[Isosceles, Inscribed, and Circumscribed Trapezoids] Area of a Trapezoid

In trapezoid $A B C D$ with bases $B C$ and $A D$ and legs $A B$ and $C D$, a circle with center $O$ is inscribed. Find the area of the trapezoid if angle $D A B$ is a right angle, $O C=2$, and $O D=4$.
Triangle $C O D$ is a right triangle.

## Solution

Let $M$ be the point of tangency of the circle with the lateral side $C D, r--$ the radius of the circle. Since $\angle C O D=90^{\circ}$, then

$$
C D=\sqrt{O C^{2}+O D^{2}}=\sqrt{4+16}=2 \sqrt{5}, O C \cdot O D=C D \cdot O M
$$

From this, we find that

$$
r=O M=\frac{O C \cdot O D}{C D}=\frac{4}{\sqrt{5}}
$$

The height of the trapezoid: $A B=2 R=\frac{8}{\sqrt{5}}$. Therefore,

$$
S_{\mathrm{ABCD}}=\frac{1}{2}(A D+B C) \cdot A B=\frac{1}{2}(C D+A B) \cdot A B=\frac{72}{5}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_cb0f0d605674d837f9abg-21.jpg?height=363&width=532&top_left_y=1948&top_left_x=769)

## Answer

$\frac{72}{5}$.