problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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class | __index_level_0__ int64 0 742k |
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1.51. The side of a regular triangle is equal to $a$. Determine the area of the part of the triangle that lies outside a circle of radius $\frac{a}{3}$, the center of which coincides with the center of the triangle. | 1.51. The desired area $S=S_{1}-S_{2}+3 S_{3}$, where $S_{1}=\frac{a^{2} \sqrt{3}}{4}-$ area of the triangle, $S_{2}=\frac{\pi a^{2}}{9}$ - area of the circle, $S_{3}$ - area of the segment cut off by the triangle from the circle. The chord of this segment is $\frac{a}{3};$ therefore,
$$
S_{3}=\frac{1}{6} \cdot \frac{... | \frac{^{2}(3\sqrt{3}-\pi)}{18} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,080 |
1.52. An isosceles triangle with an acute angle of $30^{\circ}$ at the base is circumscribed around a circle of radius 3. Determine the sides of the triangle. | 1.52. Let's draw the radius $O E \perp B C$ (Fig. 1.40). Since $\angle O B E = \frac{1}{2} \angle A B C = 60^{\circ}$, then $\angle B O E = 30^{\circ}$, i.e., $B E = \frac{1}{2} B O$. Then from
(Fig. 1.41), then $\frac{AD}{DB} = \frac{AC}{BC} = \frac{18}{12} = \frac{3}{2}$, from which $AD = 9 \text{ cm}, DB = 6 \text{ cm}$. Next, we find the length o... | 2:1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,082 |
1.54. Find the area of a triangle inscribed in a circle of radius 2 cm, if two angles of the triangle are $\frac{\pi}{3}$ and $\frac{\pi}{4}$. | 1.54. Since $\angle A C B=\frac{\pi}{4}$ (Fig.
1.42), then $A B=2 \sqrt{2}$ cm (side of the inscribed square). Draw $B D \perp A C$.
Then, considering that $\angle A B D=\frac{\pi}{6}$, we find $A D=\frac{1}{2} A B=\sqrt{2}$ (cm); further,
; therefore, the area of half a petal is $\frac{\pi a^{2}}{16}-\frac{a^{2}}{8}=\frac{a^{2}(\pi-2)}{16}$. Consequently, the desired area is
$$
S=8 \cdot \frac{a^{2}(\pi-2)}{16}=\frac{a^{2}(\pi-2)}{2}
$$
}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,084 |
1.56. Prove that among all rectangles inscribed in the same circle, the one with the largest area is a square. | 1.56. 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^{4} \ge... | proof | Geometry | proof | Yes | Yes | olympiads | false | 50,085 |
1.57. The vertices of a rectangle inscribed in a circle divide it into four arcs. Find the distance from the midpoint of one of the larger arcs to the vertices of the rectangle, if its sides are 24 and 7 cm. | 1.57. Since $A C$ is the diameter of the circle (Fig. 1.44), then $R=0.5 \sqrt{24^{2}+7^{2}}=12.5 \quad$ (cm). In $\triangle B O F$ we have $O F=$ $=\sqrt{O B^{2}-B F^{2}}=\sqrt{12.5^{2}-12^{2}}=3.5$ (cm); hence, $M F=12.5-3.5=9$ (cm), $M K=12.5+3.5=16$ (cm). From $\triangle M B F$ and $\triangle M A K$ we find the req... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,086 |
1.58. Perpendiculars are drawn from the vertex of the acute angle of a rhombus to the lines containing the sides of the rhombus to which this vertex does not belong. The length of each perpendicular is 3 cm, and the distance between their bases is $3 \sqrt{3}$ cm. Calculate the lengths of the diagonals of the rhombus. | 1.58. Since $\triangle A E F$ is isosceles (Fig. 1.45), the bisector $A M$ is perpendicular to $E F$ and lies on the diagonal of the rhombus. We find
Fig. 1.44
 is $R=\frac{a \sqrt{3}}{4}$, since $\angle A = 60^{\circ}$. The quadrilateral $KLMN$ is a rectangle because
Fig. 1... | \frac{3^2\sqrt{3}}{16} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,088 |
1.60. Given a rhombus $A B C D$, the diagonals of which are 3 and 4 cm. From the vertex of the obtuse angle $B$, heights $B E$ and $B F$ are drawn. Calculate the area of the quadrilateral $B F D E$. | 1.60. The area of the rhombus $S=0.5 \times 3 \cdot 4=6=AD \cdot BE$ (Fig. 1.47). Next, from $\triangle AOD$ we find $AD=\sqrt{2^{2}+1.5^{2}}=2.5$ (cm) and, therefore, $BE=6: 2.5=2.4$ (cm).
... | 4.32\mathrm{~}^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,089 |
1.61. Calculate the area of the common part of two rhombi, 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. | 1.61. The desired area $S$ is equal to $4\left(S_{\triangle A O B}-S_{\triangle A F K}\right)$ (Fig. 1.48). We find $S_{\triangle A O B}=0.5 \cdot 3 \cdot 2=3$ ( $\left.\mathrm{cm}^{2}\right)$. 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;... | 9.6\mathrm{~}^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,090 |
1.62. The area of a quadrilateral is $S$. Find the area of the parallelogram whose sides are equal and parallel to the diagonals of the quadrilateral. | 1.62. Since $A L \| B O$ and $L B \| A O$ (Fig. 1.49), then $A L B O$ is a parallelogram and, therefore, $S_{\triangle A L B}=S_{\triangle A O B}$. Similarly, we obtain
Fig. 1.49 $S_{\trian... | 2S | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,091 |
1.63. The entire arc of a circle with radius $R$ is divided into four large and four small segments, alternating with each other. The large segment is twice as long as the small one. Determine the area of the octagon whose vertices are the points of division of the circle's arc. | 1.63. Let the small arc contain $x$ degrees. Then $4 x+8 x=2 \pi$, from which $x=\frac{\pi}{6}$. Therefore, 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... | R^{2}(\sqrt{3}+1) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,092 |
1.64. Find the radius of the circle circumscribed around an isosceles trapezoid with bases 2 and 14 and a lateral side of 10. | 1.64. Note that the circle circumscribed around trapezoid $ABCD$ is also circumscribed around $\triangle ACD$ (Fig. 1.50), and such a circle is unique. We will find its radius using the formula $R=\frac{abc}{4S}$, where $S=S_{\triangle ACD}$. We have $S=\frac{1}{2} AD \cdot BE$, where $BE=\sqrt{AB^2 - AE^2} = \sqrt{10^... | 5\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,093 |
1.65. A circle of radius $R$ is inscribed in some angle, and the length of the chord connecting the points of tangency is $a$. Two tangents parallel to this chord are drawn, resulting in a trapezoid. Find the area of this trapezoid. | 1.65. Let $L$ and $M$ be the points of tangency (Fig. 1.51); then $N L = N M$, from which $A B = C D$, since $A D \parallel B C \parallel L M$. Draw $O K \perp L M$ and $B H \perp A D$. Then the required area $S = \frac{1}{2}(A D + B C) B H$. For the circumscribed trapezoid, we have $A D + B C = A B + C D = 2 A B$; the... | \frac{8R^{3}}{} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,094 |
1.66. A circle of radius $r$ is inscribed in a rectangular trapezoid.
Find the sides of the trapezoid if its smaller base is equal to $\frac{4 r}{3}$. | 1.66. First, let's find the side $AB = 2r$ (Fig. 1.52). Let $ED = x$. Then, using the equality $AB + CD = BC + AD$, we get
Fig. 1.52
$2r + CD = \frac{4r}{3} + \frac{4r}{3} + x$, from which... | 2r,\frac{4r}{3},\frac{10r}{3},4r | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,095 |
1.67. 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 at point $D$. The line passing through point $B$ and perpendicular to the diameter $A M$ intersects it at point $... | 1.67. By the condition, $\angle B O C=45^{\circ}$ and, consequently, $B C=O C=\frac{R}{\sqrt{2}}$, $A C=R-\frac{R}{\sqrt{2}}=\frac{R(\sqrt{2}-1)}{\sqrt{2}}$ (Fig. 1.53).
