_{1}=2$ cm of the circle mentioned in the problem satisfies the inequality $r<r_{1}<R$, where $r=\sqrt{3}$ cm is the radius of the inscribed circle, and $R=2\sqrt{3}$ cm is the radius of the circumscribed circle o... | 2(3\sqrt{3}-\pi)^2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,945 |
10.340. A circle of radius $R$ is inscribed in a rhombus. Find the area of the rhombus if its larger diagonal is 4 times the radius of the inscribed circle.
Fig. 10.125 | ## Solution.
Let $O$ be the center of the circle inscribed in the rhombus $ABCD$ (Fig. 10.125), $M$ be the point of tangency of the circle with the side of the rhombus. Then $OM=R, AO=2R$, $\angle AMO=90^{\circ}$. Therefore, $\angle OAM=30^{\circ}, \angle BAD=60^{\circ}$. Let $BE$ be the height of the rhombus. Then $B... | \frac{8R^2\sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,946 |
10.341. The hypotenuse of a right triangle is equal to $c$. The projection of the vertex of the right angle onto the hypotenuse divides it into two segments, of which the smaller is to the larger as the larger is to the entire hypotenuse. Determine the area of the triangle. | Solution.
Let $x$ be the larger segment of the hypotenuse. Then, by the condition, $\frac{c-x}{x}=\frac{x}{c}$, $x^{2}+c x-c^{2}=0$, from which $x=\frac{c(\sqrt{5}-1)}{2}$ (the second root of the equation does not fit). Hence, $x^{2}=\frac{c^{2}(3-\sqrt{5})}{2}$. Denoting the height to the hypotenuse by $h$, we have $... | \frac{^{2}\sqrt{\sqrt{5}-2}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,947 |
10.342. The lengths of the sides and diagonals of a parallelogram are equal to $a, b, c$ and $f$ respectively. Find the angles of the parallelogram if $a^{4}+b^{4}=c^{2} f^{2}$. | Solution.
Let $a$ and $\beta$ be the angles of the given parallelogram, opposite to the diagonals $c$ and $f$ respectively, and $0^{\circ}<\alpha<90^{\circ}$. Then $c^{2}=a^{2}+b^{2}-2 a b \cos \alpha$; $f^{2}=a^{2}+b^{2}-2 a b \cos \beta=a^{2}+b^{2}+2 a b \cos \alpha$. Multiplying the equations of the system $\left\{... | 45;135 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,948 |
10.343. Determine the area of a triangle if two of its sides are 35 and 14 cm, and the bisector of the angle between them is 12 cm. | Solution.
Let $B D$ be the bisector of $\triangle A B C$ (Fig. 10.126), $A B=14$ cm, $B C=35$ cm, $B D=12$ cm, $\angle A B D=\angle C B D=\alpha$. $S_{\triangle A B C}=S_{\triangle A B D}+S_{\triangle C B D}$.
Then $\frac{1}{2} A B \cdot B C \sin \angle A B C=\frac{1}{2} A B \cdot B D \sin \angle A B D+\frac{1}{2} B ... | 235.2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,949 |
10.344. Calculate the area of the common part of two rhombuses, the lengths of the diagonals of the first of which are 4 and $6 \mathrm{~cm}$, and the second is obtained by rotating the first by $90^{\circ}$ around its center. | Solution.
The required area $S$ is equal to $4\left(S_{\triangle A O B}-S_{\triangle A F K}\right)$ (Fig. 10.127). We find $S_{\triangle A O B}=0.5 \cdot 3 \cdot 2=3 \, \text{cm}^{2}$. The side of the rhombus is $\sqrt{2^{2}+3^{2}}=\sqrt{13}$ (cm). In $\triangle A O B$, the segment $O K$ is the angle bisector; then, u... | 9.6\, | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,950 |
10.345. The radius of the circle inscribed in a triangle is 2 cm. A point of tangency of this circle divides one of the sides into segments of length 4 and $6 \mathrm{~cm}$. Determine the type of the triangle and calculate its area. | ## Solution.
Let $O$ be the center of the circle with radius $r=2$ cm, inscribed in $\triangle ABC$ (Fig. 10.128), $M, N, K$ be the points of tangency of this circle with the sides $AB, BC, AC$ of the triangle, respectively, $AK=4 \text{ cm}, KC=6 \text{ cm}$. Then $AM=AK=4 \text{ cm}, CN=CK=6 \text{ cm}, BM=BN=x \tex... | 24^2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,951 |
10.346. A circle with center $O_{1}$ is divided by the diameter $A B$ into two semicircles. In one of them, two new semicircles are constructed, based on $O_{1} A$ and $O_{1} B$ as diameters. A circle is inscribed in the curvilinear figure bounded by the contours of these three semicircles. How many times smaller is it... | ## Solution.
Let $O_{2}$ be the midpoint of $A O_{1}$ - the center of one semicircle, $O_{3}$ be the midpoint of $B O_{1}$ - the center of the second semicircle, $O_{4}$ - the center of the circle inscribed in the curvilinear figure mentioned in the problem (Fig. 10.129).
, $S_{\triangle AOB}=10 \text{ cm}^2, S_{\triangle AOD}=20 \text{ cm}^2, S_{\triangle BOC}=30 \text{ cm}^2$.
Drop perpendiculars $AM$ and $CN$ to $BD$. We have
\[
\frac{S_{\triangle AOB}}{S_{\triangl... | 120^2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,953 |
10.349. A circle of radius $R$ is divided into 4 large and 4 small parts, which alternate with each other. The large part is twice as long as the small one. Determine the area of the octagon whose vertices are the points of division of the circle. | ## Solution.
