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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11.003. The side of the base of a regular triangular pyramid is equal to $a$, and the dihedral angle at the base is $45^{\circ}$. Determine the volume and the total surface area of the pyramid. | Solution.
For a regular triangular pyramid $S A=S B=S C$ (Fig. 11.5); $S H$ is the height, and $H$ is the center of the circumscribed circle; $D$ is the midpoint of $A C$,
Fig. 11.5
}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,300 |
11.004. Determine the volume of an oblique triangular prism, where the area of one of the lateral faces is $S$, and the distance from the plane of this face to the opposite edge is $d$. | ## Solution.
Through point $K$ on edge $A A^{\prime}=l$ (Fig. 11.6), we draw a section perpendicular to this edge. Then $K M \perp B B^{\prime}$, since $B B^{\prime} \| A A^{\prime} ; K N \perp C C^{\prime}$, since $C C^{\prime} \| A A^{\prime}$. In $\triangle K M N$, we draw the altitude $K D . K D \perp M N$ and $K ... | \frac{1}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,301 |
11.005. The plane angle at the vertex of a regular triangular pyramid is $90^{\circ}$. Find the ratio of the lateral surface area of the pyramid to the area of its base. | Solution.
For a regular pyramid $AB=BC=AC=a \quad$ (Fig. 11.7). $\angle ASC=90^{\circ}, SA=SC=SB$. This means that $\angle SAC=45^{\circ}, AD=DC$.
Fig. 11.7 Then from $\triangle SAC: SA=\f... | \sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,302 |
11.006. The diagonal of a rectangular parallelepiped is 13 cm, and the diagonals of its lateral faces are $4 \sqrt{10}$ and $3 \sqrt{17}$ cm. Determine the volume of the parallelepiped. | Solution.
Let the sides of the base be $a$ and $b$, the height of the parallelepiped be $h$, and let $b > a$. The diagonal of the base of the parallelepiped is $d_{0} = \sqrt{a^{2} + b^{2}}$. The diagonal of the lateral face with edges $b$ and $h$ is $d_{1} = \sqrt{b^{2} + h^{2}} = 4 \sqrt{10}$. The diagonal of the la... | 144\mathrm{~}^{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,303 |
11.007. Find the ratio of the volume of a cube to the volume of a regular tetrahedron, the edge of which is equal to the diagonal of the face of the cube. | ## Solution.
For a tetrahedron where all edges are equal, denote the edges as $a$. The radius of the circumscribed circle around the base of the tetrahedron is $R=\frac{a \sqrt{3}}{3}$. The foot of the height $H$ is the center of the circumscribed circle around $\triangle A B C$ (see Fig. 11.5). Therefore, $H=\sqrt{a^... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,304 |
11.008. In a right parallelepiped, the sides of the base are equal to $a$ and $b$, and the acute angle between them is $60^{\circ}$. The larger diagonal of the base is equal to the smaller diagonal of the parallelepiped. Find the volume of the parallelepiped. | ## Solution.
The larger diagonal of the base lies opposite the larger angle, i.e., $180^{\circ}-60^{\circ}=120^{\circ}$. From this, by the cosine theorem, $d_{1}=\sqrt{a^{2}+b^{2}-2 a b \cos 120^{\circ}}=\sqrt{a^{2}+b^{2}+a b}$. The smaller diagonal of the base of the parallelepiped $d_{2}=\sqrt{a^{2}+b^{2}-2 a b \cos... | \frac{\sqrt{6}\sqrt{}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,305 |
11.010. In a cube with an edge length of $a$, the center of the upper face is connected to the vertices of the base. Find the total surface area of the resulting pyramid. | ## Solution.
From the condition, it is clear that the height of the pyramid is equal to the height of the cube, i.e., $a$ (Fig. 11.9). $K$ is the midpoint of $AD$, $HH_{1}$ is the height. Then $HK=\frac{a}{2}$; $H_{1}K=\sqrt{HH_{1}^{2}+HK^{2}}=\sqrt{a^{2}+\left(\frac{a}{2}\right)^{2}}=\frac{a \sqrt{5}}{2}$. The semi-p... | ^{2}(1+\sqrt{5}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,307 |
11.011. The base of a regular pyramid is a polygon, the sum of the interior angles of which is $720^{\circ}$. Determine the volume of the pyramid if its lateral edge, equal to $l$, forms an angle of $30^{\circ}$ with the height of the pyramid. | ## Solution.
From the formula for the sum of the interior angles of a polygon, we determine the type of the polygon: $720^{\circ}=180^{\circ}(n-2)$. Hence, $n=6$, i.e., we have a hexagon; since the angle between the lateral edge and the height is $30^{\circ}$, the height $h=l \cos 30^{\circ}=\frac{\sqrt{3}}{2} l$. The... | \frac{3^{3}}{16} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,308 |
11.012. The diagonal of the square base of a regular quadrilateral pyramid is equal to its lateral edge and is equal to $a$. Find the total surface area of the pyramid and its volume. | ## Solution.
Let the side of the square be denoted by $b$. The diagonal of the square $d = b \sqrt{2}$. It is equal to the lateral edge of the pyramid $a = b \sqrt{2}$ (Fig. 11.10). The area of the base $S_{\text{base}} = b^2 = \frac{a^2}{2}$. The height of the pyramid, as seen from the diagram, is $h = \sqrt{a^2 - \l... | \frac{^2(1+\sqrt{7})}{2};\frac{\sqrt{3}^3}{12} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,309 |
11.013. 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. | ## Solution.
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. 11.11). Considering that $O K=\frac{1}{2} A D=\frac{a \sqrt{2}}{4}$, we find the slant height of the pyramid: $S K=\sqrt{S O^{2}+O K^{2}}=\sqrt{a^{2}+\frac{a^{2}}{8}}=\frac{3 a \sqrt{2}}{4}$.... | 2^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,310 |
11.014. The apothem of a regular hexagonal pyramid is $h$, and the dihedral angle at the base is $60^{\circ}$. Find the total surface area of the pyramid. | Solution.
Since $\angle S K O=60^{\circ}$ (Fig. 11.12), 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 $K D=\frac{1}{2} O D$. Then $O K^{2}=O D^{2}-K D^{2}=4 K D^{2}-K D^{2}=3 K D^{2}$, i.e., $K D=\frac{h \sqrt{3}}{6}, D E=2 K D=\frac{h \sqrt{3}}{... | \frac{3^{2}\sqrt{3}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,311 |
11.015. Find the total surface area of a regular triangular pyramid, the side of the base of which is equal to $a$, and the dihedral angle at the base is $60^{\circ}$. | ## Solution.
The radius of the circle inscribed in the equilateral
$\triangle A B C$ is $r=\frac{a \sqrt{3}}{6}$
Fig. 11.13
(Fig. 11.13); $S H$ is the height of the pyramid, $S K$ is the... | \frac{3^{2}\sqrt{3}}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,312 |
11.016. 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. | ## Solution.
Given $B D=b, \angle A O B=60^{\circ}$ (Fig. 11.14); hence $A B=\frac{b}{2}$, $A D=\frac{b \sqrt{3}}{2}$. 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 of the pyramid and $S O=\frac{b}{2}$.... | \frac{b^{3}\sqrt{3}}{24} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,313 |
11.017. The side of the base of a regular triangular pyramid is $1 \mathrm{~cm}$, and its lateral surface area is $3 \mathrm{~cm}^{2}$. Find the volume of the pyramid. | Solution.
Let the side of the base of a regular triangular pyramid be denoted by $a$, i.e., $a=1(\mathrm{cm}) ; S_{\text {side }}=3\left(\mathrm{~cm}^{2}\right)$. Since $S_{\text {side }}=p l$, where $p=\frac{3 a}{2}$ is the semi-perimeter of the base, and $l$ is the apothem, then $l=\frac{S_{\text {side }}}{p}=\frac{... | \frac{\sqrt{47}}{24}\mathrm{~}^{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,314 |
11.018. The base of the pyramid is a triangle with sides equal to $a$, $a$, and $b$. All lateral edges are inclined to the base plane at an angle of $60^{\circ}$. Determine the volume of the pyramid. | Solution.