Since $\triangle A B C \sim \triangle A D O$, then $\frac{D O}{B C}=\frac{A O}{A C}$,
 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,096 |
1.68. Through the intersection point of the diagonals of a trapezoid, a line parallel to the bases is drawn, intersecting the lateral sides at points $M$ and $N$. Prove that $M N=\frac{2 a b}{a+b}$, where $a$ and $b$ are the lengths of the bases. | 1.68. Let $A D=a, B C=b$ (Fig. 1.54). Since $\triangle M B O \sim \triangle A B D$, we have $\frac{M O}{a}=\frac{B O}{B D}$, i.e., $M O=a \cdot \frac{B O}{B D}$. Similarly, $\triangle O N D \sim \triangle B C D$, so $\frac{O N}{b}=\frac{O D}{B D}$, i.e., $O N=b \cdot \frac{O D}{B D}$.
Then
$$
M N=M O+O N=\frac{a \cdo... | \frac{2}{+b} | Geometry | proof | Yes | Yes | olympiads | false | 50,097 |
1.69. In trapezoid $ABCD$, the lengths of the bases $AD=24 \, \text{cm}$, $BC=8 \, \text{cm}$, and the diagonals $AC=13 \, \text{cm}$, $BD=5 \sqrt{17} \, \text{cm}$ are known. Calculate the area of the trapezoid. | 1.69. Draw $B H \perp A D$ and $C F \perp A D$ (Fig. 1.55). 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
$$
A C^{2}-A F^{2}=B D^{2}-D H^{2}
$$
Or
$13^{2}-(8+x)^{2}=(5 \sqrt{17})^{2}-(24-x)^{2}$,
from which $64 x=256$, i.e., $x=4$ (cm).... | 80\,^2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,098 |
1.70. Given a square with side $a$. On each side of the square, outside it, a trapezoid is constructed such that the upper bases of these trapezoids and their lateral sides form a regular dodecagon. Calculate its area. | 1.70. The desired area $S=12 S_{\triangle A O C}$, where $O A$ and $O C$ are radii of the circle circumscribed around the square and the dodecagon (Fig.
1.56). Since the side of the square is $a$, then $O A=O C=\frac{a}{\sqrt{2}}$. We have
; therefore, $B C=\frac{c}{2}$, $A C=\frac{c \sqrt{3}}{2}$, from which $S_{\text {base }}=\frac{1}{2} A C \cdot B C=\frac{c^{2} \sqrt{3}}{8}$. Draw $S O$ such that $A O=O B$. Then $O$ is the center of the circumscribed circle around $\triangle A B C$, and $S O$... | \frac{^{3}\sqrt{3}}{48} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,100 |
2.2. The base of the pyramid is a triangle with sides $a, a$ and $b$. All lateral edges are inclined to the base plane at an angle of $60^{\circ}$. Determine the volume of the pyramid. | 2.2. According to the condition, $A B = B C = a$ (Fig. 2.2), therefore
$$
B H = a^{2} - \frac{b^{2}}{4} = \frac{1}{2} \sqrt{4 a^{2} - b^{2}}
$$
From this
$$
S_{\mathrm{OCH}} = \frac{1}{2} A C \cdot B H = \frac{1}{4} b \sqrt{4 a^{2} - b^{2}}
$$
Draw the height SO. Since all the edges of the pyramid are equally incli... | \frac{^{2}b\sqrt{3}}{12} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,101 |
2.3. Find the volume of a regular triangular pyramid, the height of which is equal to $h$, and the planar angles at the vertex are right angles. | 2.3. Let's draw $B D \perp A C$ (Fig. 2.3). Denote the side of the base by $a$. Since the pyramid is regular, $O D=\frac{a \sqrt{3}}{6}$, $A D=D C=\frac{a}{2}$, and $S_{\mathrm{och}}=\frac{a^{2} \sqrt{3}}{4}$. Express $a$ in terms of $h$. In $\triangle S A C$ we have
$S A=S C, \angle C S A=90^{\circ}$, hence $\angle A... | \frac{^{3}\sqrt{3}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,102 |
2.4. The base of a quadrilateral pyramid is a rectangle with a diagonal equal to $b$, and the angle between the diagonals is $60^{\circ}$. Each of the lateral edges forms an angle of $45^{\circ}$ with the base plane. Find the volume of the pyramid. | 2.4. According to the condition, $B D=b, \angle A O B=60^{\circ}$ (Fig.2.4); from this, it is easy to find that $A B=\frac{b}{2}, A D=\frac{b \sqrt{3}}{2}$. Therefore, $S_{\text {base }}=A B \cdot A D=\frac{b^{2} \sqrt{3}}{4}$. Since $\angle S A O=\angle S B O=\angle S C O=\angle S D O=45^{\circ}$, $S O$ is the height ... | \frac{b^{3}\sqrt{3}}{24} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,103 |
2.5. Find the lateral surface area of a regular triangular pyramid if the plane angle at its vertex is $90^{\circ}$, and the area of the base is $S$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 2.5. Let $a$ be the side of the base -
Fig. 2.4
Then $S=\frac{a^{2} \sqrt{3}}{4}$, from which $a=\frac{2 \sqrt{S}}{\sqrt[4]{3}}$. The lateral surface area is expressed as: $S_{\text {later... | S\sqrt{3} | Number Theory | proof | Yes | Yes | olympiads | false | 50,104 |
2.6. The center of the upper base of a cube with an edge equal to $a$ is connected to the midpoints of the sides of the lower base, which are also connected in sequential order. Calculate the total surface area of the resulting pyramid. | 2.6. Since the edge of the cube is $a$, the side of the base of the pyramid $S A B C D$ is $\frac{a \sqrt{2}}{2}$ (Fig. 2.6). Considering that $O K=\frac{1}{2} A D=\frac{a \sqrt{2}}{4}$, we will find the apothem of the pyramid:
, then $O K=\frac{1}{2} S K=\frac{h}{2}$. The base of the pyramid is a regular hexagon, so $\angle K O D=30^{\circ}$ and $K D=\frac{1}{2} O D$. Therefore, $O K^{2}=O D^{2}-K D^{2}=4 K D^{2}-K D^{2}=3 K D^{2}$,
Fig. 2.7 i.e., $K D=\frac{h \sqrt{3}}{6}, D E=2 K D=\frac{h \s... | 1.5^{2}\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,106 |
2.8. At the base of the pyramid lies a square. Two lateral faces are perpendicular to the base plane, while the other two are inclined to it at an angle of $45^{\circ}$. The middle-sized lateral edge is equal to $l$. Find the volume and the total surface area of the pyramid. | 2.8. By the condition, $S C=l, \angle S B C=90^{\circ}$, $\angle S C B=45^{\circ}$ (Fig. 2.8). Hence, $S B=B C=\frac{l}{\sqrt{2}}$. We find $V=\frac{1}{3} B C^{2} \cdot S B=\frac{l^{3} \sqrt{2}}{12}$. The total surface area can be expressed as: $S_{\text {full }}=S_{\text {base }}+2 S_{\triangle S A B}+2 S_{\triangle S... | \frac{^{3}\sqrt{2}}{12};0.5^{2}(2+\sqrt{2}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,107 |
2.9. In a regular tetrahedron $S A B C$, a section is constructed by a plane passing through the edge $A C$ and a point $K$ on the edge $S B$, such that $B K: K S=2: 1$. Find the volume of the truncated pyramid $K A B C$, if the edge of the tetrahedron is $a$. | 2.9. The desired volume is found using the formula $V_{K A B C}=\frac{1}{3} S_{\triangle A B C} \cdot K N$ (Fig. 2.9), where $S_{\triangle A B C}=\frac{a^{2} \sqrt{3}}{4}$. Since $S O \| K N$, then $B N: N O=B K: K S=2: 1$, from which $B N=\frac{2}{3} B O$. But $B O=\frac{a \sqrt{3}}{3}$, i.e.
