10,11

The base of a right prism is an isosceles trapezoid $ABCD$, where $AB=CD=13, BC=11, AD=$ 21. The area of the diagonal section of the prism is 180. Find the area of the complete surface of the prism.
Let $A B C D A 1 B 1 C 1 D 1$ be the given prism, $S$ - the area of its total surface. The diagonal section of the prism is a rectangle $A C C 1 A 1$. Dropping a perpendicular $C K$ from vertex $C$ of the smaller base of the trapezoid to the larger base $A D$. Then

$$
D K=\frac{1}{2}(A D-B C)=\frac{1}{2}(21-11)=5
$$

$$
A K=\frac{1}{2}(A D+B C)=\frac{1}{2}(21+11)=16
$$

By the Pythagorean theorem from right triangles $C K D$ and $А K C$ we find that

$$
C K=\sqrt{C D^{2}-D K^{2}}=\sqrt{169-25}=12
$$

$$
A C=\sqrt{A K^{2}+C K^{2}}=\sqrt{256+144}=20
$$

According to the problem, $S_{A C C} 1 A 1=A C \cdot A A 1=180$. Therefore,

$$
A A 1=\frac{180}{A C}=\frac{180}{20}=9
$$

Thus,

$$
\begin{gathered}
S=2 S_{A B C D}+S_{A B B} 1 A 1+S_{B C C} 1 B 1+S_{C D D} 1 C 1+S_{A D D} 1 A 1= \\
=2 \cdot \frac{1}{2}(A D+B C) C K+(A B+B C+C D+A D) A A 1= \\
=2 \cdot 16 \cdot 12+58 \cdot 9=906
\end{gathered}
$$

## Answer

906.00