Given $A B=B C=a$ (Fig. 11.15); therefore, $B H=a^{2}-\frac{b^{2}}{4}=$ $=\frac{1}{2} \sqrt{4 a^{2}-b^{2}}$. Hence, $S_{\text {Base }}=\frac{1}{2} A C \cdot B H=\frac{1}{4} b \sqrt{4 a^{2}-b^{2}}$. Draw the height $SO$. Since all the edges of the pyramid are equally inclined to the base plane, point $O$ is t... | \frac{^{2}b\sqrt{3}}{12} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,315 |
11.019. The lateral edge of a regular triangular pyramid is equal to $l$, and the height is equal to $h$. Determine the volume of the pyramid. | ## Solution.
Let $A S=l, S H=h$ (see Fig. 11.5). Then
$A H=\sqrt{A S^{2}-S H^{2}}=\sqrt{l^{2}-h^{2}}$. In a regular pyramid, $A H$ is the radius of the circumscribed circle around the equilateral $\triangle A B C$, i.e., $A H=R=\frac{a \sqrt{3}}{3}$ or $\sqrt{l^{2}-h^{2}}=\frac{a \sqrt{3}}{3}$, from which $a=\frac{3 ... | \frac{\sqrt{3}(^{2}-^{2})}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,316 |
11.020. The base of an oblique prism is a parallelogram with sides of 3 dm and 6 dm and an acute angle of $45^{\circ}$. The lateral edge of the prism is 4 dm and is inclined to the base plane at an angle of $30^{\circ}$. Find the volume of the prism. | ## Solution.
Let $A B=3$ (dm), $A D=6$ (dm), $A A_{1}=4$ (dm) (Fig. 11.16); $A_{1} H-$ height. Then $A_{1} H=A A_{1} \sin 30^{\circ}=2$ (dm). The area of the base $S_{\text {base }}=A B \cdot A D \sin 45^{\circ}=9 \sqrt{2}\left(\right.$ dm $\left.{ }^{2}\right)$. Volume
$V=S_{\text {base }} \cdot A_{1} H=9 \sqrt{2} \... | 18\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,317 |
11.021. Each of the lateral edges of the pyramid is $269 / 32$ cm. The base of the pyramid is a triangle with sides 13, 14, and 15 cm. Find the volume of the pyramid. | Solution.
Let $S A=S B=S C=\frac{269}{32}$ (cm) (Fig. 11.4), and $A B=c=15$ (cm), $B C=a=13$ (cm), $A C=b=14$ (cm), $S H$ - height. This means that $\angle S H A=\angle S H B=\angle S H C$. Then $A H=B H=C H$ (from the equality of $\triangle A S H$,
. Denote $A B_{1}=b$. Then $b=\frac{a}{\operatorname{tg} 30^{\circ}}=a \sqrt{3}$. The height
$h=B B_{1}=\sqrt{b^{2}-a^{2}}=a \sqrt{2}$. Volume $V=S_{\text {base }} \cdot h$. Since $A B C D$ is a square, t... | ^{3}\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,319 |
11.023. In a regular quadrilateral pyramid, the side of the base is 6 dm, and the height is 4 dm. Find the lateral surface area of the truncated pyramid cut off from the given one by a plane parallel to its base and at a distance of 1 dm from it. | Solution.
Let $A D=D C=B C=A B=a_{1}$ (6 dm), $S H$-height, $S H=4$ (dm) (Fig. 11.18). By the condition $H H_{1}=1$ (dm). From the similarity of $\triangle H E S$ and $\Delta H_{1} E_{1} S$ (where $E$ and $E_{1}$ are the midpoints of $C D$ and $C_{1} D_{1}$ respectively): $\frac{H E}{H_{1} E_{1}}=\frac{S H}{S H_{1}}$,... | 26.25 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,320 |
11.024. The bases of a regular truncated pyramid are squares with sides $a$ and $b(a>b)$. The lateral edges are inclined to the plane of the base at an angle of $45^{\circ}$. Determine the volume of the truncated pyramid. | Solution.
Let $\quad A D=D C=B C=A B=a, \quad A_{1} B_{1}=B_{1} C_{1}=C_{1} D_{1}=D_{1} A_{1}=h$ (Fig. 11.19), $\angle D_{1} D H=45^{\circ}$, where $D_{1} H$ is the height of the truncated pyramid.
 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,321 |
11.025. The lateral edges of a regular truncated triangular pyramid are inclined to the base plane at an angle of $60^{\circ}$. The sides of the lower and upper bases are equal to $a$ and $b(a>b)$, respectively. Find the volume of the truncated pyramid. | Solution.
Let $A B=B C=A C=a, A_{1} B_{1}=B_{1} C_{1}=C_{1} A_{1}=b, \angle A_{1} A H=60^{\circ}$ (Fig. 11.20). Since $\triangle A B C$ and $A_{1} B_{1} C_{1}$ are equilateral, $O$ and $O_{1}$ are the centers
 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,322 |
11.026. The base of a right parallelepiped is a rhombus. A plane, passing through one side of the lower base and the opposite side of the upper base, forms an angle of $45^{\circ}$ with the base plane. The area of the resulting section is equal to $Q$. Determine the lateral surface area of the parallelepiped. | ## Solution.
Draw $B K \perp D C$ (Fig. 11.21), then $B_{1} K \perp D C$ as well. Hence, $\angle B K B^{\prime}=45^{\circ}$ and $B K=B B^{\prime}=h, B^{\prime} K=\sqrt{2} h$. Let the sides of the rhombus be $a$. Then the lateral surface area
$S_{6}=4 a \cdot B B_{1}=4 a h=\frac{4}{\sqrt{2}} a \cdot \sqrt{2} h=$
$=2 ... | 2Q\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,323 |
11.029. The volume of a regular triangular pyramid, whose lateral face is inclined to the base plane at an angle of $45^{\circ}$, is $9 \mathrm{~cm}^{3}$. Find the total surface area of the pyramid. | ## Solution.
Let $PABC$ be the given pyramid (Fig. 11.24), $PH$ the height, where $H$ is the center of the circle inscribed in $\triangle ABC$; $HK \perp AC$, hence $PK \perp AC$, $HK = r = \frac{a \sqrt{3}}{6}, \angle PKH = 45^{\circ}$. Further, $PH = HK \tan 45^{\circ} = \frac{a \sqrt{3}}{6}, PK = \frac{HK}{\cos 45^... | 9\sqrt{3}(1+\sqrt{2})^2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,326 |
11.030. The base of a right parallelepiped is a parallelogram with sides 1 and 4 cm and an acute angle of $60^{\circ}$. The larger diagonal of the parallelepiped is $5 \mathrm{~cm}$. Determine its volume. | Solution.
Let $AB=1$ (cm), $BC=4$ (cm) (Fig. 11.25); $\angle BAD=60^{\circ}$, then $\angle ADC=120^{\circ}$. Therefore, $AC$ can be found using the cosine theorem: $AC^{2}=$ $=AD^{2}+DC^{2}-2AD \cdot DC \cos 120^{\circ}=1+16+2 \cdot 4 \cdot \frac{1}{2}=21, AC_{1}=5$ (cm). Therefore, $CC_{1}^{2}=AC_{1}^{2}-AC^{2}=25-21... | 4\sqrt{3}\mathrm{~}^{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,327 |
11.031. The center of a cube with edge length $a$ is connected to all its vertices. Determine the volume and surface area of each of the resulting pyramids. | ## Solution.