Fig. 2.10
$\left.+\sqrt{S_{1} S_{2}}\right)$, where $S_{1}=196 \mathrm{~cm}^{2}, S_{2}=100 \ma... | 872\mathrm{~}^{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,109 |
2.11. Find the volume of a cube if the distance from its diagonal to a non-intersecting edge is $d$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 2.11. The distance from edge $A A_{1}$ to diagonal $B_{1} D$ is equal to the distance from this edge to the plane $B B_{1} D_{1} D$, i.e., the length of segment $A_{1} E$ (Fig. 2.11). Let the edge of the cube be $a$; then from $\triangle A_{1} E D_{1}$ we find $2 d^{2}=a^{2}$, from which $a=d \sqrt{2}$. Therefore, $V=a... | 2^{3}\sqrt{2} | Number Theory | proof | Yes | Yes | olympiads | false | 50,110 |
2.12. The sides of the base of a rectangular parallelepiped are in the ratio $m: n$, and the diagonal section is a square with an area equal to $Q$. Determine the volume of the parallelepiped. | 2.12. The required volume $V=S_{\text {osn }} h$, where $S_{\text {osn }}=A B \cdot A D$ (Fig.2.12), $h$ - the height of the parallelepiped. According to the condition, $B B_{1} D_{1} D$ is a square, and thus, $h=\sqrt{Q}$. Let's find $A B$ and $A D$. Since $A B: A D=m: n$, then $A D=\frac{n}{m} A B . \mathrm{B} \trian... | \frac{nQ\sqrt{Q}}{^{2}+n^{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,111 |
2.14. The base of a right parallelepiped is a rhombus with an area of $Q$. The areas of the diagonal sections are $S_{1}$ and $S_{2}$. Determine the volume and lateral surface area of the parallelepiped. | 2.14. We have $V=S_{\text {osn }} h$, where $S_{\text {osn }}=Q$ (by condition); thus, we need to find $h$. Since $A B C D$ is a rhombus, $S_{\text {osn }}=\frac{1}{2} A C \cdot B D$
Fig. 2... | \sqrt{\frac{S_{1}S_{2}Q}{2}};2\sqrt{S_{1}^{2}+S_{2}^{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,113 |
2.15. Find the lateral surface of a regular triangular prism with height $h$, if a line passing through the center of the upper base and the midpoint of the side of the lower base is inclined to the plane of the base at an angle of $60^{\circ}$. | 2.15. Let the side of the base be denoted by $a$. Then $S_{\text {6o }}=3 a h$. Draw the height $O_{1} O$ (Fig. 2.14). In $\triangle D O O_{1}$, we have $\angle O O_{1} D=30^{\circ}$, so $O_{1} D=2 O D$ and $4 O D^{2}-O D^{2}=h^{2}$, from which $O D=\frac{h \sqrt{3}}{3}$. On the other hand, $O D=\frac{a \sqrt{3}}{6}$, ... | 6^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,114 |
2.16. Find the volume of an oblique triangular prism, the base of which is an equilateral triangle with side $a$, if the lateral edge of the prism is equal to the side of the base and is inclined to the base plane at an angle of $60^{\circ}$. | 2.16. Draw $A_{1} K$ perpendicular to the plane $A B C$ (Fig. 2.15); then $V=S_{\triangle A B C} \cdot A_{1} K$, where $S_{\triangle A B C}=\frac{a^{2} \sqrt{3}}{4}$. Considering that $\angle A_{1} A K=60^{\circ}$, we find $A_{1} K=\frac{a \sqrt{3}}{2}$. Therefore,
$$
V=\frac{a^{2} \sqrt{3}}{4} \cdot \frac{a \sqrt{3}}... | \frac{3^{3}}{8} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,115 |
2.17. The base of a right prism is a rhombus. The areas of the diagonal sections of this prism are $P$ and $Q$. Find the lateral surface area of the prism. | 2.17. Let the height of the prism be $h$. Since $A B C D$ is a rhombus, then $S_{\text {side }}=4 A B \cdot h$ (Fig. 2.16). Further, $A C \perp B D$ (as diagonals of a rhombus) and,
therefore, $\quad A B=\sqrt{\frac{B D^{2}}{4}+\frac{A C^{2}}{4}}=\frac{\sqrt{B D^{2}+A C^{2}}}{2}$. Then $S_{\text {side }}=2 \sqrt{B D^{... | 2\sqrt{P^{2}+Q^{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,116 |
2.18. A regular hexagonal prism, whose lateral edges are equal to $3 \mathrm{~cm}$, is cut by a diagonal plane into two equal quadrilateral prisms. Determine the volume of the hexagonal prism if the lateral surface area of the quadrilateral prism is 30 cm $^{2}$. | 2.18. Let $AB = a$ (Fig. 2.17); then the desired volume $V = 6 \cdot \frac{a^2 \sqrt{3}}{4} \cdot h$, where $h = 3$ cm is the height of the prism. It remains to find $a$. According to the problem, the lateral surface area of the prism $ABCD A_1 B_1 C_1 D_1$ is $30 \, \text{cm}^2$. But $S_{\text{lat}} = (3 AB + AD) h$; ... | 18\sqrt{3}\,^3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,117 |
2.19. The lateral surface of a cone is unfolded on a plane into a sector with a central angle of $120^{\circ}$ and an area of $S$. Find the volume of the cone. | 2.19. Let $r$ be the radius of the base of the cone, and $l$ be its slant height. Then the area of the lateral surface $S=\frac{1}{2} l^{2} \cdot \frac{2 \pi}{3}=\frac{\pi l^{2}}{3}$, from which $l=\sqrt{\frac{3 S}{\pi}}$. But $S=\pi r l=\pi r \sqrt{\frac{3 S}{\pi}}$ and, therefore, $r=\sqrt{\frac{S}{3 \pi}}$. The volu... | \frac{2S\sqrt{6\piS}}{27\pi} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,118 |
2.20. Prove that if two equal cones have a common height and parallel bases, then the volume of their common part is $\frac{1}{4}$ of the volume of each of them. | 2.20. Let the radius of the base of each of the given cones be $R$, and the height be $H$ (Fig. 2.18). Then the volume of each cone $V_{1}=\frac{1}{3} \pi R^{2} H$. The common part consists of two cones; its volume $V_{2}=\frac{2}{3} \pi r^{2} h$, where $r$ is the radius of the base, and $h$ is the height. Consider the... | proof | Geometry | proof | Yes | Yes | olympiads | false | 50,119 |
2.21. The height of the cylinder is $H$, and the radius of its base is $R$. Inside the cylinder, a pyramid is placed, the height of which coincides with the generatrix $A A_{1}$ of the cylinder, and the base is an isosceles triangle $A B C(A B=A C)$ inscribed in the base of the cylinder. Find the lateral surface area o... | 2.21. Draw $A D \perp B C$ and connect points $A_{1}$ and $D$ (Fig. 2.19). According to the theorem of three perpendiculars, we have $A_{1} D \perp B C$. Since arc $C A B$ contains $120^{\circ}$, and arcs $A C$ and $A B$ each contain $60^{\circ}$, then $B C=R \sqrt{3}$, $A B=R$. In $\triangle A B D$ we have $A D=\frac{... | 0.25R(4H+\sqrt{3R^{2}+12H^{2}}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,120 |
2.22. A metal sphere of radius $R$ is melted and recast into a cone, the lateral surface area of which is three times the area of the base. Calculate the height of the cone. | 2.22. The $z$ a n i e. Use the fact that the volumes of a sphere and a cone are $\frac{4}{3} \pi R^{3}$, as well as the equality $3 \pi r^{2}=\pi r l$, where $r$ is the radius of the base of the cone, and $l$ is its slant height.
