Question

The figure shows the rectangle $ABCD$ with a semicircle and a circle inscribed inside in it as shown. What is the ratio of the area of the circle to that of the semicircle?

Answer 1

Let us reconstruct and rename the given figure as shown.

Here, $\square OEBC$ is a square as $\angle B={90}^o$

Also, $OE=OC=Oq=r$ ....(radii of the same circle)

$\therefore OB=\sqrt{2}r$ and $Bq=(\sqrt{2}r-r)=(\sqrt{2}-1)r$ ......(i).

Now $pq=OB-Oq-Bp=\sqrt{2}r-r-Bp=(\sqrt{2}-1)r-Bp$ .....(ii).

To find $Bp$, let us take O' as the centre of the full circle.

Then, $O^{\prime }q=O^{\prime }m=O^{\prime }n=r$ ...say

Obviously, $O^{\prime }mBn$ is a square by argument same as (i)

$\therefore O^{\prime }B=\sqrt{2}{O}^{\prime }m=\sqrt{2}{O}^{\prime }n=\sqrt{2}{O}^{\prime }p=\sqrt{2}\frac{pq}{2}=\frac{pq}{\sqrt{2}}$

From here, we cannot proceed due to lac of data.

We need either $pq$ or $Bq$ or $O^{\prime }m$ or any suitable data to evaluate $Bp.$