Let the small arc contain $x$ radians. Then $4 x+8 x=2 \pi, x=\frac{\pi}{6}$. Thus, the octagon contains four triangles with a central angle of $\frac{\pi}{3}$ (their total area is $4 \cdot \frac{R^{2} \sqrt{3}}{4}$) and four triangles with a central angle of $\frac{\pi}{6}$ (their total area is $4 \cdot ... | R^{2}(\sqrt{3}+1) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,954 |
10.351. Let $B D$ be the height of triangle $A B C$, and point $E$ be the midpoint of $B C$. Calculate the radius of the circle circumscribed around triangle $B D E$, if $A B=30 \text{ cm}, B C=26 \text{ cm}$ and $A C=28 \text{ cm}$. | Solution.
Let's find the area of the given triangle using Heron's formula:
$S_{\triangle A B C}=336$ cm $^{2}$. Then $B D=\frac{2 S_{\triangle A B C}}{A C}=24$ cm (Fig. 10.133). $D E$ is the median of the right $\triangle B D C$, drawn to the hypotenuse $B C$. Then $D E=\frac{1}{2} B C=$ $=13$ cm, $\cos \angle C B D=... | 16.9 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,956 |
10.353. On each median of a triangle, a point is taken that divides the median in the ratio $1: 3$, counting from the vertex. By what factor is the area of the triangle with vertices at these points smaller than the area of the original triangle | ## Solution.
Let $O$ be the point of intersection of the medians $A N, B K, C M$ of triangle $A B C$ (Fig. 10.134),
Fig. 10.134
$$
\frac{A D}{D N}=\frac{B E}{E K}=\frac{C F}{F M}=\frac{... | 2.56 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,958 |
10.354. Point $M$ lies inside an equilateral triangle $A B C$. Calculate the area of this triangle, given that $A M=B M=$ $=2 \text{ cm}, \text{a} C M=1 \text{ cm}$. | Solution.
Let the side of the triangle be $a$ and draw $M D \perp A B$ (Fig. 10.135). Since $A M=B M$, points $C, M, D$ lie on the height $C D$. In $\triangle A C D$ and $\triangle A M D$, we have $(1+M D)^{2}=a^{2}-\frac{a^{2}}{4}, M D^{2}=4-\frac{a^{2}}{4}$. Then $a^{2}=4\left(4-M D^{2}\right)$ and we get the quadra... | \frac{9\sqrt{3}+3\sqrt{15}}{8}\approx3.4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,959 |
10.355. An isosceles triangle with sides 8.5 and 5 is divided into three equal areas by perpendiculars drawn from a certain point to its sides.
Find the distance from this point to each side of the given triangle. | ## Solution.
Let $B K$ be the height of $\triangle A B C, A B=B C$ (Fig. 10.136.1). Points $D, E$, $F_{1}$ are the bases of the perpendiculars dropped from point $F$ to the sides of the triangle, dividing it into three equal areas. We will prove, using proof by contradiction, that point $F$ lies on $B K$, and point $F... | \frac{4\sqrt{3}}{3};\frac{4\sqrt{3}}{3};\frac{9-5\sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,960 |
10.356. Prove that among all rectangles inscribed in the same circle, the one with the largest area is a square. | ## Solution.
Let $x$ be the side of a rectangle inscribed in a circle of radius $R$. Then its area is $x \sqrt{4 R^{2}-x^{2}}$. The area of the inscribed square is $2 R^{2}$. We will show that $2 R^{2} \geq x \sqrt{4 R^{2}-x^{2}}$. Indeed, from the obvious inequality $\left(2 R^{2}-x^{2}\right)^{2} \geq 0$ we get $4 R... | proof | Geometry | proof | Yes | Yes | olympiads | false | 48,961 |
10.357. In trapezoid $ABCD$, the lengths of the bases $AD=24$ cm and $BC=8$ cm, and the diagonals $AC=13$ cm, $BD=5\sqrt{17}$ cm are known. Calculate the area of the trapezoid. | ## Solution.
Draw $B H \perp A D$ and $C F \perp A D$ (Fig. 10.137). Let $A H=x$; then $A F=8+x, D H=24-x$. Considering that $B H=C F$, in $\triangle A F C$ and $\triangle B H D$ we have $A C^{2}-A F^{2}=B D^{2}-D H^{2} . \Rightarrow 64 x=256, x=4$ cm.
Then $C F=\sqrt{A C^{2}-A F^{2}}=\sqrt{13^{2}-12^{2}}=5 \text{ cm... | 80 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,962 |
10.358. In trapezoid $ABCD$, the bases $AD=a, BC=b$ are given. On the extension of $BC$, a point $M$ is chosen such that the line $AM$ cuts off $1/4$ of the area of the trapezoid. Find the length of the segment $CM$. | Solution.
Draw the height $K N$ of the trapezoid $A B C D$ through the point $E$ of intersection of $A M$ and $C D$ (Fig. 10.138). Let $K N=h, K E=y, C M=x$. Then $E N=h-y$. Consider the case when $S_{\triangle A E D}=\frac{1}{4} S_{A B C D}$. Then $\frac{1}{2} A D \cdot E N=\frac{1}{4} \cdot \frac{A D+B C}{2} \cdot K... | \frac{(3-b)}{+b} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,963 |
10.359. In trapezoid $ABCD$ with base lengths $AD=12 \mathrm{~cm}, BC=8 \mathrm{~cm}$, a point $M$ is taken on ray $BC$ such that $AM$ divides the trapezoid into two figures of equal area. Find $CM$. | ## Solution.