Answer: $2 R \sqrt[3]{4}$. | 2R\sqrt[3]{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,121 |
2.23. A cone and a hemisphere have a common base, the radius of which is equal to $R$. Find the lateral surface area of the cone, if its volume is equal to the volume of the hemisphere. | 2.23. Since the volumes of the cone and the hemisphere are equal, then \( V = \frac{2}{3} \pi R^{3} \); on the other hand, \( V = \frac{1}{3} \pi R^{2} h \), where \( h \) is the height of the cone, i.e., \( h = 2 R \). We have \( S_{\text {side }} = \pi R l \), where \( l = \sqrt{R^{2} + h^{2}} = R \sqrt{5} \). Theref... | \piR^{2}\sqrt{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,122 |
2.24. An isosceles trapezoid with bases 2 and 3 cm and an acute angle of $60^{\circ}$ rotates around the smaller base. Calculate the surface area and volume of the resulting solid of revolution. | 2.24. The desired volume $V=V_{\text{cyl}}-2 V_{\text{cone}}$, and the surface area $S=S_{\text{cyl}}+2 S_{\text{cone}}$, where $V_{\text{cyl}}, V_{\text{cone}}$ and $S_{\text{cyl}}, S_{\text{cone}}$ are the volumes and lateral surfaces of the cylinder and cone, respectively (Fig. 2.20). We have $V_{\text{cyl}}=\pi r^{... | 4\pi\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,123 |
2.25. A rhombus rotates around its larger diagonal, and then around its smaller diagonal. Prove that the ratio of the volumes of the resulting figures of revolution is equal to the ratio of the areas of their surfaces. | ### 2.25. The side of the rhombus is equal to
$a$, and its diagonals are equal to $2 d_{1}$ and $2 d_{2}$ (Fig. 2.21). When rotated, a body consisting of two cones is formed. Let the volume and surface area of the solid of revolution around diagonal $A C$ be denoted by $V_{A C}$ and $S_{A C}$, and around diagonal $B D... | proof | Geometry | proof | Yes | Yes | olympiads | false | 50,124 |
2.26. A cone with base radius $R$ is inscribed in an arbitrary pyramid, the perimeter of the base of which is $2 p$. Determine the ratio of the volumes and the ratio of the lateral surface areas of the cone and the pyramid. | 2.26. Let the total height of the cone and the pyramid be $H$ (Fig. 2.22). Denote the volumes of the cone and the pyramid by $V_{1}$ and $V_{2}$, and their lateral surfaces by $S_{1}$ and $S_{2}$; then $V_{1}=\frac{1}{3} \pi R^{2} H, S_{1}=\pi R l$, where $l-$
. Let the radius of the sphere be \( R \): then \( AB = R \sqrt{3}, ... | 9:16 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,126 |
2.28. Find the ratio of the surface area and volume of a sphere to the surface area and volume of an inscribed cube.
## Group 5 | 2.28. Let the radius of the sphere be $R$, and the edge of the cube be $a$; then $R^{2}-\left(\frac{a}{2}\right)^{2}=\frac{a^{2}}{2}$, from which $a=\frac{2 R}{\sqrt{3}}$. Denote the volumes and surfaces of the sphere and the cube by $V_{1}, V_{2}$ and $S_{1}, S_{2}$, respectively. We have $V_{1}=\frac{4}{3} \pi R^{3},... | \pi:2;\pi\sqrt{3}:2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,127 |
2.29. The lateral surface area of a regular triangular pyramid is 3 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. | 2.29. The volume of the pyramid $V=\frac{1}{3} S_{\text {base }} \cdot S O$ (Fig. 2.24). Let the side of the base be $a$, then
$$
S_{\text{base}}=\frac{a^{2} \sqrt{3}}{4}, O D=\frac{a \sqrt{3}}{6} \text{. By the condition, }
$$
Fig. 2.24
$\pi O D^{2}=O D$, or $O D=\frac{1}{\pi}$, hence $\frac{a \sqrt{3}}{6}=\frac{1}{... | \frac{2\sqrt{6}}{\pi^{3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,128 |
2.30. Find the volume of a regular triangular pyramid, where the dihedral angle at the vertex is $90^{\circ}$, and the distance between a lateral edge and the opposite side of the base is $d$. | 2.30. According to the condition, $B S \perp S A$ and $B S \perp S C$ (Fig. 2.25), i.e., $B S$ is perpendicular to the face $S A C$ and $S D=d$. Therefore, the required volume $V=\frac{1}{3} S_{\triangle A C S} \cdot B S$. In $\triangle S A D$ we have $\angle S D A=90^{\circ}, \angle A S D=45^{\circ}$, from which $A D=... | \frac{^{3}\sqrt{2}}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,129 |
2.31. The side of the base of a regular triangular pyramid is equal to $a$. A plane is drawn through one of the sides of the base, perpendicular to the opposite lateral edge and dividing this edge in the ratio $\boldsymbol{m}: \boldsymbol{n}$, counting from the vertex of the base. Determine the total surface area of th... | 2.31. The full surface area of the pyramid can be found using the formula $S_{\text {full }}=\frac{a^{2} \sqrt{3}}{4}+\frac{1}{2} \cdot 3 a \cdot S D$ (Fig. 2.26). Since $\triangle B O S \sim \triangle B K D$ (right triangles with a common angle), we have
$$
\frac{B D}{B S}=\frac{B K}{B O}, \text { or } \frac{a \sqrt{... | \frac{^{2}\sqrt{3}}{4}(1+\sqrt{\frac{3(+2n)}{}}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,130 |
2.32. The centers of the faces of a regular tetrahedron serve as the vertices of a new tetrahedron. Find the ratio of their surface areas and the ratio of their volumes. | 2.32. Let the volumes and surface areas of the given and new tetrahedra be denoted by \( V_{1}, S_{1} \) and \( V_{2}, S_{2} \) respectively. Let \( AB = a \) (Fig. 2.27); then \( DK = \frac{a}{2} \). Since \( \triangle DSK \sim \triangle MSN \), we have \( DK : MN = SD : SM = \frac{3}{2} \), from which \( MN = \frac{a... | 9:1;27:1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,131 |
2.33. 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 $S$. | ### 2.33. The sought lateral surface is expressed as: $S_{\text {side}}=2 A B \cdot S M$ (Fig. 2.28). Let $A B=a, S O=h$; we need to find the relationship between $a$ and $h$. Draw $A K \perp S B$, $C K \perp S B$; since $\angle A K C=120^{\circ}$, then $\angle A K O=60^{\circ}$ and from $\triangle A K O$ we get $O K=A... | 4S | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,132 |
2.34. The base of the pyramid is a rhombus with diagonals $d_{1}$ and $d_{2}$. The height of the pyramid passes through the vertex of the acute angle of the rhombus. The area of the diagonal section, passing through the smaller diagonal, is $Q$. Calculate the volume of the pyramid given that $d_{1}>d_{2}$. | 2.34. The desired volume $V=\frac{1}{3} S_{\text {osn }} \cdot S A$, where $S_{\text {osn }}=\frac{1}{2} d_{1} d_{2}$ (Fig.
