Bakayev E.V.

In a convex quadrilateral, two opposite sides are equal and perpendicular, and the other two are equal to $a$ and $b$. Find its area.
Let $b > a$.

First method. Denote the length of the two equal sides by $x$. Extend them until they intersect and denote the lengths of the two resulting short segments by $y$ and $z$ (see the left figure).

The area $S$ of the original quadrilateral is the difference in the areas of two right triangles: one with legs $x+y$ and $x+z$, and the other with legs $y$ and $z$. Therefore, $2S = (x+y)(x+z) - yz = x^2 + xy + xz$.

By the Pythagorean theorem, $y^2 + z^2 = a^2$, and $(x+y)^2 + (x+z)^2 = b^2$. Therefore, $b^2 - a^2 = 2x^2 + 2xy + 2xz = 4S$.
![](https://cdn.mathpix.com/cropped/2024_05_06_9fc5d17caeb61bf6ddabg-16.jpg?height=440&width=1128&top_left_y=249&top_left_x=473)

Second method. From four such polygons, a square with side $b$ can be formed, from which a square with side $a$ is cut out (see the right figure). Therefore, the area of one polygon is $1 / 4\left(b^2 - a^2\right)$.

## Answer

$1 / 4\left|b^2 - a^2\right|$.