Draw the height $K N$ of the trapezoid $A B C D$ through the point $E$ of intersection of $A M$ and $C D$ (Fig. 10.138). Let $K N=h, K E=y, C M=x$. Then $E N=h-y$.
Since $S_{\triangle A E D}=\frac{1}{2} S_{A B C D}$, we have $\frac{1}{2} A D \cdot E N=\frac{1}{2} \cdot \frac{A D+B C}{2} \cdot K N$, $12(h... | 2.4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,964 |
10.360. The center of the circle circumscribed around an isosceles trapezoid divides its height in the ratio 3:4. Find the bases of the trapezoid, if the radius of the circle is 10 and its midline is equal to the height. | ## Solution.
Let $O K=x, A L=y, B K=z$ (Fig. 10.139).
Then $M N=z+y=x+\frac{3}{4} x=\frac{7 x}{4} \cdot$ In $\triangle O K B$ and $\triangle O L A$ we have $O B^{2}=O K^{2}+B K^{2}, O A^{2}=O L^{2}+A L^{2}$. We arrive at the system of equations
$$
\left\{\begin{array}{l}
x^{2}+z^{2}=100, \\
\frac{9 x^{2}}{16}+y^{2}=... | 1216 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,965 |
11.106. The side of the base of a regular hexagonal pyramid is equal to $a$. Calculate the volume of the pyramid if it is known that its lateral surface area is 10 times greater than the area of the base. | Solution.
$S O$ is the height of the regular pyramid SABCDEF (Fig. 11.3), $A B=B C=F A=a$. Then $C O=O D=C D=a$, and $S_{A B C D E F}=6 S_{\triangle C O D}=$ $=6 \frac{a^{2} \sqrt{3}}{4}=3 \frac{a^{2} \sqrt{3}}{2}=S_{1} . S M-$ is the apothem of the lateral face CSD. Then $O M \perp C D, O M=\frac{a \sqrt{3}}{2}, \ang... | \frac{9^{3}\sqrt{11}}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,966 |
11.107. The volume of a regular octagonal prism is $8 \mathrm{~m}^{3}$, and its height is 2.2 m. Find the lateral surface area of the prism. | Solution.
Let $O$ be the center of the regular octagon $A B C D E F K N$, which is the base of the given regular prism (Fig. 11.4). The volume of the prism $V=8 \mathrm{M}^{3}$, height $H=2.2$ m, then the area of the base of the prism $S_{\text {base}}=\frac{8}{2.2}=\frac{40}{11}\left(\mathrm{m}^{2}\right)$. Let $A N=... | 16\sqrt{2.2(\sqrt{2}-1)}(\mathrm{M}^{2}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,967 |
11.108. The bases of the truncated pyramid are two regular octagons. The side of the lower base of the pyramid is 0.4 m, and the upper base is 0.3 m; the height of the truncated pyramid is 0.5 m. The truncated pyramid is completed to a full pyramid. Determine the volume of the complete pyramid. | Solution.
Let $AB$ be the side of the lower base, $A_{1}B_{1}$ be the corresponding side of the upper base of the given truncated octagonal pyramid, $CO$ be the height of the complete pyramid, $O_{1}$ be the intersection point of $CO$ and the plane of the upper base (Fig. 11.5). Then $OO_{1}$ is the height of the trun... | \frac{16}{75}(\sqrt{2}+1){}^{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,968 |
11.109. Find the volume of a regular quadrilateral pyramid with the side of the base equal to $a$, and the planar angles at the vertex equal to the angles of inclination of the lateral edges to the base. | ## Solution.
Draw the height of the regular pyramid $M A B C D-M O$ and the height of the lateral face $A M D$ (Fig. 11.7). Let $A D=a$. Since $\angle M C O=\angle A M D$, the right-angled triangles $M O C$ and $M K D$ are equal (by hypotenuse and acute angle) and, consequently, $M O=D K$. Let $D K=x, M E$ be the heig... | \frac{^{3}\sqrt{\sqrt{5}+1}}{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,969 |
11.111. In a regular quadrilateral pyramid, the dihedral angle at the lateral edge is $120^{\circ}$. Find the lateral surface area of the pyramid if the area of its diagonal section is $S$. | ## Solution.
$E O$ is the height of the regular pyramid $E A B C D, \angle B K D$ is the linear angle of the dihedral angle at the lateral edge $E C$ (Fig. 11.9), $\angle B K D=120^{\circ}$. Then $O K \perp E C, S_{\triangle A E C}=2 S_{\triangle E O C}=E C \cdot O K, D K \perp E C, S_{\triangle D E C}=\frac{1}{2} E C... | 4S | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,970 |
11.112. The base of the pyramid is a parallelogram $ABCD$ with an area of $m^2$ and such that $BD \perp AD$; the dihedral angles at the edges $AD$ and $BC$ are $45^{\circ}$, and at the edges $AB$ and $CD$ are $60^{\circ}$. Find the lateral surface area and the volume of the pyramid. | Solution.
Let $MO$ be the height of the pyramid $MABCD$ (Fig. 11.10). Through point $O$ in the plane of the base, draw a line $EF$ ($E$ belongs to $AB$, $F$ belongs to $CD$), perpendicular to the parallel lines $AB$ and $CD$. $OF$ and $OE$ are the projections of $MF$ and $ME$ on the plane of the base, respectively. Th... | \frac{^3\sqrt[4]{2}}{6},\frac{^2(\sqrt{2}+2)}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,971 |
11.113. In a slanted parallelepiped, the projection of a lateral edge onto the base plane is 5 dm, and the height is 12 dm. A section perpendicular to the lateral edge is a rhombus with an area of 24 dm² and a diagonal of 8 dm. Find the lateral surface area and the volume of the parallelepiped. | Solution.