2.29). In $\triangle S A O$ we have $S A=\sqrt{S O^{2}-A O^{2}}$, and $A O=\frac{d_{1}}{2}$, while
. From the equality $6^{2} + 8^{2} = 10^{2}$, it follows that $\triangle A B D$ is a right triangle and $S_{\text{base}} = 8 \cdot 6 = 48 \left(\mathrm{M}^{2}\right)$. Since $B O \perp A B$, then $S B \perp A B$, and
\mathrm{M}^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,134 |
2.36. 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 volume of the pyramid. | 2.36. Let $A D$ be denoted as $x$, and $B D$ as $y$ (Fig. 2.31). Then $S_{\text {base }}=x y=m^{2}$. Since $B D \perp A D$, it follows that $S D \perp A D$, i.e., $\angle S D B$ is the dihedral angle and, by the condition, $\angle S D B=45^{\circ}$. Through point $O$, draw $M K \perp A B$; we have $\angle S M K=\angle ... | \frac{^{2}(2+\sqrt{2})}{2};\frac{^{3}\sqrt[4]{2}}{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,135 |
2.37. The base of the pyramid is a regular hexagon with a side equal to $a$. 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. | 2.37. We have \( S_{\text {full }}=S_{\text {base }}+2 S_{\triangle A S F}+2 S_{\triangle F S E}+2 S_{\triangle E S D} \) (Fig. 2.32), where \( S_{\text{base}}=6 \cdot \frac{a^{2} \sqrt{3}}{4}=\frac{3 a^{2} \sqrt{3}}{2}, S_{\triangle A S F}=\frac{a^{2}}{2} \), since \( S A \perp A F \) and \( S A=a \). Further, since \... | 0.5^{2}(6+3\sqrt{3}+\sqrt{7}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,136 |
2.38. Through the median $B E$ of the base $A B C$ of the pyramid $A B C D$ and the midpoint $F$ of the edge $D C$, a plane is drawn. Find the volume of the figure $A D B F E$, if the volume of the pyramid $A B C D$ is $40 \mathrm{~cm}^{3}$. | 2.38. The volume of the figure $A D B F E$ is equal to the difference in volumes of pyramids $A B C D$ and $E C B F$ (Fig. 2.33). To find the volume of pyramid $E C B F$, we compare it with the volume of pyramid $A B C D$. For this, it is sufficient to find the ratios of the areas of their bases and corresponding heigh... | 30\mathrm{~}^{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,137 |
2.39. In a truncated triangular pyramid, the height is $10 \mathrm{M}$, the sides of one base are 27, 29, and 52 m, and the perimeter of the other base is 72 m. Determine the volume of the truncated pyramid. | 2.39. The desired volume $V=\frac{1}{3} H\left(S_{1}+S_{2}+\sqrt{S_{1} S_{2}}\right)$, where $S_{1}$ is found using Heron's formula. We have
Fig. 2.34. Thus, $2 p_{1}=27+29+52=108$ (m), from... | 1900\mathrm{M}^{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,138 |
2.40. In a truncated triangular pyramid, a plane parallel to the opposite lateral edge is drawn through the side of the upper base. In what ratio is the volume of the truncated pyramid divided if the corresponding sides of the bases are in the ratio $1: 2$? | 2.40. Since the sides of the bases are in the ratio $1: 2$, the areas of the bases are in the ratio $1: 4$ (Fig. 2.35). Then the volume of the truncated pyramid is
Fig. 2.35
$$
V=\frac{1}{3} h\left(S_{1}+S_{2}+\sqrt{S_{1} S_{2}}\right)=\frac{1}{3} h\left(4 S_{2}+S_{2}+2 S_{2}\right)=\frac{7}{3} S_{2} h
$$
where $S_{... | 3:4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,139 |
2.41. The sides of the bases of a regular truncated quadrilateral pyramid are 2 and 1 cm, and the height is 3 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 them. | 2.41. Let $O_{1} O_{2}=x$ (Fig. 2.36); then $O O_{2}=3-x$. Since $\Delta B_{1} O_{2} D_{1} \sim \Delta B O_{2} D$, we have $B_{1} D_{1}: B D = O_{1} O_{2}: O O_{2}$, or $1: 2 = x: (3-x)$, from which $x=1$ (cm). Further, $\Delta B_{1} D_{1} B \sim \Delta L O_{2} B$ and, therefore, $B_{1} D_{1}: L O_{2} = O O_{1}: O O_{2... | \frac{152}{27} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,140 |
2.42. 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. | 2.42. Instruction. Use the equality $\frac{S_{1}}{S_{2}}=\frac{x^{2}}{H^{2}}$, where $\boldsymbol{x}=\boldsymbol{H}-\boldsymbol{h}(\boldsymbol{H}-$ the height of the full pyramid, and $h$ - the height of the truncated pyramid).
