Task A-1.6. (10 points)

The sum of the lengths of the legs of a trapezoid is $4 \sqrt{10}$, and the height is 6. The area of the trapezoid is 72. If the trapezoid is inscribed in a circle, determine the radius of that circle.
## Solution.

Given the height $v=6$ and the area $P=72$, from $P=\frac{1}{2}(a+c) v$ we get $a+c=24$, where $a$ is the longer and $c$ is the shorter base.

(1 point)

The trapezoid is inscribed in a circle, so it must be isosceles.

Therefore, the length of the leg $b=2 \sqrt{10}$.

From $v^{2}+\left(\frac{a-c}{2}\right)^{2}=b^{2}$ we get $\frac{a-c}{2}=\sqrt{(2 \sqrt{10})^{2}-6^{2}}=2$.

Now, $a+c=24, a-c=4$, so $a=14$ and $c=10$.

(1 point)

Let $M$ and $N$ be the midpoints of the bases of the trapezoid $ABCD$ as shown in the figure. To determine the radius of the circumscribed circle, observe the right triangles $AMS$ and $DNS$.

We have $|AM|^{2}+|MS|^{2}=|AS|^{2}$ and $|DN|^{2}+|NS|^{2}=|DS|^{2}$.
![](https://cdn.mathpix.com/cropped/2024_05_30_d6d90a155ebaa296cbbeg-04.jpg?height=668&width=1556&top_left_y=1182&top_left_x=250)

Let the radius of the circle be $R$ and let $|MS|=x$.

We are not sure whether the center of the circle lies inside or outside the trapezoid. Therefore, we have two possibilities:

$$
|NS|=6-x \quad \text{or} \quad|NS|=6+x
$$

![](https://cdn.mathpix.com/cropped/2024_05_30_d6d90a155ebaa296cbbeg-04.jpg?height=117&width=1033&top_left_y=2186&top_left_x=183)

By subtracting, we get: $\left(49+x^{2}\right)-\left(25+36-12 x+x^{2}\right)=0$,

or $12 x=12$ so $x=1$ and finally $R^{2}=50$.

(2 points)

The radius of the circle is $5 \sqrt{2}$.

The second case leads to the system $\left\{\begin{array}{l}7^{2}+x^{2}=R^{2}, \\ 5^{2}+(6+x)^{2}=R^{2},\end{array}\right.$ from which we get $x=-1<0$, so this case is invalid.