Let $C O_{1}$ be the height of the oblique parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$ (Fig. 11.11), $C_{1} O=12$ dm, then $C O$ is the projection of $C C_{1}$ onto the base plane, $C O=5$ dm, and the rhombus $M N K L$ is the perpendicular cross-section of the parallelepiped.
. Determine the volume of the truncated pyramid. | Solution.
Let $S_{1}$ and $S_{2}$ be the areas of the bases of the given truncated pyramid, and $P_{1}$ and $P_{2}$ be the perimeters of the bases, with $P_{2}=72 \text{ m}$. Then $P_{1}=27+29+52=108 \text{ (m)}$, from which, using Heron's formula, $S_{1}=270 \text{ m}^{2}$.
$ is a perpendicular cross-section of prism $A B C D A_{1} B_{1} C_{1} D_{1}$ (Fig. 11.13). Then the height of the trapezoid is $h$. Let $S_{1}$ be the area of face $A A_{1} D_{1} D$, $S_{2}$ be the area of face $B B_{1} C_{1} C$, $l$ be the length of the lateral edge of the... | \frac{(S_{1}+S_{2})}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,974 |
11.117. The base of a right prism is an isosceles trapezoid $A B C D ; A B=C D=13$ cm, $B C=11$ cm, $A D=21$ cm. The area of its diagonal section is $180 \mathrm{~cm}^{2}$. Calculate the total surface area of the prism. | Solution.
$C K$ is the height of the base of the right prism $A B C D A_{1} B_{1} C_{1} D_{1}$ (Fig. 11.14). Given $A B=C D$, then $K D=\frac{A D-B C}{2}=5 \text{ cm}, A K=\frac{A D+B C}{2}=16 \text{ cm}$.
From $\triangle C K D\left(\angle C K D=90^{\circ}\right): C K=\sqrt{C D^{2}-K D^{2}}=12 \text{ cm}$.
From $\tr... | 906 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,975 |
11.118. The base of the parallelepiped is a rhombus with side $a$ and an acute angle of $30^{\circ}$. The diagonal of one lateral face is perpendicular to the base plane, and the lateral edge forms an angle of $60^{\circ}$ with the base plane. Find the total surface area and volume of the parallelepiped. | Solution.
In the rhombus $ABCD$, which is the base of the parallelepiped $ABCD A_1 B_1 C_1 D_1$ (Fig. 11.15), $CD = a$, $\angle BCD = 30^\circ$. Then its area $S_1 = a^2 \sin 30^\circ = \frac{a^2}{2}$. The diagonal $C_1 B$ of the lateral face $BB_1 C_1 C$, perpendicular to the plane of the base, is the height of the p... | ^2(1+2\sqrt{3}+\sqrt{13});\frac{^3\sqrt{3}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,976 |
11.119. The side of the base of a regular triangular pyramid is equal to $a$, and the height dropped from the vertex of the base to the opposite lateral face is equal to $b$. Determine the volume of the pyramid. | ## Solution.
Let $M O$ be the height of the regular pyramid $M A B C$ (Fig. 11.16). $P$ is the midpoint of $B C$. Then the plane $M P A$ is perpendicular to the plane $B M C$, and the perpendicular $A K$, dropped onto the plane $B M C$, lies in the plane $M P A$, and the point $K$ belongs to $M P$. Let $M O=h$.
Then ... | \frac{^{3}b}{12\sqrt{3^{2}-4b^{2}}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,977 |
11.120. The lateral surface area of a regular triangular pyramid is three times the area of the base. The area of the circle inscribed in the base is numerically equal to the radius of this circle. Find the volume of the pyramid. | Solution.
Let $DO$ be the height of the regular pyramid $DABC$, $K$ be the midpoint of $AB$ (Fig. 11.17). Let the radius of the circle inscribed in $\triangle ABC$, $OK=r$.
Then, by the condition, $\pi r^{2}=r$, from which $r=\frac{1}{\pi}$.
$AB=2 r \sqrt{3}, S_{\triangle ABC}=\frac{AB^{2} \sqrt{3}}{4}=3 r^{2} \sqrt... | \frac{2\sqrt{6}}{\pi^{3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,978 |
11.121. A regular triangular pyramid is intersected by a plane passing through the vertex of the base and the midpoints of two lateral edges. Find the ratio of the lateral surface area of the pyramid to the area of the base, given that the intersecting plane is perpendicular to one of the lateral faces (specify which o... | ## Solution.
Let $DO$ be the height of the regular pyramid $DABC$, $M$ be the midpoint of $DB$, $N$ be the midpoint of $DC$, and $\triangle MAN$ be the section of the pyramid mentioned in the problem (Fig. 11.18). Suppose the plane $MAN$ is perpendicular to the plane of the face $ADB$. $\triangle ADM = \triangle ADN$ ... | \sqrt{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,979 |
11.122. The sides of the bases of a regular truncated quadrilateral pyramid are 2 and 1 cm, and the height is $3 \mathrm{~cm}$. A plane is drawn through the point of intersection of the diagonals of the pyramid, parallel to the bases of the pyramid, dividing the pyramid into two parts. Find the volume of each of the re... | Solution.