Answer: $\frac{V S_{2} \sqrt{S_{2}}}{S_{2} \sqrt{S_{2}}-S_{1} \sqrt{S_{1}}}$. | \frac{VS_{2}\sqrt{S_{2}}}{S_{2}\sqrt{S_{2}}-S_{1}\sqrt{S_{1}}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,141 |
2.43. Find the distance between the midpoints of two skew edges of a cube, the total surface area of which is $36 \mathrm{~cm}^{2}$. | 2.43. Since the total surface area of the cube $S_{\text {full }}=36 \mathrm{~cm}^{2}$, the area of one face $S=6 \mathrm{~cm}^{2}$ and the edge of the cube
Fig. 2.37
$A D=a=\sqrt{6}$ cm (... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,142 |
2.45. 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. | 2.45. Since $C D=3 \text{ cm}, A D=$ $=4 \text{ cm}$ (Fig. 2.39), from $\triangle A D C$ we find $\quad A C=\sqrt{3^{2}+4^{2}}=5 \quad$ (cm). Draw $D K \perp A C$, then $D_{1} K \perp A C$ and $\angle D K D_{1}=60^{\circ}$ (by condition). But $\triangle A D C \sim \triangle C D K$, hence $\frac{A C}{C D}=\frac{A D}{D K... | \frac{144\sqrt{3}}{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,144 |
2.46. 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 dm ${}^{2}$ and a diagonal of 8 dm. Find the lateral surface area and the volume of the parallelepiped. | 2.46. According to the condition, $A_{1} K=12$ dm, $A K=5$ dm, $A_{1} K \perp A K$ (Fig. 2.40); therefore, $A A_{1}=\sqrt{A_{1} K^{2}+A K^{2}}=13$ (dm). Since $S_{A_{1} L M N}=$ $=24$ dm $^{2}$ and $A A_{1}$ is perpendicular to the section, the required volume is
$$
V=S_{A_{1} L M N} \cdot A A_{1}=24 \cdot 13=312\left... | 260 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,145 |
2.47. 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}$ is projected 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. | 2.47. Let's draw a section through $B C$ perpendicular to $A A_{1}$ (Fig. 2.41); then $S_{\text {side }}=(B C+M C+$ $+M B) A A_{1}=(B C+2 M C) A A_{1}$. Since $\angle A_{1} A O=60^{\circ}$ (by the problem statement), then
}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,146 |
2.48. In an oblique triangular prism, the distances between the lateral edges are equal to $a, b$ and $c$. The lateral edge is equal to $l$, and the height of the prism is $h$. Determine the total surface area of the prism. | 2.48. The total surface area of the prism $S_{\text {full }}=2 S_{\text {base }}+S_{\text {lateral }}$. Since $a$, $b, c$ are the distances between the lateral edges of the prism, $a+b+c$ is the perimeter of the section perpendicular to the edge. Therefore, $S_{\text {lateral }}=$ $=(a+b+c) l=2 p l$, where $p=\frac{1}{... | \frac{2}{}(\sqrt{p(p-)(p-b)(p-)}+p) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,147 |
2.50. 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... | 2.50. The desired volume $V=\frac{1}{3} S_{\text {base }} \cdot C C_{1}$ (Fig. 2.42). Considering that $\angle A B C=30^{\circ}$, we find $A C=\frac{c}{2}, B C=\frac{c \sqrt{3}}{2}$, and thus, $S_{\text {base }}=\frac{c^{2} \sqrt{3}}{8}$. On the other hand, $S_{\text {base }}=\frac{1}{2} A B \cdot C D$, where $C D \per... | \frac{^{3}}{32} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,149 |
2.51. A plane passing through the vertex of a cone intersects the base along a chord, the length of which is equal to the radius of this base. Determine the ratio of the volumes of the resulting parts of the cone. | 2.51. Let $A O=r, S O=h$ (Fig. 2.43), $V$ - the volume of the cone, $V_{1}$ and $V_{2}$ the volumes of its parts. We will find $V_{1}$ as the difference between the volumes of the part of the cone whose base is the sector $A O B$, and the pyramid whose base is $\triangle A O B$. According to the condition, $A B=r$, i.e... | (2\pi-3\sqrt{3}):(10\pi+3\sqrt{3}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,150 |
2.52. The radius of the base of the cone is $R$, and the angle of the sector of its lateral surface is $90^{\circ}$. Determine the volume of the cone. | 2.52. Let $l$ be the slant height of the cone. Since the length of the arc of the lateral surface development is equal to the circumference of the base, we have $2 \pi R=\frac{2 \pi l}{4}$ (by the condition, the development is a quarter circle), from which $l=4 R$. Let's find the height of the cone: $h=\sqrt{l^{2}-R^{2... | \frac{\piR^{3}\sqrt{15}}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,151 |
2.54. A parallelogram with a perimeter of $2 p$ rotates around an axis perpendicular to a diagonal of length $d$ and passing through its end. Find the surface area of the solid of revolution. | 2.54. The surface $S$ of the body of revolution consists of the lateral surfaces of two truncated cones obtained by rotating segments $B C$ and $C D$ (Fig. $2.44$), and two cones obtained by rotating segments $A B$ and $A D$. Thus,
$$
\begin{gathered}
S=\pi(K B+A C) B C+ \\
+\pi(M D+A C) C D+ \\
+\pi K B \cdot A B+\pi... | 2\pi | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,153 |
2.55. The base of the pyramid is an equilateral triangle with a side length of $a$. One of the lateral edges is perpendicular to the base plane and equals $b$. Find the radius of the sphere circumscribed around the pyramid. | 2.55. Let $O$ be the center of the sphere circumscribed around the pyramid $ABCD$ (Fig. 2.45). Then $OA = OB = OC = OD$. Drop the perpendicular $OK$ to the plane $ABC$ and draw $OE \perp DB$. Since point $O$ is equidistant from the vertices of triangle $ABC$, point $K$ is the
. The radius of the circle in the obtained section is equal to the radius of the sphere. Since all the edges of the pyramid are equal to $a$, then
$$
S D=\frac{a \sqrt{3}}{2}, O D=\frac{a \sqrt{3}}... | \frac{\pi^{2}}{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,155 |
2.57. A regular truncated hexagonal pyramid is inscribed in a sphere of radius $R$, with the plane of the lower base passing through the center of the sphere, and the lateral edge forming an angle of $60^{\circ}$ with the base plane. Determine the volume of the pyramid. | 2.57. According to the condition, $\angle O A A_{1}=60^{\circ}$ (Fig. 2.47); hence, $\angle O_{1} O A_{1}=$ $30^{\circ}$ and $A_{1} O_{1}=\frac{1}{2} A_{1} O=\frac{R}{2}, O O_{1}=\frac{R \sqrt{3}}{2}$. We find
; then $A C=\boldsymbol{r}$.
Fig. 2.48
$B C=h$. By... | SL | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,158 |
2.60. Determine the lateral surface area and volume of a truncated cone with slant height $l$, circumscribed about a sphere of radius $r$. | 2.60. To find the lateral surface area of a truncated cone, we use the formula \( S_{\text {bok }}=\pi\left(r_{1}+r_{2}\right) l \), where \( r_{1} \) and \( r_{2} \) are the radii of the bases of the truncated cone. We draw a plane through the height of the cone. In the cross-section, we get an isosceles trapezoid cir... | \pi^{2};\frac{2}{3}\pir(^{2}-r^{2}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,159 |
3.1. The sum of two unequal heights of an isosceles triangle is equal to $l$, the angle at the vertex is $\alpha$. Find the lateral side. | 3.1. Given $A B=A C$, $A A_{1} \perp B C$, $B B_{1} \perp A C$, $\angle B A C=\alpha$, $A A_{1} + B B_{1}=l$ (Fig. 3.6). Let $B C=a$. From $\triangle A A_{1} C$ we find $A A_{1}=\frac{a}{2} \operatorname{ctg} \frac{\alpha}{2}$, $A C=\frac{a}{2 \sin \frac{\alpha}{2}}$, and from $\triangle B B_{1} C$ we get $B B_{1}=a \s... | \frac{}{\cos\frac{\alpha}{2}+\sin\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,160 |
3.2. In an isosceles triangle, the base $a$ and the angle $\alpha$ at the base are given. Find the length of the median drawn to the lateral side. | 3.2. I n d i c a t i o n. Express the lateral sides of the triangle in terms of $a$ and $\alpha$, and then use formula (1.35), which relates the length of the median of a triangle to the lengths of its sides.
