$$
[\quad \text { Inscribed Quadrilaterals (etc.). }]
$$

A quadrilateral $A B C D$ is inscribed in a circle with diameter 1, where angle $D$ is a right angle, and $A B = B C$.

Find the area of the quadrilateral $A B C D$ if its perimeter is $\frac{9 \sqrt{2}}{5}$.
$A B C$ - is an isosceles right triangle.

## Solution

Since $\angle D=90^{\circ}$, then $A C$ is the diameter of the circle. Therefore, $\angle B=90^{\circ}$. Consequently, $A B=B C=\frac{A C}{\sqrt{2}}=$

$\frac{1}{\sqrt{2}}, A D+C D=\frac{9 \sqrt{2}}{5}-A B-B C=\frac{4 \sqrt{2}}{5}$.

Since $A D^{2}+C D^{2}=1$, then $2 A D \cdot C D=(A D+C D)^{2}-\left(A D^{2}+C D^{2}\right)={ }^{32} / 25-1=7 / 25, S_{A D C}=1 / 2 A B \cdot C D=7 / 100$.

Since $S_{A B C}=1 / 2 A B \cdot B C=1 / 4$, then $S_{A B C D}=S_{A D C}+S_{A B C}=7 / 100+1 / 4=8 / 25$.

![](https://cdn.mathpix.com/cropped/2024_05_06_a7c46f4bbe36942e4ae1g-36.jpg?height=440&width=489&top_left_y=1301&top_left_x=790)

Answer

$8 / 25$.