Since the construction resulted in 6 identical pyramids (Fig. 11.26), each of them has a volume $V=\frac{a^{3}}{6}$ and a total surface area $S_{\text {full }}=a^{2}+4 \cdot \frac{1}{2} a \cdot O K$. But $O K=\sqrt{O P^{2}+P K^{2}}=\frac{a \sqrt{2}}{2}$, from which $S_{\text {full }}=a^{2}+a^{2} \sqrt{2}=... | \frac{^{3}}{6};^{2}(\sqrt{2}+1) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,328 |
11.033. In a triangular pyramid, the lateral edges are mutually perpendicular and have lengths $\sqrt{70}, \sqrt{99}$, and $\sqrt{126}$ cm. Find the volume and the area of the base of the pyramid. | Solution.
Given $\angle A P C=\angle A P B=\angle B P C=90^{\circ}, A P=\sqrt{70}, P B=\sqrt{99}$, $P C=\sqrt{126}$ (Fig. $11.28, a$ ). By the Pythagorean theorem, $A B=\sqrt{P A^{2}+P B^{2}}=\sqrt{70+99}=13, B C=\sqrt{P B^{2}+P C^{2}}=\sqrt{99+126}=15$,
. 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... | 2^{3}\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,330 |
11.036. Determine the volume of an octahedron (a regular octahedron), the edge of which is equal to $a$. | Solution.
$$
V=\frac{1}{3} S H, \text { where } H=2 \cdot h=2 \cdot S O, S=a^{2}
$$
(Fig. 11.31). In $\triangle S O C \angle S O C=90^{\circ}$, then $S O^{2}=S C^{2}-O C^{2}$, where $O C=\frac{a \sqrt{2}}{2}$; $S O^{2}=\frac{2 a^{2}-a^{2}}{2}=\frac{a^{2}}{2} ; h=\frac{a}{\sqrt{2}}, H=\frac{2 a}{\sqrt{2}}$, and thus,
... | \frac{^{3}\sqrt{2}}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,331 |
11.037. The base of the prism is a square with a side equal to $a$. One of the lateral faces is also a square, the other is a rhombus with an angle of $60^{\circ}$. Determine the total surface area of the prism. | Solution.
$S_{\text {full }}=4 \cdot S_{A B C D}+2 \cdot S_{B C C_{1} B_{1}}$ (Fig. 11.32), $S_{A B C D}=a^{2}$ (since $A B C D-$ is a square and $A B=a), \quad S_{B C C_{1} B_{1}}=\frac{1}{2} a^{2} \sin 60^{\circ}=\frac{a^{2} \sqrt{3}}{2}\left(B B_{1} C_{1} C-\right.$ is a rhombus, $\left.B B_{1}=a, \angle B_{1} C_{1... | ^{2}(4+\sqrt{3}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,332 |
11.039. In a cube, the centers of the bases are connected to the centers of the side faces. Calculate the surface area of the resulting octahedron if the edge of the cube is $a$. | Solution.
$$
S=8 \cdot S_{\triangle B E C}, \text { where } C E=\sqrt{\frac{a^{2}}{4}+\frac{a^{2}}{4}}=\frac{a \sqrt{2}}{2} ; C E=B E=B C \text { (Fig. 11.34), }
$$
therefore $\angle B E C=\angle E C B=\angle C B E=60^{\circ}$ and
$$
S_{\triangle B E C}=\frac{1}{2} \frac{a \sqrt{2}}{2} \cdot \frac{a \sqrt{2}}{2} \si... | ^{2}\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,333 |
11.040. The base of the pyramid is a triangle with side lengths of 6, 5, and $5 \mathrm{~cm}$. The lateral faces of the pyramid form equal dihedral angles with its base, each containing $45^{\circ}$. Determine the volume of the pyramid. | ## Solution.
$$
V=\frac{1}{3} S \cdot H, S=\sqrt{p(p-6)(p-5)(p-5)} ; p=\frac{16}{2}=8, \text { therefore } S=12
$$
$S=p \cdot r ;$ hence $r=\frac{12}{8}=\frac{3}{2}$. In $\triangle S O E, \angle S O E=90^{\circ}$ (Fig. 11.35),
 } \angle A_{1} B C=90^{\circ} ; \\
& a=B C=A_{1} C \cdot \sin 30^{\circ}=\frac{l}{2} ; A_{1} B=\frac{l \sqrt{3}}{2} \cdot \mathrm{B} \Delta A_{1} C_{1} C, \angle A_{1} C_{1} C=90^{\circ} ; \\
& C_{1} C... | \frac{^{3}\sqrt{2}}{8} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,335 |
11.042. Determine the volume of a regular truncated square pyramid if its diagonal is 18 cm, and the lengths of the sides of the bases are 14 and 10 cm. | ## Solution.
The desired volume is expressed by the formula $V=\frac{h}{3}\left(S_{1}+S_{2}+\sqrt{S_{1} S_{2}}\right)$, where $S_{1}=196\left(\mathrm{~cm}^{2}\right), S_{2}=100\left(\mathrm{~cm}^{2}\right)$. Let's find $h=B_{1} K$ (Fig. 11.37). We have $B_{1} K=$ $=\sqrt{B_{1} D^{2}-K D^{2}}$. Since $B B_{1} D_{1} D$ ... | 872\mathrm{~}^{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,336 |
11.043. The base of a right parallelepiped is a rhombus with an area of $Q$. The areas of the diagonal sections are $S_{1}$ and $S_{2}$. Determine the volume and the lateral surface area of the parallelepiped. | Solution.
We have $V=S_{\text {base }} h$, where $S_{\text {base }}=Q$ (by
Fig. 11.38 condition); thus, we need to find $h$. Since $A B C D$ is a rhombus, then $S_{\text {base }}=0.5 A C \... | \sqrt{\frac{S_{1}S_{2}Q}{2}};2\sqrt{S_{1}^{2}+S_{2}^{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,337 |
11.045. The largest diagonal of a regular hexagonal prism is $d$ and forms an angle of $30^{\circ}$ with the lateral edge of the prism. Find the volume of the prism. | Solution.
$$
V=S H . \mathrm{B} \triangle A D D_{1} \text { (Fig. 11.40) } \angle A D D_{1}=90^{\circ}, \angle A D_{1} D=30^{\circ}, A D_{1}=d
$$
then $H=D_{1} D=d \cdot \cos 30^{\circ}=d \sqrt{3} / 2 ; A D=d \sin 30^{\circ}=d / 2$. Let $B C=x$,
. Hence $A D_{1}=a \cdot \operatorname{ctg} 30^{\circ}=a \sqrt{3}, H=\sqrt{A D_{1}^{2}-A_{1} D_{1}^{2}}=\sqrt{3 a^{2}-b^{2}}$;
$$
V=S \cdot H=a b... | \sqrt{3^{2}-b^{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,340 |
11.048. Find the volume of an oblique triangular prism, the base of which is an equilateral triangle with a side equal to $a$, 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 $60^{\circ}$. | ## Solution.
Draw $A_{1} K$ perpendicular to the plane $A B C$ (Fig. 11.43); 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 \sq... | \frac{3^{3}}{8} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,341 |
11.050. Find the lateral surface area of a regular hexagonal pyramid, the height of which is equal to $h$, and the lateral edge is equal to $l$. | Solution.
$$
S_{60 \mathrm{~K}}=6 \cdot S_{\triangle F S A}=6 \cdot \frac{1}{2} \cdot A F \cdot S K=3 A F \cdot S K \text { (Fig. 11.45); } \triangle S O A
$$
$\angle S O A=90^{\circ}, \quad O A=\sqrt{l^{2}-h^{2}}=A F, \quad S K=\sqrt{l^{2}-\frac{l^{2}-h^{2}}{4}}=\frac{\sqrt{3 l^{2}+h^{2}}}{2} ;$
then $S_{\text {sid... | \frac{3}{2}\sqrt{(^{2}-^{2})(3^{2}+^{2})} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,343 |
11.051. Find the volume of a regular triangular pyramid, where the plane angle at the vertex is $90^{\circ}$, and the side of the base is 3 cm. | Solution.