$O$ and $O_{1}$ are the centers of the lower and upper bases of the right truncated pyramid $A B C D A_{1} B_{1} C_{1} D_{1}, O_{2}$ is the point of intersection of its diagonals $A C_{1}$ and $A_{1} C$ (Fig. 11.19), $O O_{1}=3 \text{ cm}, A D=2 \text{ cm}$, $A_{1} D_{1}=1$ cm. The square $A_{2} B_{2} C_{2} ... | \frac{37}{27} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,980 |
11.124. The base of a right prism is an isosceles triangle, the base of which is equal to $a$, and the angle at it is $45^{\circ}$. Determine the volume of the prism if its lateral surface area is equal to the sum of the areas of the bases. | Solution.
The base of the prism is an isosceles right triangle, its area \( S = \frac{a^2}{4} \), perimeter \( P = a(\sqrt{2} + 1) \), and the lateral surface area \( S_6 = P H = a H (\sqrt{2} + 1) \), where \( H \) is the height of the prism.
Since \( S_6 = 2 S \), then \( a H (\sqrt{2} + 1) = \frac{a^2}{2} ; H = \f... | \frac{^3(\sqrt{2}-1)}{8} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,982 |
11.125. The base of the prism $A B C A_{1} B_{1} C_{1}$ is an equilateral triangle $A B C$ with side $a$. The vertex $A_{1}$ projects onto the center of the lower base, and the edge $A A_{1}$ is inclined to the base plane at an angle of $60^{\circ}$. Determine the lateral surface area of the prism.
, $O$ is the projection of $A_{1}$ onto the plane $ABC, \angle OAB = \frac{1}{2} \angle BAC = 30^{\circ}, \angle A_{1}AO = 60^{\circ}$.
Then $\cos \angle A_{1}AB = \cos \angle A_{1}AO \cdot \cos \angle OAB = \frac{\sqrt{3}}{4}$;
$$
\sin \angle A_{1}AB... | \frac{^2\sqrt{3}(2+\sqrt{13})}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,983 |
11.126. The side of the base of a regular hexagonal pyramid is equal to $a$. All diagonal sections of it are equal. Find the volume and lateral surface area of the pyramid. | ## Solution.
Let's draw SO - the height of the regular pyramid $S A B C D E F$ (Fig. 11.22). Let $S O=H$. Then $S_{\triangle A S D}=\frac{A D \cdot S O}{2}=\frac{2 a H}{2}=a H$.
$K$ is the intersection point of the diagonals $B O$ and $A C$ of the rhombus $A B C O$.
$$
O K=\frac{1}{2} B O=\frac{a}{2}, A K=\frac{a \s... | \frac{3^{3}}{4};\frac{3^{2}\sqrt{6}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,984 |
11.127. A cube with an edge length of $a$ is cut at the corners by planes such that a regular octagon remains from each face. Determine the volume of the resulting polyhedron. | ## Solution.
Let the segment cut off by each of the specified planes on the edge of the cube be $x: A P=A M=E A_{1}=x$ (Fig. 11.23). Then $M P=E P=x \sqrt{2}$; $A P+P E+E A_{1}=A A_{1}, 2 x+x \sqrt{2}=a, \quad x=\frac{a}{2+\sqrt{2}}=\frac{a(2-\sqrt{2})}{2}$. We will calculate the volume of each of the cut-off pyramids... | \frac{7}{3}^{3}(\sqrt{2}-1) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,985 |
11.128. A cube is inscribed in a regular quadrilateral pyramid such that four of its vertices are on the apothems of the pyramid and four are in the plane of the base. All edges of the pyramid are equal, each having a length of $a$. Calculate the total surface area and volume of the cube. | Solution.
Let the edge of the cube $E F=x$ (Fig. 11.24). Consider $\triangle B S C$ (equilateral $): S E=\frac{a \sqrt{3}}{2} \cdot$ From $\triangle S O E\left(\angle S O E=90^{\circ}\right)$ :
$$
S O=\sqrt{S E^{2}-O E^{2}}=\sqrt{\frac{3 a^{2}}{4}-\frac{a^{2}}{4}}=\frac{a \sqrt{2}}{2} ;
$$
$F E=O E-O F=\frac{a}{2}-\... | \frac{3}{4}^{2};\frac{^{3}\sqrt{2}}{32} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,986 |
11.129. The height of a regular truncated quadrilateral pyramid is 3 cm, its volume is $38 \mathrm{~cm}^{3}$, and the areas of the bases are in the ratio $4: 9$. Determine the lateral surface area of the truncated pyramid. | Solution.
Let the area of the upper base be $S_{1}=4 x$ cm $^{2}$. Then the area of the lower base $S_{2}=9 x \mathrm{~cm}^{2}$, and since the volume of the truncated pyramid
Fig. 11.25 $... | 10\sqrt{19}\mathrm{~}^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,987 |
11.130. Find the ratio of the volumes of a regular tetrahedron and an octahedron, given that their total surface areas are equal. | Solution.
$P O$ is the height of the regular tetrahedron $P M N K$ (Fig. 11.26). Let $M N=a$, then $M O=\frac{a \sqrt{3}}{3}, P O=\sqrt{M P^{2}-M O^{2}}=\sqrt{a^{2}-\frac{a^{2}}{3}}=\frac{a \sqrt{2}}{\sqrt{3}}$, the volume of the tetrahedron $V_{1}=\frac{1}{3} S_{\triangle M N K} \cdot P O=\frac{1}{3} \frac{a^{2} \sqr... | \frac{1}{\sqrt{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,988 |
11.131. At the base of an oblique prism lies an equilateral triangle with a side equal to $a$. One of the lateral faces of the prism is perpendicular to the base plane and is a rhombus, the diagonal of which is equal to $b$. Find the volume of the prism. | Solution.
