8,9 |

The center of a circle with a radius of 5, circumscribed around an isosceles trapezoid, lies on the larger base, and the smaller base is equal to 6. Find the area of the trapezoid.
Let the center of the circle circumscribed around the isosceles trapezoid $ABCD$ with bases $AB$ and $CD$ lie on the larger base $AD$. Then $AD$ is the diameter of the circle, so $AD=10$ and $\angle ACD=90^{\circ}$. Let $CH$ be the height of the trapezoid. Then

$$
AH=\frac{1}{2}(AD+BC)=\frac{1}{2}(10+6)=8, DH=\frac{1}{2}(AD-BC)=\frac{1}{2}(10-6)=2 .
$$

By the property of the height of a right triangle drawn from the vertex of the right angle

$$
CH=\sqrt{AH \cdot DH}=\sqrt{8 \cdot 2}=4
$$

Therefore,

$$
S_{ABCD}=\frac{1}{2}(BC+AD) \cdot CH=8 \cdot 4=32
$$

## Answer

32.00