An inscribed circle of radius $R$ is drawn in an isosceles trapezoid. The upper base of the trapezoid is half the height of the trapezoid. Find the area of the trapezoid.
Find the segments into which the point of tangency of the inscribed circle divides the lateral side.

## Solution

The upper base of the given trapezoid is equal to $R$. The lateral side is divided by the point of tangency into segments, one of which is equal to $\frac{R}{2}$, and the second is $-\frac{R^{2}}{\frac{R}{2}}=2 R$. Therefore, the lower base is equal to $4 R$. Consequently, the area of the trapezoid is $(4 R+R) R=5 R^{2}$.

The upper base of the given trapezoid is equal to $R$. The lateral side is divided by the point of tangency into segments, one of which is equal to $\frac{R}{2}$, and the second is $-\frac{R^{2}}{\frac{R}{2}}=2 R$. Therefore, the lower base is equal to $4 R$. Consequently, the area of the trapezoid is $(4 R+R) R=5 R^{2}$.

The upper base of the given trapezoid is equal to $R$. The lateral side is divided by the point of tangency into segments, one of which is equal to $\frac{R}{2}$, and the second is $-\frac{R^{2}}{\frac{R}{2}}=2 R$. Therefore, the lower base is equal to $4 R$. Consequently, the area of the trapezoid is $(4 R+R) R=5 R^{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_1b17058527caf5b5d058g-07.jpg?height=435&width=646&top_left_y=214&top_left_x=709)

## Answer

$5 R^{2}$