A n s w e r: $\frac{a}{4} \sqrt{9+\operatorname{tg}^{2} \alpha}$. | \frac{}{4}\sqrt{9+\operatorname{tg}^{2}\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,161 |
3.3. The base of an isosceles triangle is equal to $a$, and the angle at the vertex is $\alpha$. Find the length of the bisector drawn to the lateral side. | 3.3. By the condition, $A B=A C, B C=a, \angle B A C=\alpha, \angle A B B_{1}=\angle B_{1} B C$ (Fig. 3.7). We have $\angle A B C=\angle A C B=\frac{\pi}{2}-\frac{\alpha}{2}, \angle B_{1} B C=\frac{\pi}{4}-\frac{\alpha}{4}$,
} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,162 |
3.4. The angle at the base of an acute isosceles triangle \(ABC (AB = BC)\) is \(\alpha\). In what ratio, counting from vertex \(A\), does the height \(BD\) divide the height \(AE\) ? | 3.4. According to the condition, $\triangle ABC$ is acute-angled; hence, the point $K$ of intersection of the altitudes lies inside the triangle (Fig. 3.8). Let $AD = a$. From $\triangle AEC$ and $\triangle AKD$, we find $AE = 2a \sin \alpha$, $AK = \frac{a}{\sin \alpha} (\angle AKD = \angle C$, since both angles compl... | -\frac{1}{\cos2\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,163 |
3.5. The area of an isosceles triangle is $S$, and the angle between the medians drawn to its lateral sides, opposite the base, is $\alpha$. Find the base. | 3.5. By the condition, $A B=A C, S_{\triangle A B C}=S$, $A B_{1}=B_{1} C, A C_{1}=C_{1} B, B B_{1} \cap C C_{1}=O$, $\angle B O C=\alpha$ (Fig.3.9). We have $S_{\triangle B O C}=\frac{1}{3} S_{\triangle A B C}=\frac{1}{3} S$, since the height of $\triangle B O C$, drawn from $O$, is $\frac{1}{3}$ of the height of $\tr... | BC=\frac{2\sqrt{3S\operatorname{tg}\frac{\alpha}{2}}}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,164 |
3.6. In a right-angled triangle, the area $S$ and an acute angle $\alpha$ are given. Find the distance from the point of intersection of the medians of the triangle to the hypotenuse. | 3.6. By the condition, $\angle A C B=90^{\circ}, \angle C A B=\alpha$, $S_{\triangle A B C}=S$ (Fig. 3.10). Let $A B=2 c$; then $C D=A D=D B=c$. Further, let $A C=b$, O - the point of intersection of the medians; then $O D=\frac{1}{3} C D=\frac{c}{3}$. Draw $C F \perp A B$ and $O E \perp A B$ (segment $O E$ is the desi... | \frac{1}{3}\sqrt{S\sin2\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,165 |
3.7. Find the angle of a triangle if it is known that the sides enclosing this angle are equal to $a$ and $b$, and the bisector of the angle is equal to $l$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 3.7. I n d i c a t i o n. Use formula (3.3), expressing the length of the bisector of a triangle.
A n s w e r: $\alpha=2 \arccos \frac{l(a+b)}{2 a b}$. | Combinatorics | MCQ | Yes | Yes | olympiads | false | 50,166 | |
3.8. Show that if in a triangle the ratio of the tangents of two angles is equal to the ratio of the squares of the sines of these same angles, then the triangle is isosceles or right-angled. | 3.8. Given that in $\triangle A B C$ we have $\frac{\operatorname{tg} A}{\operatorname{tg} B}=\frac{\sin ^{2} A}{\sin ^{2} B}$; we need to prove that the triangle is either isosceles or right-angled. From the given equality, we obtain
$$
\sin A \sin ^{2} B \cos B-\cos A \sin B \sin ^{2} A=0
$$
or
$$
\sin A \sin B(\s... | proof | Geometry | proof | Yes | Yes | olympiads | false | 50,167 |
3.9. In the square $A B C D$, a line is drawn through the midpoint $M$ of side $A B$, intersecting the opposite side $C D$ at point $N$. In what ratio does the line $M N$ divide the area of the square if the acute angle $A M N$ is equal to $\alpha$? Specify the possible values of $\alpha$. | 3.9. Given that $ABCD$ is a square, $M \in AB, MA=MB, \angle AMN=\alpha, N \in CD$ (Fig. 3.10); we need to find $S_{AMND} : S_{BMNC}$. Let $AB=a$; then
Fig. 3.11 $S_{AMND}=\frac{MA+ND}{2} ... | \operatorname{tg}(\alpha-\frac{\pi}{4}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,168 |
3.10. In the square $A B C D$, an isosceles triangle $A E F$ is inscribed; point $E$ lies on side $B C$, point $F$ lies on side $C D$, and $A E=A F$. The tangent of angle $A E F$ is 3. Find the cosine of angle $F A D$. | 3.10. Given: $ABCD-$ is a square, $AEF$ is an isosceles triangle, $AE=AF, E \in BC, F \in CD, \operatorname{tg} \angle AEF = 3$ (Fig. 3.12); we need to find $\cos \angle FAD$. Let $\angle AEF = \alpha, \angle FAD = \beta$; then $\angle EAF = 180^\circ - 2\alpha$. Since $\triangle ABE = \triangle ADF$ (by leg and hypote... | \frac{2\sqrt{5}}{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,169 |
3.11. The diagonal of a rectangle is equal to $d$ and divides the angle of the rectangle in the ratio $m: n$. Find the perimeter of the rectangle. | 3.11. Given that $ABCD$ is a rectangle, $AC = d$, $\frac{\angle ACB}{\angle ACD} = \frac{m}{n}$ (Fig. 3.13). Since $\angle BCD = \frac{\pi}{2}$, then $mx + nx = \frac{\pi}{2}$,
Fig. 3.13
$... | 2\sqrt{2}\cos\frac{\pi(-n)}{4(+n)} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,170 |
3.12. Two heights of a parallelogram, drawn from the vertex of the obtuse angle, are equal to $h_{1}$ and $h_{2}$, and the angle between them is $\alpha$. Find the larger diagonal of the parallelogram. | 3.12. According to the condition, in parallelogram $ABCD$ we have: $BF \perp AD, BE \perp CD, BF = h_{1}, BE = h_{2}, \angle FBE = \alpha$ (Fig. 3.14). Then $\angle ABE = 90^{\circ}$, and $\angle BAF = \angle FBE$ as the acute angles with mutually perpendicular sides. Draw CKLAD;
, have equal lengths and $S_{\triangle C E D} = S_{\triangle C F B}$ (by condition); therefore, $D E = F B$, and thus $C E = C F$ and $A E = A F$. We draw the diagonal $A C$, which divides the quadrilateral $A E C F$ into two equal tri... | \frac{8a}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,172 |
3.14. In a rhombus, a line is drawn through the vertex of the acute angle, equal to $\alpha$, dividing this angle in the ratio $1: 2$. In what ratio does this line divide the side of the rhombus that it intersects? | 3.14. Given that in rhombus $ABCD$ we have: $\angle BAD = \alpha$, $\angle EAD = \frac{\alpha}{3}$, $\angle BAE = \frac{2\alpha}{3}$ (Fig. 3.16). Let $CE = m$, $DE = n$; we need to find $\frac{m}{n}$. In $\triangle AED$, $\angle AED = \frac{2\alpha}{3}$ and by the Law of Sines, $n: \sin \frac{\alpha}{3} = AD: \sin \fra... | \frac{\cos\frac{\alpha}{2}}{\cos\frac{\alpha}{6}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,173 |
