10.410 A circle is inscribed in a trapezoid, where the smaller base is equal to $a$. One of the lateral sides of the trapezoid is divided by the point of tangency into segments of length $\boldsymbol{m}$ and $n$, measured from the larger base. Determine the area of the trapezoid.
Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-248.jpg?height=270&width=417&top_left_y=1070&top_left_x=90)

Let $O$ be the center of the circle with radius $r$ inscribed in the given trapezoid $ABCD$. $M, N, K, L$ are the points of tangency of the trapezoid with the circle. $BC = a, AM = m, BM = n$. $\angle AOB = \angle COD = 90^\circ$ (Theorem 9) $\Rightarrow$

$\Rightarrow r^2 = OM^2 = BM \cdot AM = m \cdot n = CK \cdot DK$ (Theorem 5). Further, $n = BM = BN, m = AM = AL$ (Theorem 3) $\Rightarrow CK = CN = a - n$,

$$
DK = DL = AD - m \Rightarrow m \cdot n = (a - n)(AD - m) \Rightarrow AD = \frac{a \cdot m}{a - n}
$$

$$
S_{ABCD} = 2r \cdot \frac{BC + AD}{2} = \frac{\sqrt{m \cdot n} \cdot (a^2 - a \cdot n + a \cdot m)}{a - n}.
$$

Answer: $\frac{a \sqrt{m \cdot n} \cdot (a - n + m)}{a - n}$.