$V=\frac{1}{3} S H \quad$ (Fig. 11.46); $S=\frac{9}{4} \sqrt{3} ; O C=\frac{a \sqrt{3}}{3}=\frac{3 \sqrt{3}}{3}=\sqrt{3} \quad$ (since $A B=B C=A C=a=3 \mathrm{~cm}) ; E O=\frac{a \sqrt{3}}{6}=\frac{\sqrt{3}}{2} . \mathrm{B} \triangle E S C \angle E S C=90^{\circ}, H^{2}=E O \times$ $\times O C=\frac{\sqrt{3... | \frac{9\sqrt{2}}{8}\mathrm{~}^{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,344 |
11.052. In a regular triangular prism, the area of the section passing through a lateral edge perpendicular to the opposite lateral face is $Q$. The side of the base of the prism is $a$. Find the total surface area of the prism. | Solution.
$$
S=2 S_{\mathrm{oCH}}+3 S_{B B_{1} A_{i} A} \text { (Fig. 11.47); } C D=\frac{a \sqrt{3}}{2} ; A B=B C=C A=a \text {; }
$$
$$
\begin{aligned}
& Q=C D \cdot C_{1} C=\frac{a \sqrt{3}}{2} \cdot C_{1} C ; C_{1} C=\frac{2 Q}{a \sqrt{3}} ; S_{\mathrm{och}}=\frac{a^{2} \sqrt{3}}{4} ; \\
& S_{B B_{1} A_{1} A}=A B... | \sqrt{3}(0.5^{2}+2Q) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,345 |
11.053. The height of a regular tetrahedron is $h$. Calculate its total surface area. | Solution.
$$
S=S_{\mathrm{och}}+3 S_{\triangle B S A}=4 S_{\triangle A B C}=4 \frac{A B^{2} \sqrt{3}}{4}=A B^{2} \sqrt{3} \text { (Fig. 11.48); }
$$
$A O=A B \sqrt{3} / 3$, since $S A=A B=S B=B C=S C=A C$ (by condition); $\triangle A O S, \angle A O S=90^{\circ} ; A S^{2}=O A^{2}+S O^{2} A S^{2}=\frac{A B^{2}}{3}+h^{... | \frac{3^{2}\sqrt{3}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,346 |
11.054. Each of the lateral edges of the pyramid is equal to $b$. Its base is a right triangle, the legs of which are in the ratio $m: n$, and the hypotenuse is equal to $c$. Calculate the volume of the pyramid. | Solution.
$V=\frac{1}{3} S \cdot H$. Let $A B=m x, B C=n x$ (Fig. 11.49). By the condition $A C=c, S C=b$. In $\triangle A B C, \angle A B C=90^{\circ}$; thus, $c^{2}=m^{2} x^{2}+n^{2} x^{2}$, $x=\sqrt{\frac{c^{2}}{m^{2}+n^{2}}}=\frac{c}{\sqrt{m^{2}+n^{2}}}$ and $S=\frac{1}{2} m n x^{2}=\frac{m n c^{2}}{2\left(m^{2}+n... | \frac{n^{2}\sqrt{4b^{2}-^{2}}}{12(^{2}+n^{2})} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,347 |
11.055. The center of the upper base of a cube is connected to the midpoints of the sides of the lower base. A quadrilateral angle is formed, each plane angle of which is equal to $\alpha$. Prove that $30^{\circ}<\alpha<45^{\circ}$. | Solution.
We have $AB = \sqrt{AM^2 + MB^2} = \sqrt{\frac{a^2}{4} + \frac{a^2}{4}} = \frac{a \sqrt{2}}{2}$ (Fig. 11.50). From $\triangle SOB$, we find $SB = SA = \sqrt{a^2 + \frac{a^2}{4}} = \frac{a \sqrt{5}}{2}$. Let $\angle BSA = \alpha$; then in $\triangle SAB$, by the cosine rule, we get $AB^2 = 2SB^2 - 2SB^2 \cos ... | 30<\alpha<45 | Geometry | proof | Yes | Yes | olympiads | false | 51,348 |
11.056. The diagonal of a rectangular parallelepiped is 10 cm and forms an angle of $60^{\circ}$ with the base plane. The area of the base is $12 \mathrm{~cm}^{2}$. Find the lateral surface area of the parallelepiped. | Solution.
From $\triangle D_{1} D B$ (Fig. 11.51) we find $B D=5 \text{ cm}, D D_{1}=5 \sqrt{3} \text{ cm}$. The lateral surface area $S_{\text {lateral }}=2\left(D D_{1} \cdot D C+D D_{1} \cdot A D\right)=2 D D_{1}(D C+A D)$. But $D C^{2}+A D^{2}=$ $=B D^{2}=25$, and by the condition $D C \cdot A D=12$. Therefore, $(... | 70\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,349 |
11.057. Determine the volume of a rectangular parallelepiped if its diagonal is equal to $d$, and the lengths of the edges are in the ratio $m: n: p$. | Solution.
$V=a b c, D_{1} B=d$. Let $A B=m x, B C=n x, B_{1} B=p x$ (since $m: n: p=A B: B C: B_{1} B$) (Fig. 11.52). In $\triangle D_{1} D B \angle D_{1} D B=90^{\circ}$, then $D_{1} B^{2}=D_{1} D^{2}+D B^{2}$. In $\triangle D A B \angle D A B=90^{\circ}$ and $D B^{2}=A D^{2}+A B^{2}$, so $D_{1} B^{2}=D_{1} D^{2}+A D... | \frac{np\cdot^{3}}{(^{2}+n^{2}+p^{2})^{3/2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,350 |
11.058. Determine the volume of a regular triangular pyramid if the height of the triangle serving as its base is $h$, and the apothem of the pyramid is $m$. | Solution.
$V=\frac{1}{3} S H, M C=h, S M=m$ (Fig. 11.53). $A B=B C=A C$, therefore
$C M=\frac{A B \sqrt{3}}{2} \Rightarrow A B=\frac{2 C M}{\sqrt{3}}=\frac{2 h}{\sqrt{3}}$ and $S=\frac{h^{2}}{\sqrt{3}} . M O=\frac{A B \sqrt{3}}{6}=\frac{h}{3}$. In $\triangle S O M \angle S O M=90^{\circ} \Rightarrow H=\sqrt{S M^{2}-M... | \frac{\sqrt{3}}{27}^{2}\sqrt{9^{2}-^{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,351 |
11.059. The areas of the lateral faces of a right triangular prism are $M, N$ and $P$. The lateral edge of the prism is $l$. Determine the volume of the prism. | Solution.
According to the condition $S_{A^{\prime} B_{B} B A}=M, S_{B_{1} C_{1} B C}=N, S_{A_{1} A C C_{1}}=P, C C_{1}=l$
(Fig. 11.54). Then $B C=\frac{N}{l} ; A B=\frac{M}{l} ; A C=\frac{P}{l}$;
$$
V=S \cdot H=S \cdot l=l \cdot \sqrt{\frac{N+M+P}{2 l} \cdot\left(\frac{N+P-M}{2 l}\right) \cdot\left(\frac{N+M-P}{2 l... | \frac{1}{4}\sqrt{(N+M+P)(M+N-P)(M+P-N)(N+P-M)} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,352 |
11.060. The base area $P$ and volume $V$ of a regular quadrilateral prism are known. Calculate its total surface area. | Solution.
