Area of Rectangle Inscribed in Circle

Let the side lengths of the rectangle be $a$ and $b$. Since the rectangle is inscribed in a circle of radius $R$, the diagonal of the rectangle is a diameter of the circle: $a^2 + b^2 = (2R)^2 = 4R^2$

Perimeter $P = 2(a + b) \implies a + b = \frac{P}{2}$.

Squaring both sides: $(a + b)^2 = \frac{P^2}{4}$ $a^2 + b^2 + 2ab = \frac{P^2}{4}$ $4R^2 + 2ab = \frac{P^2}{4}$ $2ab = \frac{P^2}{4} - 4R^2$ $ab = \frac{P^2}{8} - 2R^2$

Area of rectangle $= ab = \frac{P^2}{8} - 2R^2$.