Around a circle with radius 3, an isosceles trapezoid $A B C D$ ($B C \| A D$) is described, the area of which is 48. The circle touches the sides $A B$ and $C D$ at points $K$ and $L$. Find $K L$.

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Let $B H$ be the height of the trapezoid, $O P$ be the perpendicular dropped from the center $O$ of the circle to the chord $K L$, and $K$ be the point of tangency of the circle with the lateral side $A B$. Then triangles $O P K$ and $A H B$ are similar.

## Solution

Let the circle with radius $r=3$ touch the lateral sides $A B$ and $C D$ of the trapezoid $A B C D$ at points $K$ and $L$ respectively; $B H=2 r$ is the height of the trapezoid. By symmetry, $K L \| B C$. Since the trapezoid $A B C D$ is circumscribed, $B C + A D = A B + C D = 2 A B$.

According to the problem, $1 / 2 (B C + A D) \cdot 6 = 48$, from which $B C + A D = 16$. Therefore, $A B = 8$.

The perpendicular $O P$, dropped from the center $O$ of the circle, bisects the chord $K L$. From the similarity of the right triangles $O P K$ and $A H B$, it follows that

$P K : O K = B H : A B$. Hence, $K L = 2 P K = 2 B H \cdot O K / A B = 9 / 2$.

![](https://cdn.mathpix.com/cropped/2024_05_06_4c124ad32a76c9faf91dg-26.jpg?height=520&width=1066&top_left_y=743&top_left_x=496)

## Answer

4.5.