$S_{\pi}=2 S_{\text {base }}+4 S_{6 \text { side }}$ (Fig. 11.55), where $P=S_{\text {base }}$ and $P=a^{2}(a-$ side of the square). $V=P \cdot H \Rightarrow H=\frac{V}{P}$;
Fig.... | 2P+\frac{4V}{\sqrt{P}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,353 |
11.062. The base of the pyramid is 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. | Solution.
Given $S C=l, \angle S B C=90^{\circ}, \angle S C B=45^{\circ}$ (Fig. 11.57), hence $S B=B C=\frac{l}{\sqrt{2}}$. We have $V=\frac{1}{3} B C^{2} \cdot S B=\frac{l^{3} \sqrt{2}}{12}$. The total surface area $-S_{\text {full }}=S_{\text {base }}+2 S_{\triangle S A B}+2 S_{\triangle S A D}$, since $S_{\triangle... | \frac{^{3}\sqrt{2}}{12},\frac{^{2}(2+\sqrt{2})}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,355 |
11.064. Find the volume of a regular triangular pyramid, the height of which is $h$, and all plane angles at the vertex are right angles. | ## Solution.
Draw $B D \perp A C$ (Fig. 11.59). Let the side of the base be $a$: Since the pyramid is regular, then $O D=\frac{a \sqrt{3}}{6}, A D=D C=\frac{a}{2}$, $S_{\text {base }}=\frac{a^{2} \sqrt{3}}{4}$. Express $a$ in terms of $h$. In $\triangle S A C$ we have $S A=S C$,
. Since $B D=D C=\frac{a}{2}$, then $S_{6 \text { lateral }}=\frac{3 a^{2}}{4}=\f... | S\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,357 |
11.069. The lateral edge of a regular triangular prism is equal to the height of the base, and the area of the section made through this lateral edge and the height of the base is $Q$. Determine the volume of the prism. | Solution.
The volume of the prism is found as $V=S \cdot H$ (Fig. 11.64). The area of the section $D D_{1} C_{1} C \quad Q=D C \cdot C C_{1}$. Since by condition $D C=C C_{1}$, then $D C=C C_{1}=\sqrt{Q}$. The area of $\triangle A B C: S=\frac{x^{2} \sqrt{3}}{4}$. From $\triangle C D B Q+\frac{x^{2}}{4}=x^{2}$. Hence ... | Q\sqrt{Q/3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,359 |
11.071. Find the ratio of the volume of a regular hexagonal pyramid to the volume of a regular triangular pyramid, given that the sides of the bases of these pyramids are equal, and their apothems are twice the length of the sides of the base. | Solution.
The volume of the hexagonal pyramid: $V_{1}=\frac{1}{3} S_{1} \cdot H_{1}$. The area of the base of the hexagonal pyramid $S_{1}=\frac{3}{2} a^{2} \sqrt{3}$, and the height is $H_{1}=\sqrt{l^{2}-r_{1}^{2}}$, where the slant height $l=2 a$, and the radius of the inscribed circle $r_{1}=\frac{3 a}{2 \sqrt{3}}$... | \frac{6\sqrt{1833}}{47} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,360 |
11.072. 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. | Solution.
The required volume $V=S_{\text {base }} h$, where $S_{\text {base }}=A B \cdot A D$ (Fig. 11.66), $h$ is 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 $A D=\frac{n}{m} A B$. In $\t... | \frac{n}{^{2}+n^{2}}\cdotQ\sqrt{Q} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,361 |
11.073. The measurements of a rectangular parallelepiped are 2, 3, and 6 cm. Find the length of the edge of a cube such that the volumes of these bodies are in the same ratio as their surface areas. | Solution.
The volume of a rectangular parallelepiped is $V_{\text {par }}=2 \cdot 3 \cdot 6=36 \mathrm{~cm}^{3}$. The volume of a cube is $V_{\mathrm{x}}=a^{3}$. The area of the complete surface of a rectangular parallelepiped is calculated as follows: $S_{\text {par }}=S_{\text {side }}+2 S_{\text {base }}=P H+2 S_{\... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,362 |
11.074. The height of the pyramid is 8 m. A plane parallel to the base is drawn at a distance of 3 m from the vertex. The area of the resulting section is $4 \mathrm{~m}^{2}$. Find the volume of the pyramid. | Solution.
The volume of the pyramid $V=\frac{1}{3} S H$ (Fig. 11.67). Pyramid $S A B C D E F$ is similar to pyramid $S A_{1} B_{1} C_{1} D_{1} E_{1} F_{1}$ with a similarity coefficient $k=\frac{S O}{S O_{1}}=\frac{8}{3}$.
Then the areas of their bases are in the ratio $\frac{S_{A B C D E F}}{S_{A_{1} B_{1} C_{1} D_{... | 75.85\mathrm{~}^{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,363 |
11.076. The height of the cone is equal to the diameter of its base. Find the ratio of the area of its base to the lateral surface area. | ## Solution.
The area of the base of the cone: $S_{\text {osn }}=\pi R^{2}$ (Fig. 11.68). The lateral surface area of the cone $S_{\text {bok }}=\pi R l$, where
$l=\sqrt{H^{2}+R^{2}}=\sqrt{4 R^{2}+R^{2}}=R \sqrt{5}$. Therefore, $S_{\text {bok }}=\pi R^{2} \sqrt{5}$. Finally, we get $\frac{S_{\text {osn }}}{S_{\text {... | \frac{\sqrt{5}}{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,365 |
11.078. A cylinder is formed by rotating a rectangle around one of its sides. Express the volume $V$ of the cylinder in terms of the area $S$ of the rectangle and the length $C$ of the circumference described by the point of intersection of its diagonals. | ## Solution.
The volume of the cylinder is found as $V=\pi R^{2} H$ (Fig. 11.70). By construction, $R=a, H=b$. The area of the rectangle $S=a \cdot b=R H$.
Fig. 11.71
. Find the sum of the total surface areas and the sum of the volumes of the cones, if the height of the cylinder is $2 a$. | ## Solution.
By construction $H=2 R$ (Fig. 11.72). These cones are identical. The area of the total surface of one cone: $S_{1}=S_{\text {lat }}+S_{\text {base }}=\pi R l+\pi R^{2}$. From $\triangle A B C \quad 2 l=2 \sqrt{2} R$. Hence $l=\sqrt{2} R$. Since $2 R=2 a$, then $S_{1}=\pi a^{2} \sqrt{2}+\pi a^{2}$. And the... | 2\pi^{2}(\sqrt{2}+1);\frac{2}{3}\pi^{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,368 |
11.081. A cone with base radius $R$ is inscribed in an arbitrary pyramid, where the perimeter of the base of the pyramid is $2 p$. Determine the ratio of the volumes and the ratio of the lateral surface areas of the cone and the pyramid. | ## Solution.
Let the total height of the cone and the pyramid be $H$ (Fig. 11.73). 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 $l$ is the slant height of the cone. Let's find $V_{2}$... | \frac{\piR}{p} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,369 |
11.082. The height of the cone and its slant height are 4 and 5 cm, respectively. Find the volume of a hemisphere inscribed in the cone, with its base lying on the base of the cone. | Solution.
From $\triangle S O A: A O=R=\sqrt{25-16}=3$ cm (Fig. 11.74). The volume of the desired hemisphere is $V=\frac{2}{3} \pi r^{3}$. From $\triangle S B O: r^{2}+(5-x)^{2}=16$. From $\triangle A B O: x^{2}+r^{2}=9$. Solving the system $\left\{\begin{array}{l}r^{2}+(5-x)^{2}=16, \\ r^{2}+x^{2}=9,\end{array}\right... | \frac{1152}{125}\pi\mathrm{}^{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,370 |
11.083. Determine the volume of a sphere inscribed in a regular pyramid, where the height is equal to $h$, and the dihedral angle at the base is $60^{\circ}$. | Solution.
Consider the section of a regular pyramid through the apothem $AB$ and the height $BC = h$, on which lies the center of the sphere inscribed in the pyramid (Fig. 11.75). The sphere touches the lateral face at point $K$ on the apothem $AB$.
