# find the area of the largest rectangle that can fit inside a semi circle of radius 2 cm

find the area of the largest rectangle that can fit inside a semi circle of radius 2 cm I have absolutely no idea where to get started on this...What I did do is $A=(\pi r^2)/2$ (its a semi circle)

that gave me $6.28\ \mathrm{cm}^2$ now what do I do?

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Let's agree that we only have to consider rectangles with two vertices on the diameter of the semicircle. This means that the four vertices have coordinates $\pm(r\cos\phi,0)$ and $\pm(r\cos\phi, r\sin\phi)$ for some $\phi\in\bigl[0,{\pi\over2}\bigr]$. The area of such a rectangle is $2r^2\cos\phi\sin\phi=r^2 \sin(2\phi)$, and this is maximal when $\phi={\pi\over4}$. Therefore the maximal possible area of such a rectangle is $r^2$.
Hint: Imagine the semicircle being above the $x$ axis, so it is $y=\sqrt{4-x^2}$. If you are given the width (which lies along $x$), can you calculate the height and then the area? Then take the derivative of the area with respect to width, set to zero, etc.