## Maximum Area of Rectangle Inscribed in Semicircle

What is the maximum area of a rectangle that can be inscribed in a semicircle of radius
$r$
?
Let the triangle OBC subtend an angle
$\theta$
at the centre of the circle.
The triangles ABO abd CDO have the same area by symmetry, and together subtend an angle of
$\pi - \theta$
. We can consider them to form a triangle with sides
$r$
from the centre of the circle to the edge, and subtending an angle of
$\pi - \theta$
at the centre.
The area of the rectangle is then
\begin{aligned} \frac{1}{2} r^2 sin \theta + \frac{1}{2} sin (\pi - \theta) &= \frac{1}{2} r^2 (sin \theta + sin (\pi - \theta )) \\ &= \frac{1}{2}r^2( sin \theta + sin \theta ) \\ &= r^2 sin \theta \end{aligned}

The maximum value of
$sin \theta$
is 1, when
$\theta = \frac{ \pi}{2}$
so the maximum area of the rectangle is
$r^2$
.