. From this, $r=2$ cm. The volume of the desired cone is found as $V_{\mathbf{k}}=\frac{1}{3} \pi R^{2} H$. Since $\triangle A S B$ is equilateral, the radius of the inscribed circle $r=\frac{a \sqrt{3}}{6}$. From... | 24\pi\mathrm{}^{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,373 |
11.088. A sphere, a cylinder with a square axial section, and a cone are given. The cylinder and the cone have the same bases, and their heights are equal to the diameter of the sphere. How do the volumes of the cylinder, the sphere, and the cone relate? | Solution.
$$
S_{\mathrm{ocH}_{\mathrm{u}}}=S_{\mathrm{och}_{\mathrm{k}}} ; \quad H_{\mathrm{k}}=H_{\mathrm{u}}=2 R_{\mathrm{u}} ; \quad R_{\mathrm{k}}=R_{\mathrm{u}}=R_{\mathrm{w}} . \quad \text { Volume } \quad \text { of } \quad \text { the } \quad \text { sphere }
$$
$V_{\mathrm{wl}}=\frac{4}{3} \pi R_{\mathrm{w}}... | 3:2:1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,374 |
11.091. On the segment $A B$ as a diameter, a semicircle is constructed with the center at point $O$, and on the segments $O A$ and $O B$, two semicircles are constructed, located in the same half-plane with boundary $A B$ as the first one. Find the surface area and volume of the figure formed by rotating around $A B$ ... | ## Solution.
The volume of the desired figure is: $V=V_{\boldsymbol{w}_{1}}-2 V_{\boldsymbol{w}_{2}}$, where $V_{\boldsymbol{w}_{1}}=\frac{4}{3} \pi R_{1}^{3}$ (Fig. 11.79). Since $AB=20 \text{ cm}$, then $R_{1}=10 \text{ cm}$. Thus, we get $V_{\boldsymbol{w}_{1}}=\frac{4000}{3} \pi$. Since $R_{1}=10 \text{ cm}$, then... | 600\pi | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,375 |
11.092. A triangle with sides 10, 17, and 21 cm is rotated around the larger side. Calculate the volume and surface area of the resulting solid of revolution. | ## Solution.
From $\triangle A O B(O B=21-x) R^{2}=17^{2}-(21-x)^{2}$ (Fig. 11.80). From $\triangle A O C$ $(O C=x) R^{2}=10^{2}-x^{2}$. Solving the system: $\left\{\begin{array}{l}R^{2}=17^{2}-(21-x)^{2}, \\ R^{2}=10^{2}-x^{2},\end{array}\right.$ we get $O C=x=6 \text{ cm}$, and $O B=21-6=15 \text{ cm}$. Then $R=\sqr... | 448\pi | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,376 |
11.095. A sphere is circumscribed around a regular triangular prism, the height of which is twice the side of the base. How does its volume relate to the volume of the prism? | ## Solution.
The volume of the sphere is: $V_{s}=\frac{4}{3} \pi R^{3}$ (Fig. 11.82). The volume of the prism is: $V_{p}=S \cdot H$. The area of the base of the prism is: $S=\frac{a^{2} \sqrt{3}}{4}$, and the height of the prism $H=2 a$. Therefore, $V_{p}=\frac{a^{3} \sqrt{3}}{2}$. Consider $\triangle A O B: O B=\frac... | \frac{64\pi}{27} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,379 |
11.096. Determine the surface area of the sphere circumscribed around a cone with a base radius of $R$ and height $h$. | Solution.
Fig. 11.83
The surface area of the sphere is: \( S_{\mathrm{u}} = 4 \pi r^{2} \) (Fig. 11.83). The cross-section of the cone is an isosceles triangle. For this triangle, the radi... | \frac{\pi(R^{2}+^{2})^{2}}{^{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,380 |
11.099. 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. | Solution.
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.
 ; \quad S_{1}=\pi R_{1}^{2}, S_{2}=\pi R_{2}^{2} \text {, then we get, }
$$
that $V_{\text {truncated cone }}=\frac{1}{3} h \pi\left(R_{1}^{2}+R_{2}^{2}+R_{1} R_{2}\right)$ (Fig. 11.87). According to the condition, we ha... | \frac{7V}{27} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,382 |
11.101. The lateral surface of a cone is unfolded onto a plane as a sector with a central angle of $120^{\circ}$ and an area of $S$. Find the volume of the cone. | ## Solution.
Let the radius of the base of the cone be $r$, and its slant height be $l$. Then the area of the development $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 ... | \frac{2S\sqrt{6\piS}}{27\pi} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,383 |
11.102. A copper blank in the form of a rectangular parallelepiped with dimensions $80 \times 20 \times 5 \mathrm{~cm}$ is rolled into a sheet with a thickness of 1 mm. Determine the area of this sheet. | ## Solution.
The required area can be found using the formula $S=V / h$, where $V$ is the volume of the sheet, and $h$ is its thickness (the shape of the sheet does not matter). Since the volumes of both bodies are equal, then $V=80 \cdot 20 \cdot 5=8000$ (cm³). Therefore, $S=8000 / 0.1=80000 \mathrm{~cm}^{2}=8 \mathr... | 8\mathrm{~}^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,384 |
11.103. A metal sphere of radius $R$ is recast into a cone, the lateral surface area of which is three times the area of the base. Calculate the height of the cone. | Solution.
Since the radius of the sphere is $R$, the volumes of the bodies are $\frac{4}{3} \pi R^{3}$. Let $r$ be the radius of the base of the cone; then, by the condition, $S_{\text {side }}=3 \pi r^{2}$. On the other hand, $S_{\text {side }}=\pi r l$, where $l$ is the slant height of the cone. Therefore, $l=3 r$, ... | 2R\sqrt[3]{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,385 |
11.105. 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. | Solution.
Let the side of the rhombus be $a$, and its diagonals be $2 d_{1}$ and $2 d_{2}$ (Fig. 11.89). When rotated, a body consisting of two cones is formed. Let the volume and surface area of the body of revolution around diagonal $A C$ be denoted by $V_{A C}$ and $S_{A C}$, and around diagonal $B D$ by $V_{B D}$ ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 51,386 |
12.001. The sum of two unequal heights of an isosceles triangle is $l$, the angle at the vertex is $\alpha$. Find the lateral side. | Solution.
By the condition $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. 12.4). Let $B C=a$. From $\triangle A A_{1} C$ we have $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 find $B B_{... | \frac{}{2\sin\frac{\pi+\alpha}{4}\cos\frac{\pi-3\alpha}{4}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,387 |
12.002. The angle at the base of an isosceles triangle is equal to $\alpha$. Find the ratio of the radii of the inscribed and circumscribed circles. | ## Solution.
Given $A B=B C, \angle B A C=\angle B C A=\alpha$. Let $A C=x$, then from $\triangle A E O, \angle A E O=90^{\circ}$ we have $O E=r=\frac{x}{2} \operatorname{tg} \frac{\alpha}{2}$ (Fig. 12.5), where $r$ is the radius of the inscribed circle. From $\triangle A E B \quad \angle A E B=90^{\circ}$ we have tha... | \operatorname{tg}\frac{\alpha}{2}\cdot\sin2\alpha | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,388 |
12.003. 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? | Solution.
Let $\angle A B C=\alpha$ and $\frac{\angle A B E}{\angle E B C}=\frac{1}{2}$, then $\angle E B C=\frac{2}{3} \alpha, \angle A B E=\frac{\alpha}{3}$, $\angle D B C=\frac{\angle A B C}{2}=\frac{\alpha}{2}, \angle E B M=\angle E B C-\angle D B C=\frac{2 \alpha}{3}-\frac{\alpha}{2}=\frac{\alpha}{6}$ (Fig. 12.6)... | \cos\frac{\alpha}{6}:\cos\frac{\alpha}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,389 |
12.004. In square $ABCD$, a line is drawn through the midpoint $M$ of side $AB$, intersecting the opposite side $CD$ at point $N$. In what ratio does the line $MN$ divide the area of the square if the acute angle $AMN$ is equal to $\alpha$? Specify the possible values of $\alpha$. | ## Solution.