By condition
the face $B B_{1} C_{1} C$ is a rhombus
Fig. 11.28
(Fig. 11.28) and is perpendicular to the base. Then the height $B_{1} D$ of the rhombus is the height of the pr... | \frac{\sqrt{12^{2}-3b^{2}}}{8} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,989 |
11.132. The base of a quadrilateral pyramid is a rectangle with an area of $S$; the lateral edges of the pyramid are equal and form an angle of $45^{\circ}$ with the base plane. The angle between the diagonals of the base is $60^{\circ}$. Find the volume of the pyramid. | Solution.
Given that $ABCD$ is a rectangle, draw $EO$ as the height of the pyramid $EABCD$ (Fig. 11.29). Then $\angle EAO = 45^\circ$, and since the lateral edges of the pyramid are equal, point $O$ is the center of the circle circumscribed around the rectangle $ABCD$ - the point of intersection of the diagonals. Let ... | \frac{S\sqrt{S}\cdot\sqrt[4]{27}}{9} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,990 |
11.133. The base of the pyramid is an equilateral triangle with a side equal to $a$. One of the lateral faces is also an equilateral triangle and is perpendicular to the plane of the base. Determine the total surface area of the pyramid. | ## Solution.
Let the face $A M C$ of the pyramid $M A B C$ be an equilateral triangle and perpendicular to the plane of the base (Fig. 11.30). Then its height and median $M O$ is the height of the pyramid. Drop a perpendicular $O E$ to $A B$. Then $A E=\frac{A B}{4}=\frac{a}{4}, M E \perp A B$.
From $\triangle A E M\... | \frac{^{2}\sqrt{3}(2+\sqrt{5})}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,991 |
11.134. A regular triangular pyramid is cut by a plane perpendicular to the base and dividing two sides of the base in half. Determine the volume of the cut-off pyramid if the side of the base of the original pyramid is $a$, and the dihedral angle at the base is $45^{\circ}$. | Solution.
Let $MO$ be the height of the regular pyramid $MABC$, $\triangle ETF$ be the section of the pyramid mentioned in the problem (Fig. 11.31), and $D$ be the midpoint of
Fig. 11.31 ... | \frac{^3}{128} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,992 |
11.135. Determine the volume of a regular truncated quadrilateral pyramid if the side of the larger base is $a$, the side of the smaller base is $b$, and the acute angle of the lateral face is $60^{\circ}$. | Solution.
Let $D_{1} T$ be the height of the isosceles trapezoid $D D_{1} C_{1} C$, which is a lateral face of the regular truncated pyramid $A B C D A_{1} B_{1} C_{1} D_{1}$ (Fig. 11.32). Then $D T=\frac{C D-C_{1} D_{1}}{2}=\frac{a-b}{2} ; \quad D_{1} T=D T \operatorname{tg} \angle D_{1} D T=\frac{a-b}{2} \operatorna... | \frac{(^{3}-b^{3})\sqrt{2}}{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,993 |
11.136. The side of the base of a regular triangular pyramid is equal to $a$. A plane is drawn through one of the base edges, perpendicular to the opposite lateral edge and dividing this edge in the ratio $m: n$, counting from the vertex of the base. Determine the total surface area of the pyramid. | Solution.
The full surface area of the pyramid can be found using the formula
$$
S_{\text {full }}=\frac{a^{2} \sqrt{3}}{4}+\frac{1}{2} \cdot 3 a \cdot S D \text { (Fig. 11.33). Since } \triangle B O S \sim \triangle B K D \text { (right-angled }
$$
triangles with a common angle), we have $\frac{B D}{B S}=\frac{B K}... | \frac{^{2}\sqrt{3}}{4}(1+\sqrt{\frac{3(+2n)}{}}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,994 |
11.137. Through the vertices $A, C$ and $D_{1}$ of a rectangular parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$, a plane is drawn, forming a dihedral angle of $60^{\circ}$ with the base plane. The sides of the base are 4 and 3 cm. Find the volume of the parallelepiped. | ## Solution.
Drop a perpendicular $D K$ from point $D$ to $A C$ (Fig. 11.34). $D K$ is the projection of $D_{1} K$ on the base plane of the parallelepiped. Then $D_{1} K \perp A C, \angle D_{1} K D$ is the linear angle of the dihedral angle, which by condition is $\angle D_{1} K D=60^{\circ}$. From $\triangle A D C\le... | \frac{144\sqrt{3}}{5}(\mathrm{~}^{3}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,995 |
11.138. The base of the pyramid is a parallelogram with sides equal to 10 and $18 \mathrm{~cm}$, and the area is $90 \mathrm{~cm}^{2}$. The height of the pyramid passes through the point of intersection of the diagonals of the base and is 6 cm. Determine the lateral surface area of the pyramid. | ## Solution.
Draw $S O$ - the height of the pyramid $S A B C D$ (Fig. 11.35). Drop perpendiculars $S M, S L, S N, S K$ to $A B, B C, C D$, and $A D$ respectively. By the condition, $A B C D$ is a parallelogram, so point $O$, the intersection of the diagonals, is the midpoint of segments $M N$ and $K L$, which are the ... | 192\mathrm{~}^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,996 |
11.139. A cube is inscribed in a regular octahedron such that its vertices lie on the edges of the octahedron. By what factor is the surface area of the octahedron greater than the surface area of the inscribed cube? | Solution.
