8,9

A circle inscribed in trapezoid $ABCD$ touches the lateral side $AB$ at point $F$. Find the area of the trapezoid if $AF=m, FB=n$, and the smaller base of the trapezoid $BC$ is equal to $b$.
Draw a radius to the point of tangency with side $A B$ and apply the theorem about the height dropped from the vertex of a right angle to the hypotenuse.

## Solution

Let $O$ be the center of the circle; $M, K, N$ be the points of tangency with sides $B C, C D, A D$ respectively, $R$ be the radius of the circle. Then

$$
O F^{2}=B F \cdot A F=m n, R=O F=\sqrt{m n}, C K=C M=C B-B M=b-n
$$

Since $O K^{2}=C K \cdot K D$, then

$$
K D=\frac{m n}{b-n}, A D=A N+N D=m+\frac{m n}{b-n}
$$

Therefore,

$$
S_{\mathrm{ABCD}}=\frac{B C+A D}{2} \cdot 2 R=\frac{1}{2}\left(b+m+\frac{m n}{b-n}\right) \cdot 2 \sqrt{m n}=\left(b+m+\frac{m n}{b-n}\right) \sqrt{m n}
$$

Let $O$ be the center of the circle; $M, K, N$ be the points of tangency with sides $B C, C D, A D$ respectively, $R$ be the radius of the circle. Then

$$
O F^{2}=B F \cdot A F=m n, R=O F=\sqrt{m n}, C K=C M=C B-B M=b-n
$$

Since $O K^{2}=C K \cdot K D$, then

$$
K D=\frac{m n}{b-n}, A D=A N+N D=m+\frac{m n}{b-n}
$$

Therefore,

$$
S_{\mathrm{ABCD}}=\frac{B C+A D}{2} \cdot 2 R=\frac{1}{2}\left(b+m+\frac{m n}{b-n}\right) \cdot 2 \sqrt{m n}=\left(b+m+\frac{m n}{b-n}\right) \sqrt{m n}
$$

Let $O$ be the center of the circle; $M, K, N$ be the points of tangency with sides $B C, C D, A D$ respectively, $R$ be the radius of the circle. Then

$$
O F^{2}=B F \cdot A F=m n, R=O F=\sqrt{m n}, C K=C M=C B-B M=b-n
$$

$$
K D=\frac{m n}{b-n}, A D=A N+N D=m+\frac{m n}{b-n}
$$

Therefore,

$$
S_{\mathrm{ABCD}}=\frac{B C+A D}{2} \cdot 2 R=\frac{1}{2}\left(b+m+\frac{m n}{b-n}\right) \cdot 2 \sqrt{m n}=\left(b+m+\frac{m n}{b-n}\right) \sqrt{m n}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_96a7587f2ba0a7b3aee2g-43.jpg?height=411&width=604&top_left_y=366&top_left_x=727)

Answer

$\left(b+m+\frac{m n}{b-n}\right) \sqrt{m n}$