In trapezoid $A B C D$, inscribed around a circle, $B C \| A D, A B=C D, \angle B A D=45^{\circ}$. The area of the trapezoid is 10.

Find $A B$.
In the given trapezoid, the midline is equal to the lateral side

## Solution

Let $r$ be the radius of the circle inscribed in the given trapezoid, $S$ be its area, $A B = C D = a$, and $B H$ be the height of the trapezoid.

Since a circle is inscribed in the trapezoid, we have $A D + B C = A B + C D = 2a$, and the height of the trapezoid is $2r$. Therefore,

$$
S = \frac{A D + B C}{2} \cdot B H = a \cdot 2r = 2ar
$$

From the right triangle $A B H$, we find that

$$
2r = B H = A B \cdot \sin 45^{\circ} = \frac{a}{\sqrt{2}}
$$

Therefore,

$$
2ar = 2a \cdot \frac{a}{2\sqrt{2}} = \frac{a^2}{\sqrt{2}} = 10
$$

Thus, $A B = a = \sqrt{10 \sqrt{2}}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_4c124ad32a76c9faf91dg-33.jpg?height=368&width=783&top_left_y=2543&top_left_x=640)