Square $ABCD$ is the common base of the regular pyramids $SABCD$ and $S_1ABCD$, which together form the given regular octahedron, $O$ is the center of this square, $MNPQM_1N_1P_1Q_1$ is the cube mentioned in the problem
}{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,997 |
11.141. The area of the section of a regular tetrahedron that has the shape of a square is $m^{2}$. Find the surface area of the tetrahedron. | Solution.
Let the square $D P Q F$ be the section of the regular tetrahedron $M A B C$ (Fig. 11.37), as mentioned in the problem statement. Then $P Q = P D = m$. Since $P Q$ is parallel to $D F$, the line $P Q$ is parallel to the plane $A B C$. Therefore, the plane $A M C$, passing through the line $P Q$, intersects t... | 4^{2}\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,998 |
11.142. In a regular triangular prism, a plane is drawn through the side of the lower base and the opposite vertex of the upper base, forming an angle of $45^{\circ}$ with the plane of the lower base. The area of the section is $S$. Find the volume of the prism. | ## Solution.
$\triangle A C_{1} B$ is the section of the regular prism $A B C A_{1} B_{1} C_{1}$ (Fig. 11.39) mentioned in the problem, and $K$ is the midpoint of $A B$. Then $C K \perp A B$, $C_{1} K \perp A B$, $\angle C_{1} K C$ is the angle of inclination of the plane $A C_{1} B$ to the plane of the base, $\angle ... | \frac{S\sqrt{S}\cdot\sqrt[4]{6}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 48,999 |
11.143. A regular triangular prism is placed inside a regular tetrahedron such that the vertices of one of its bases lie on the lateral edges of the tetrahedron, and the other base lies in the plane of the tetrahedron's base. The edge of the tetrahedron is equal to \(a\). Determine the volume of the prism, given that a... | ## Solution.
Let $SO$ be the height of the tetrahedron $SABC$, and $A_1B_1C_1A_2B_2C_2$ be the regular prism mentioned in the problem (Fig. 11.40). From $\triangle SOC (\angle SOC = 90^\circ): SO = \sqrt{SC^2 - OC^2} = \sqrt{a^2 - \frac{a^2}{3}} = \frac{a \sqrt{2}}{\sqrt{3}}$. Let the edge of the prism be $x$. Then $C... | \frac{^3(27\sqrt{2}-22\sqrt{3})}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,000 |
11.144. The base of a right prism is a right-angled triangle with a hypotenuse equal to c and an acute angle of $30^{\circ}$. A plane is drawn through the hypotenuse of the lower base and the vertex of the right angle of the upper base, forming an angle of $45^{\circ}$ with the base plane. Determine the volume of the t... | ## Solution.
The desired volume $V=\frac{1}{3} S_{\text {base }} \cdot C C_{1}$ (Fig. 11.41). Since $\angle A B C=30^{\circ}$, then $A C=\frac{c}{2}, B C=\frac{c \sqrt{3}}{2}$, hence $S_{\text {base }}=\frac{c^{2} \sqrt{3}}{8}$. On the other hand, $\quad S_{\text {base }}=\frac{1}{2} A B \cdot C D$, where $C D \perp A... | \frac{^{3}}{32} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,001 |
11.145. The lateral faces of a triangular pyramid are mutually perpendicular, and their areas are equal to $a^{2}, b^{2}$ and $c^{2}$. Determine the volume of the pyramid. | ## Solution.
Let $S A=x, S B=y, S C=z$ (Fig. 11.42). From the mutual perpendicularity of the lateral faces, it follows that these edges are mutually perpendicular, hence the desired volume $V=\frac{1}{3} S_{\triangle A S B} \cdot S C=\frac{1}{6} x y z$. By the given conditions $\frac{1}{2} x y=a^{2}, \frac{1}{2} x z=b... | \frac{1}{3}\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,002 |
11.146. The base of the pyramid is a regular hexagon with a side length of $a$. One of the lateral edges is perpendicular to the base plane and equals the side of the base. Determine the total surface area of the pyramid.
, the edge $K B$ is perpendicular to the plane of the base, $K B=A B=a$. Then $\triangle A B K=\triangle C B K$ (by two legs), $S_{\triangle A B K}=\frac{1}{2} A B \cdot B K=\frac{a^{2}}{2} \cdot \angle B A N=\angle A N M=1... | \frac{^{2}(6+\sqrt{2}+3\sqrt{3})}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,003 |
11.147. The base of the pyramid is a parallelogram with sides equal to 10 and 8 m, and one of the diagonals is 6 m. The height of the pyramid passes through the point of intersection of the diagonals of the base and is 4 m. Determine the total surface area of the pyramid. | ## Solution.
Given $A B=8 \mathrm{M}, A D=10 \mathrm{M}, B D=6 \mathrm{M}$ (Fig. 11.44). Since $6^{2}+8^{2}=10^{2}$, $\triangle A B D$ is a right triangle and $S_{\text {base }}=8 \cdot 6=48\left(\mathrm{M}^{2}\right)$.
Since $B D \perp A B$, then $S B \perp A B$, and $S_{\triangle A S B}=\frac{1}{2} A B \sqrt{S O^{2... | 8(11+\sqrt{34})(\mathrm{M}^{2}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,004 |
11.148. The areas of the bases of a truncated pyramid are $S_{1}$ and $S_{2}\left(S_{1}<S_{2}\right)$, and its volume is $V$. Determine the volume of the complete pyramid. | ## Solution.