$A B C D$ is a square, $M \in A B, M A=M B, \angle A M N=\alpha, N \in C D$ (Fig. 12.7); it is required to find $S_{A M N D}: S_{B M N C}$. Let $A B=a$; then
$$
S_{A M N D}=\frac{M A+N D}{2} \cdot a=\frac{a+2 N D}{4} \cdot a, S_{B M N C}=\frac{M B+N C}{2} \cdot a=\frac{\frac{a}{2}+a-N D}{2} \cdot a=
$$
... | \operatorname{tg}(\alpha-\frac{\pi}{4});\operatorname{arctg}2<\alpha<\pi/2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,390 |
12.005. 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. | ## Solution.
Given that in the isosceles trapezoid $ABCD$ we have: $AB = CD$, $BB_1 \perp AD, BB_1 = h, AC \cap BD = O, \angle COD = \alpha$ (Fig. 12.8). Since $\angle COD$ is the external angle of the isosceles triangle $AOD$, we have $\angle OAD = \angle ODA = \frac{\alpha}{2}$. Further, we have $B_1D = ED + B_1E = ... | \operatorname{ctg}\frac{\alpha}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,391 |
12.006. 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. | Solution.
In the right triangle $ACB$, we have: $\angle ACB=90^{\circ}, \angle CAB=\alpha$, $S_{\triangle ABC}=S, AB_1=B_1C$ and $AC_1=C_1B, BB_1 \cap CC_1=O$ (Fig. 12.9);
Fig. 12.8
, triangle $A E F$ is inscribed. Point $E$ lies on side $B C$, point $F$ - on side $C D$. Find the tangent of angle $E A F$, if $A B: B C = B E: E C = C F: F D = k$. | Solution.
By the condition $\frac{A B}{B C}=\frac{B E}{E C}=\frac{C F}{F D}=k$ (Fig. 12.10). Let $A D=B C=x$, then $A B=D C=k x \cdot B C=B E+E C=k E C+E C$, hence $E C=\frac{B C}{1+k}=\frac{x}{1+k}$,
^{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,393 |
12.010. An isosceles trapezoid with an acute angle \(\alpha\) at the base is circumscribed around a circle of radius \(R\). Find the perimeter of this trapezoid. | ## Solution.
Given that $A B C D$ is a trapezoid, $A B=C D, O$ is the center of the circle inscribed in the trapezoid, $O E \perp A D, O E=R, \angle B A D=\alpha, \alpha<\pi / 2$ (Fig. 12.13); we need to find $P_{A B C D}=A D+2 A B+B C$. By the property of a circumscribed quadrilateral, $A D+B C=2 A B$ and $P_{A B C D... | \frac{8R}{\sin\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,396 |
12.013. The ratio of the area of a right-angled triangle to the area of a square constructed on its hypotenuse is $k$. Find the sum of the tangents of the acute angles of the triangle. | ## Solution.
In $\triangle A C B \quad \angle A C B=90^{\circ}$. Let $\angle C A B=\alpha, \angle A B C=\beta$, then $\operatorname{tg} \alpha+\operatorname{tg} \beta=\frac{\sin (\alpha+\beta)}{\cos \alpha \cdot \cos \beta}=\frac{1}{\cos \alpha \cdot \cos \beta} \cdot\left(\alpha+\beta=90^{\circ}\right)$. Denote $A B=... | \operatorname{tg}\alpha+\operatorname{tg}\beta=\frac{1}{2k} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,399 |
12.014. The area of a rectangular trapezoid is $S$, the acute angle is $\alpha$. Find the height of the trapezoid if its smaller diagonal is equal to the larger base. | Solution.
Given $\angle C B A=90^{\circ}, \angle A D C=\alpha, A C=A D . S_{A B C D}=S$, then $\quad S=\frac{B C+A D}{2} \cdot h$, from which $\quad h=\frac{2 S}{B C+A D}=\frac{2 S}{A D-E D+A D}$ (Fig. 12.17). From $\triangle C E D, \angle C E D=90^{\circ}, E D=h \cdot \operatorname{ctg} \alpha$. From $\triangle A E C... | \sqrt{2S\operatorname{ctg}\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,400 |
12.016. 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. | Solution.
By the condition $B F \perp C D, B F=h_{2}, B E \perp A D, B E=h_{1}, \angle E B F=\alpha$. Then $\angle A B F=\angle C F B=90^{\circ} \quad(A B \| C D), \angle A B E=90^{\circ}-\alpha$. From $\triangle B E A$ $\angle B E A=90^{\circ}, A E=h_{1} \cdot \operatorname{tg}\left(90^{\circ}-\alpha\right)=h_{1} \cd... | \frac{\sqrt{h_{2}^{2}+h_{1}^{2}+2h_{1}h_{2}\cdot\cos\alpha}}{\sin\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,402 |
12.022. Find the angle of a triangle if it is known that the sides enclosing this angle are 1 and 3, and the bisector of the angle is \(0.75 \sqrt{3}\). | ## Solution.
In $\triangle A B C$ we have: $A C=1, B C=3, \angle A C C_{1}=\angle B C C_{1}, C C_{1}=0.75 \sqrt{3}$; we need to find $\angle A C B$ (Fig. 12.25).
We use the formula $l_{c}=\frac{2 a b \cos (C / 2)}{a+b}$ (see "Some relations between elements of figures", p. 713). Expressing $\cos \frac{C}{2}$ from thi... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,407 |
12.024. Find the ratio of the perimeter of a trapezoid circumscribed about a circle to the length of this circle, if the angles at the larger base of the trapezoid are $\alpha$ and $\beta$.
... | ## Solution.
Given that $ABCD$ is a trapezoid circumscribed around a circle, $\angle BAD = \alpha, \angle CDA = \beta$. Let $R$ be the radius of the inscribed circle. The circumference of the circle is $l = 2\pi R$ (Fig. 12.27). $P = BC + AD + AB + CD$, $BC + AD = AB + CD$, $2(AB + CD) = P$. From $\triangle BFA$, $\an... | \frac{4\cdot\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}}{\pi\cdot\sin\alpha\sin\beta} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,409 |
12.025. In a right triangle $ABC$, the acute angle $A$ is equal to $\alpha$ radians. An arc of a circle centered at the right angle vertex $C$ touches the hypotenuse at point $D$ and intersects the legs $AC$ and $BC$ at points $E$ and $F$, respectively. Find the ratio of the areas of the curvilinear triangles $ADE$ and... | ## Solution.
By the condition in $\triangle A C B \quad \angle A C B=90^{\circ}, \angle C A B=\alpha$ (Fig. 12.28). Let $A C=x$. From $\triangle A D C, \angle A D C=90^{\circ}, C D=x \sin \alpha$, then
$$
\begin{aligned}
& \left.S_{C E D}=\frac{1}{2} x^{2} \sin ^{2} \alpha \cdot\left(\frac{\pi}{2}-\alpha\right) \text... | \frac{\operatorname{tg}\alpha-\alpha}{\operatorname{ctg}\alpha-\frac{\pi}{2}+\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,410 |
12.026. A rhombus is inscribed in a parallelogram with sides $a$ and $b(a<b)$ and an acute angle $\alpha$; two of its vertices coincide with the midpoints of the larger sides of the parallelogram, the other two lie on the smaller sides (or their extensions). Find the angles of the rhombus. | Solution.