3.15. The height of an isosceles trapezoid is $h$, and the angle between its diagonals, opposite the lateral side, is $\alpha$. Find the midline of the trapezoid. | 3.15. By the condition, in the isosceles trapezoid $A B C D$ we have: $A B=C D$, $B B_{1} \perp A D, \quad B B_{1}=h$, $A C \cap B D=O, \angle C O D=\alpha$ (Fig. 3.17). Since $\angle C O D$ is the external angle of the isosceles triangle $A O D$,
then $\angle O A D=\angle O D A=\frac{\alpha}{2}$. Next, we find $B_{1}... | \operatorname{ctg}\frac{\alpha}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,174 |
3.16. The height of an isosceles trapezoid is equal to \( h \). The upper base of the trapezoid is seen from the midpoint of the lower base at an angle of \( 2 \alpha \), and the lower base from the midpoint of the upper base at an angle of \( 2 \beta \). Find the area of the trapezoid in this general case and calculat... | 3.16. Given that in trapezoid $ABCD$ we have: $AB = CD, BC \parallel AD$, $M \in AD, AM = MD, N \in BC, BN = NC, BE \perp AD, BE = h, \angle BMC = 2\alpha$,
Fig. 3.18
$\angle AND = 2\beta... | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,175 |
3.17. A circle of radius $R$ is circumscribed around an isosceles trapezoid with an acute angle $\alpha$ at the base. Find the perimeter of the trapezoid. | 3.17. Given that $ABCD$ is a trapezoid, $AB = CD$, $O$ is the center of the circle inscribed in the trapezoid, $OE \perp AD$, $OE = R$, $\angle BAD = \alpha$, $\alpha < \frac{\pi}{2}$ (Fig. 3.19), we need to find $P_{ABCD} = AD + 2AB + BC$. According to the property of a circumscribed quadrilateral, we have $AD + BC = ... | \frac{8R}{\sin\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,176 |
3.18. In an isosceles triangle, the angle at the base is equal to $\alpha$, the radius of the inscribed circle is $r$. A line is drawn through the vertex of the angle at the base and the center of the inscribed circle. Find the segment of this line that is contained within the triangle. | 3.18. By the condition, $A B=A C, O$ is the center of the inscribed circle in $\triangle A B C$, $\angle A B C=\alpha, O D \perp B C, O D=r$, $B O \cap A C=B_{1}$ (Fig. 3.20). Since $B B_{1}$ is the bisector of $\angle A B C$, then $\angle B_{1} B C=\frac{\alpha}{2}, \angle B B_{1} C=\pi-\frac{3 \alpha}{2}$. From $\tri... | \frac{4r\cos^{2}\frac{\alpha}{2}}{\sin\frac{3\alpha}{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,177 |
3.19. In the rhombus $A B C D$ and the triangle $A B C$, which contains its larger diagonal, circles are inscribed. Find the ratio of the radii of these circles if the acute angle of the rhombus is $\alpha$. | 3.19. Given that $A B C D-$ is a rhombus (Fig. 3.21), $\angle B A D=$ $=\alpha\left(\alpha<90^{\circ}\right), r_{1}-$ is the radius of the circle inscribed in the rhombus, $r_{2}$ is the radius of the circle inscribed in $\triangle A B C$. Let $A B=a$. Draw $B E \perp A D$. In $\triangle B E A$ we have $B E=$ $=2 r_{1}... | 2\cos^{2}\frac{\alpha}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,178 |
3.21. From a point taken on a circle of radius $R$, two equal chords are drawn, forming an inscribed angle equal to $\alpha$ radians. Find the part of the area of the circle; enclosed by this inscribed angle. | 3.21. By the condition, $\angle B A C=\alpha, A B$ $=A C, A O=R$ (Fig. 3.23). Then $\cup B C=2 \alpha, \cup A B=\cup A C=$ $=\frac{2 \pi-2 \alpha}{2}=\pi-\alpha$. The area of the part of the circle $S_{B A C}=S_{\text {sect } B O C}+$ $+2 S_{\triangle O A C}$. But
$$
S_{\text {sect } \cdot B O C}=\frac{1}{2} R^{2} \cd... | R^{2}(\alpha+\sin\alpha) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,180 |
3.22. In the segment, the arc of which contains $\alpha^{\circ}$, a regular triangle is inscribed such that one of its vertices coincides with the midpoint of the arc, and the other two lie on the chord. The area of the triangle is $S$. Find the radius of the arc of the segment. | 3.22. We have $\cup A C=\cup C B=\frac{\alpha}{2}$ (Fig. 3.24), $C D=C E=D E=a, S_{\triangle D C E}=$ $=S, O B=O C=R$ - the desired value.
$$
\text { From the equality } S=\frac{a^{2} \sqrt{3}}{4} \text { (see formula }
$$
. This means that $CD \perp AB$ and $CD=R$, where $R$ is the radius of the circle. We need to find
$$
\frac{S_{A D E}}{S_{B D F}}=\frac{S_{\triangle D A C}-S_{\text {cext.... | \frac{\operatorname{ctg}\alpha-\frac{\pi}{2}+\alpha}{\operatorname{tg}\alpha-\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,182 |
3.24. The plane of the square forms an angle $\alpha$ with the plane passing through one of its sides. What angle does the diagonal of the square form with the same plane? | 3.24. Figure 3.26 shows a square $ABCD$ inclined to the plane $\gamma$ passing through its side $CD$. Draw $AL \perp \gamma$; then $DL$ is the projection of $AD$ on the plane $\gamma$. Since $AD \perp DC$, it follows that $LD \perp DC$ and, therefore, $\angle ADL = \alpha$ is the linear angle of the dihedral angle betw... | \arcsin\frac{\sin\alpha}{\sqrt{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,183 |
3.25. In the face of a dihedral angle equal to $\alpha$, a line is drawn that makes an angle $\beta$ with the edge of the dihedral angle. Find the angle between this line and the other face. | 3.25. By the condition, $\gamma_{1} \cap \gamma_{2}=l$, the dihedral angle $l$ is equal to $\alpha, A B \in \gamma_{1}$, $(; ; A B)=\beta$; it is required to find $(\gamma_{2}; A B)$ (Fig. 3.27). Draw $A O \perp \gamma_{2}$, $A C \perp$; then $\angle A C O=\alpha$ (as the linear angle of the dihedral angle $A B C$), $\... | \arcsin(\sin\alpha\sin\beta) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,184 |
3.26. The diagonal of a rectangular parallelepiped is equal to $l$ and forms an angle $\alpha$ with the lateral edge. Find the volume of the parallelepiped if the perimeter of its base is $P$. | 3.26. Given that $A B C D A_{1} B_{1} C_{1} D_{1}$ is a rectangular parallelepiped, $B_{1} D=l$, $\angle B_{1} D D_{1}=\alpha, P_{A B C D}=P$ (Fig. 3.28). From $\triangle B_{1} D_{1} D$ we find $D D_{1}=l \cos \alpha, \mathrm{a} D_{1} B_{1}=D B=$ $=l \sin \alpha$. Let $A B=x$, $A D=y$; then we arrive at the following s... | \frac{(P^{2}-4^{2}\sin^{2}\alpha)\cos\alpha}{8} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,185 |
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