A circle with its center at the intersection of the diagonals $K M$ and $L N$ of an isosceles trapezoid $K L M N$ touches the smaller base $L M$ and the lateral side $M N$. Find the perimeter of trapezoid $K L M N$, given that its height is 36 and the radius of the circle is 11.
Prove that the lateral side of the trapezoid is equal to the larger base.

## Solution

Let $O$ be the point of intersection of the diagonals of the given trapezoid, $A$ be the point of tangency of the circle with the base $L M, A B$, and $L P$ be the heights of the trapezoid. Then

$A M: K B = A O: O B = 11: 25$.

Let $L M = 11 x, K N = 25 x$. Since $L N$ is the angle bisector of $\angle K L M$, then $\angle L N K = \angle M L N = \angle N L K$.

Therefore, triangle $L K N$ is isosceles, $M N = L K = K N = 25 x, K P = 1 / 2(K N - L M) = 7 x$.

By the Pythagorean theorem, $625 x^{2} = 49 x^{2} + 36^{2}$, from which $x = 3 / 2$.

Therefore, the perimeter of the trapezoid is $86 x = 129$.

![](https://cdn.mathpix.com/cropped/2024_05_06_a7c46f4bbe36942e4ae1g-30.jpg?height=646&width=774&top_left_y=1575&top_left_x=642)

Answer

129.

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