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
\[\begin{equation} \begin{aligned} \frac{1}{2} r^2 sin \theta + \frac{1}{2} sin (\pi - \theta) &= \frac{1}{2} r^2 (sin \theta + sin (\pi - \theta )) \\ &= \frac{1}{2}r^2( sin \theta + sin \theta ) \\ &= r^2 sin \theta \end{aligned} \end{equation}\]

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

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