Let $H$ be the height of the complete pyramid, $h$ be the height of the truncated pyramid, and $x=H-h$.
We have $\frac{S_{1}}{S_{2}}=\frac{x^{2}}{H^{2}}$ or $\frac{\sqrt{S_{1}}}{\sqrt{S_{2}}}=\frac{x}{x+h}$, from which $x \sqrt{S_{1}}+h \sqrt{S_{1}}=x \sqrt{S_{2}}$, $x=\frac{h \sqrt{S_{1}}}{\sqrt{S_{2}}-... | \frac{VS_{2}\sqrt{S_{2}}}{S_{2}\sqrt{S_{2}}-S_{1}\sqrt{S_{1}}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,005 |
11.149. The base of a right parallelepiped is a parallelogram, one of the angles of which is $30^{\circ}$. The area of the base is 4 dm $^{2}$. The areas of the lateral faces of the parallelepiped are 6 and 12 dm ${ }^{2}$. Find the volume of the parallelepiped. | Solution.
The volume of the parallelepiped $V=S_{\text {base }} h$, where $h$ is the height of the parallelepiped. Since the parallelepiped is a right one, the heights of the lateral faces are also equal to $h$ (Fig. 11.44). According to the problem, $\angle B A D=30^{\circ}, A B \cdot h=6$ dm $^{2}, A D \cdot h=12$ d... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,006 |
11.151. The base of an oblique parallelepiped is a rhombus $ABCD$ with a side equal to $a$ and an acute angle of $60^{\circ}$. The edge $AA_{1}$ is also equal to $a$ and forms angles of $45^{\circ}$ with the edges $AB$ and $AD$. Determine the volume of the parallelepiped. | Solution.
Let $A_{1} O$ be the height of the parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$ mentioned in the problem (Fig. 11.47). Since $\angle A_{1} A B = \angle A_{1} A D$, point $O$ lies on the bisector of $\angle B A D$ - the ray $A C$. $A A_{1}$ is an oblique to the base plane of the prism, $A O$ is its projec... | \frac{^{3}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,007 |
11.152. The centers of the faces of a regular tetrahedron serve as the vertices of a new tetrahedron (Fig. 11.48). Find the ratio of their surface areas and the ratio of their volumes. | ## Solution.
Let $M, N, K, O$ be the centers of the faces $AFC, BFC, AFB, ABC$ of the given tetrahedron, respectively, and let $D$ be the midpoint of $AC$, and $E$ be the midpoint of $BC$. Then $FM = \frac{2}{3} FD, FN = \frac{2}{3} FE$, and consequently, $MN \| DE, MN = \frac{2}{3} DE = \frac{1}{2} AB$. Let $V_1$ and... | 27:1;9:1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,008 |
11.153. In a truncated triangular pyramid, a plane is drawn through the side of the upper base parallel to the opposite lateral edge. In what ratio will the volume of the truncated pyramid be divided if the corresponding sides are in the ratio $1: 2$? | Solution.
The sides of the bases are in the ratio $1: 2$, so the areas of the bases are in the ratio $1: 4$ (Fig. 11.49). Then the volume of the truncated pyramid $V=\frac{1}{3} h\left(S_{1}+S_{2}+\sqrt{S_{1} S_{2}}\right)=\frac{1}{3} h\left(4 S_{2}+S_{2}+2 S_{2}\right)=\frac{7}{3} S_{2} h$, where $S_{2}$ is the area ... | 3:4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,009 |
11.154. The distance between any two lateral edges of an oblique triangular prism is $a$. The lateral edge is equal to $l$ and is inclined to the base plane at an angle of $60^{\circ}$. Determine the total surface area of the prism. | Solution.
Let $A_{1} O$ be the height of the oblique prism $A B C A_{1} B_{1} C_{1}$ (Fig. 11.50).
Fig. 11.50
,
Fig. 11.52
$M$ is the midpoint of $CD... | \frac{(^{2}++b^{2})}{3(+b)} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,012 |
11.158. In a triangular pyramid, each of the lateral edges of which is equal to $a$, one of the planar angles at the vertex of the pyramid is right, and each of the others is $60^{\circ}$. Calculate the volume of the pyramid. | Solution.
Let $D$ be the vertex of the pyramid $DABC$ (Fig. 11.53). According to the problem, $DA=DB=DC=a, \angle ADC=\angle BDC=60^{\circ}, \angle ADB=90^{\circ}$.
Then $\triangle ADC$ and $\triangle BDC$ are equilateral, $AC=BC=a, AB=a\sqrt{2}$. Since $AC^{2}+BC^{2}=AB^{2}$, then $\angle ACB=90^{\circ}$ and $S_{\tr... | \frac{^{3}\sqrt{2}}{12} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,013 |
11.159. The base of the pyramid is a parallelogram, the adjacent sides of which are 9 and $10 \mathrm{~cm}$, and one of the diagonals is $11 \mathrm{~cm}$. The opposite lateral edges are equal, and the length of each of the larger edges is 10.5 cm. Calculate the volume of the pyramid. | Solution.
Let the parallelogram $ABCD$ be the base of the pyramid $EABCD$ (Fig. 11.54), $AB=9 \text{~cm}, AD=10 \text{~cm}, BD=11 \text{~cm}, AE=CE, BE=DE$. $O$ is the point of intersection of the diagonals $AC$ and $BD$ of the parallelogram. Then $EO$ is the median of the isosceles triangles $AEC$ and $BED$.
Therefo... | 200~^3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 49,014 |
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