Let $D C=b, B C=a, \angle A D C=\alpha=\angle A K O ; N E F M$ be a rhombus, $K O=\frac{D C}{2}=\frac{b}{2}$ (Fig. 12.29), $O F=\frac{B C}{2}=\frac{a}{2}$. From $\triangle E O F \quad \angle E O F=90^{\circ}$, $\operatorname{tg} \angle O E F=\frac{O F}{E O} \cdot$ From $\triangle K E O \quad \angle K E O=90^... | 2\operatorname{arctg}\frac{}{b\sin\alpha};\pi-2\operatorname{arctg}\frac{}{b\sin\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,411 |
12.027. A trapezoid with angles $\alpha$ and $\beta$ at the larger base is circumscribed around a circle of radius $R$. Find the area of this trapezoid. | Solution.
Let $\angle BAD = \alpha, \angle CDA = \beta, OE = R$ (Fig. 12.30). Then $S_{ABCD} = \frac{BC + AD}{2} \cdot 2R \cdot S_{ABCD} = (BC + AD) \cdot R \cdot$ Since $BC + AD = AB + CD$, we have $BC + AD = 2R\left(\frac{1}{\sin \alpha} + \frac{1}{\sin \beta}\right)\left(AB = \frac{2R}{\sin \alpha}, CD = \frac{2R}{... | \frac{4R^2\cdot\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}}{\sin\alpha\sin\beta} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,412 |
12.030. An equilateral triangle is intersected by a line passing through the midpoint of one of its sides and forming an acute angle $\alpha$ with this side. In what ratio does this line divide the area of the triangle | Solution.
Let $A B=B C=A C=a, \angle M E C=\alpha, A E=E C$ (Fig. 12.33). From $\triangle M K E$, $\angle M K E=90^{\circ}, E K=M K \cdot \operatorname{ctg} \alpha$. In $\triangle M K C, \angle M K C=90^{\circ}, \quad K C=$ $=M K \cdot \operatorname{ctg} 60^{\circ}=\frac{M K}{\sqrt{3}}, E C=E K+K C=M K\left(\operatorn... | \frac{2\sqrt{3}\cos\alpha+\sin\alpha}{\sin\alpha} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,415 |
12.031. In the square $A B C D$, an isosceles triangle $A E F$ is inscribed; point $E$ lies on side $B C$, point $F$ - 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$. | Solution.
Given: $ABCD$ is a square, $AEF$ is an isosceles triangle, $AE = AF, E \in BC, F \in CD, \operatorname{tg} \angle AEF = 3$ (Fig. 12.34); 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... | \frac{2\sqrt{5}}{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,416 |
12.032. In an isosceles triangle, the angle between the lateral sides is $\alpha$, and the radius of the inscribed circle is $r$. Find the area of the triangle. | ## Solution.
Given in $\triangle ABC \quad AB=BC, \angle ABC=\alpha, r$ - the radius of the inscribed circle. From $\triangle OEC \quad \angle OEC=90^{\circ}$ (Fig. 12.35), $EC=r \cdot \operatorname{ctg} \angle ECO=$ $=r \cdot \operatorname{ctg}\left(\pi / 4-\frac{\alpha}{4}\right) \cdot$ From $\triangle BEC, \angle B... | r^{2}\cdot\operatorname{ctg}^{2}(\frac{\pi-\alpha}{4})\cdot\operatorname{ctg}\frac{\alpha}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,417 |
12.033. A rectangular trapezoid with an acute angle $\alpha$ is circumscribed around a circle. Find the height of the trapezoid if its perimeter is $P$. | ## Solution.
Let $\angle C D A=\alpha, \angle A B C=90^{\circ}, P-$ be the perimeter of the trapezoid (Fig. 12.36) $(B C \| A D)$. Since $B C+A D=A B+C D$, and $P=B C+A D+A B+C D$, then $B C+A D=\frac{P}{2}=D C+A B$. Denote $A B=C E=x$. From $\triangle C E D$,
$$
\angle C E D=90^{\circ}, C D=\frac{x}{\sin \alpha}, \t... | \frac{P\sin\alpha}{4\cdot\cos^{2}(\frac{\pi}{4}-\frac{\alpha}{2})} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,418 |
12.039. The height $BD$ of the equilateral triangle $ABC$ is extended beyond vertex $B$, and on this extension, a segment $BF$ equal to the side of the triangle is taken. Point $F$ is connected to vertex $C$ by a straight line segment. Using this construction, show that $\operatorname{tg} 15^{\circ}=2-\sqrt{3}$. | Solution.
By the condition $B F=A B=B C=A C \cdot$ In $\triangle F D C \quad \angle F D C=90^{\circ}$, $\angle D F C+\angle F C D=90^{\circ}, \angle B F C+\angle D C B+\angle B C F=90^{\circ}, \angle B C F=\angle B F C$ since $B F=B C$ (by the condition), then $2 \cdot \angle B F C+60^{\circ}=90^{\circ}, \angle B F C=... | 2-\sqrt{3} | Geometry | proof | Yes | Yes | olympiads | false | 51,423 |
12.040. The height of an isosceles trapezoid is $h$. 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 is seen from the midpoint of the upper base at an angle of $2 \beta$. Find the area of the trapezoid in this general case and calculate it withou... | ## Solution.
In trapezoid $ABCD$, we have: $AB=CD, BC \| AD, M \in AD, AM=MD$, $N \in BC, BN=NC, BE \perp AD, BE=h, \angle BMC=2\alpha, \angle AND=2\beta$ (Fig. 12.42). Since $\triangle ABM = \triangle CMD$ and $\triangle ABN = \triangle NCD$ (by two sides and the included angle), we have $BM=MC$ and $AN=ND$. From $\t... | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,424 |
12.041. Given two sides $b$ and $c$ of a triangle and its area, equal to $0.4 b c$. Find the third side. | ## Solution.
By the condition $S_{\triangle A B C}=0.4 \cdot b \cdot c$, where $b=A C, c=A B$ (Fig. 12.43). Let $C B=x$, then $S_{\triangle A B C}=\sqrt{p(p-c)(p-b)(p-x)}$, where $p=\frac{c+b+x}{2}$ (by Heron's formula) and $S_{\triangle A B C}=\sqrt{\frac{c+b+x}{2} \cdot \frac{b+x-c}{2} \cdot \frac{b+c-x}{2} \cdot \f... | R^{2}(\alpha+\sin\alpha) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,425 |
12.043. Through the vertex $A$ of an isosceles acute triangle $ABC$ and the center of the circumscribed circle around this triangle, a line is drawn, intersecting the side $BC$ at point $D$. Find the length of $AD$, if $AB=BC=b$ and $\angle ABC=\alpha$. | Solution.
Given $A B=B C=b, \angle A B C=\alpha, O$ is the center of the circumscribed circle. Consider $\triangle B E O, \angle B E O=90^{\circ}, B O=\frac{b}{2 \cos \frac{\alpha}{2}}=A O$, $O E=B E \cdot \operatorname{tg} \frac{\alpha}{2}=\frac{b}{2} \operatorname{tg} \frac{\alpha}{2}$ (Fig. 12.45). Since $A O=B O$,... | \frac{b\sin\alpha}{\sin\frac{3\alpha}{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,426 |
12.044. In a rectangular parallelepiped, the diagonal of the base is equal to $d$ and forms an angle $\alpha$ with a side of the base. A plane is drawn through this side and the opposite side of the upper base, forming an angle $\beta$ with the plane of the base. Find the lateral surface area of the parallelepiped. | Solution.
Given $A C=d, \angle A C D=\alpha, \angle A_{1} D A=\beta$ (Fig. 12.46).
$D C=d \cos \alpha$. From $\triangle A A_{1} D, \angle A A_{1} D=90^{\circ}, A A_{1}=d \sin \alpha \cdot... | 2\sqrt{2}^{2}\cdot\sin\alpha\cdot\operatorname{tg}\beta\cdot\sin(\alpha+\frac{\pi}{4}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 51,427 |
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