10.360. The center of the circle circumscribed around an isosceles trapezoid divides its height in the ratio 3:4. Find the bases of the trapezoid, if the radius of the circle is 10 and its midline is equal to the height.
## Solution.

Let $O K=x, A L=y, B K=z$ (Fig. 10.139).

Then $M N=z+y=x+\frac{3}{4} x=\frac{7 x}{4} \cdot$ In $\triangle O K B$ and $\triangle O L A$ we have $O B^{2}=O K^{2}+B K^{2}, O A^{2}=O L^{2}+A L^{2}$. We arrive at the system of equations

$$
\left\{\begin{array}{l}
x^{2}+z^{2}=100, \\
\frac{9 x^{2}}{16}+y^{2}=100, \\
\frac{7 x}{4}-z=y
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x^{2}+z^{2}=100 \\
\frac{9 x^{2}}{16}+\left(\frac{7 x}{4}-z\right)^{2}=100
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x^{2}+z^{2}=100 \\
\frac{27 x^{2}}{8}-\frac{7 x z}{2}+z^{2}=100
\end{array}\right.\right.\right.
$$

Subtracting the first equation from the second, we get $\frac{21 x^{2}}{8}-\frac{7 x z}{2}=0$, from which (since $x \neq 0$) $x=\frac{4 z}{3}$. Therefore, $z^{2}+\frac{16 z^{2}}{9}=100$ and $z=6$. From this, we find $x=8, y=8$. Thus, $A D=2 y=16, B C=2 z=12$.

Answer: 12 and 16.

## Solutions to Chapter 11

## PROBLEMS IN STEREOOMETRY

## BASIC FORMULAS

1. Arbitrary prism ( $l$ - lateral edge; $P$ - perimeter of the base; $S$ - area of the base; $H$ - height; $P_{\text {sec}}$ - perimeter of the perpendicular section; $S_{\text {sec}}$ - area of the perpendicular section; $S_{\text {lat}}$ - area of the lateral surface; $V$ - volume):

$$
\begin{gathered}
S_{60 \mathrm{~K}}=P_{\text {sec}} l \\
V=S H \\
V=S_{\text {sec}} l
\end{gathered}
$$

2. Right prism:

$$
S_{6 \mathrm{ok}}=P l .
$$

3. Rectangular parallelepiped ( $a, b, c-$ its dimensions; $d-$ diagonal):

$$
\begin{gathered}
S_{60 \mathrm{k}}=P H \\
V=a b c \\
d^{2}=a^{2}+b^{2}+c^{2}
\end{gathered}
$$

4. Cube ( $a$ - edge):

$$
\begin{gathered}
V=a^{3} \\
d=a \sqrt{3}
\end{gathered}
$$

5. Arbitrary pyramid ( $S$ - area of the base; $H$ - height; $V$ - volume):

$$
V=\frac{1}{3} S H
$$

6. Regular pyramid ( $P$ - perimeter of the base; $l$ - apothem; $S_{\text {lat}}$ - area of the lateral surface):

$$
\begin{gathered}
S_{6 \mathrm{ok}}=\frac{1}{2} P l \\
V=\frac{1}{3} S H
\end{gathered}
$$

7. Arbitrary truncated pyramid ( $S_{1}$ and $S_{2}$ - areas of the bases; $h$ - height; $V$ - volume):

$$
V=\frac{1}{3} h\left(S_{1}+S_{2}+\sqrt{S_{1} S_{2}}\right)
$$

8. Regular truncated pyramid ( $P_{1}$ and $P_{2}$ - perimeters of the bases; $l$ - apothem; $S_{\text {lat}}$ - area of the lateral surface):

$$
S_{6 \mathrm{ok}}=\frac{1}{2}\left(P_{1}+P_{2}\right) l
$$

9. Cylinder ( $R$ - radius of the base; $H$ - height; $S_{\text {lat}}$ - area of the lateral surface; $V$ - volume):

$$
\begin{gathered}
S_{6 \text { ok }}=2 \pi R H \\
V=\pi R^{2} H
\end{gathered}
$$

10. Cone ( $R$ - radius of the base; $H$ - height; $l$ - slant height; $S_{\text {lat}}$ - area of the lateral surface; $V$ - volume):

$$
\begin{gathered}
S_{6 \mathrm{ok}}=\pi R l \\
V=\frac{1}{3} \pi R^{2} H
\end{gathered}
$$

11. Sphere, spherical surface ( $R$ - radius of the sphere; $S$ - area of the spherical surface; $V$ - volume):

$$
\begin{aligned}
S & =4 \pi R^{2} \\
V & =\frac{4}{3} \pi R^{3}
\end{aligned}
$$

12. Spherical segment ( $R$ - radius of the sphere; $h$ - height of the segment; $S$ - area of the spherical surface of the segment; $V$ - volume):

$$
S=2 \pi R h
$$

$$
V=\pi h^{2}\left(R-\frac{1}{3} h\right)
$$

13. Spherical sector ( $R$ - radius of the sphere; $h$ - height of the segment; $V$ - volume):

$$
V=\frac{2}{3} \pi R^{2} h
$$

## ADDITIONAL RELATIONS BETWEEN ELEMENTS OF PRISMS AND PYRAMIDS

1. Suppose in a pyramid one of the following two conditions is met: a) all lateral edges form equal angles with the plane of the base; b) the lengths of all lateral edges are equal. Then the vertex of the pyramid projects onto the center of the circle circumscribed around the base of the pyramid (this point also serves as the point of intersection of the perpendicular bisectors of the sides of the base of the pyramid).
2. Suppose in a pyramid one of the following conditions is met: a) all lateral faces form equal angles with the base; b) the lengths of all apothems of the lateral faces are equal.

Then the vertex of the pyramid projects onto the center of the circle inscribed in the base of the pyramid (this point also serves as the point of intersection of the angle bisectors of the base of the pyramid).

3. If in a skewed prism the lateral edge $A_{1} B_{1}$ forms equal angles with the sides of the base forming vertex $A_{1}$ (Fig. 11.1), then the base $O$ of the height $B_{1} O$ lies on the bisector of angle $A_{1}$.

This statement can also be

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0820.jpg?height=508&width=521&top_left_y=1280&top_left_x=706)

Fig. 11.1

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0821.jpg?height=359&width=327&top_left_y=79&top_left_x=174)

Fig. 11.2

formulated as follows: if in a trihedral angle two acute dihedral angles are equal, then the projection of their common edge onto the plane of the third dihedral angle is its bisector.

4. If the height of a triangular pyramid passes through the point of intersection of the altitudes of the triangle lying in the base, then the opposite edges of the pyramid are perpendicular. The converse statement is also true.
5. If $S O$ is the height of the pyramid $S A B C$ and $S A \perp B C$, then the area $S A O \perp B C$ (Fig. 11.2).

Proofs of the indicated additional relations can be found in any edition of this problem book of recent years.