In the given figure, ABCD is a rectangle and a semicircle with CD as diameter is drawn. A circle is drawn as shown in the figure. If BC = 7 cm, then what is the radius of the small circle?

\(21-14\sqrt {2}\)

Given, BC = MB = MC = 7 cm

By pythagoras theorem,

(MB)² = BC² + MC²

MB² = 7² + 7² = 49 × 2

MB = \(7\sqrt {2}\)

Let radius of small circle = r

MB = MA + OA + OB

\(7\sqrt {2}\) = 7 + r + OB

OB = \((7\sqrt {2}-7)-r\)

In ΔOBP

OB² = OP² + BP²

\((7\sqrt {2}-7-r)^2\) = r² + r²

\(7\sqrt {2}-7-r=\sqrt {2}r\)

\(7(\sqrt {2}-1)=\sqrt {2}r+r\)

\(r(\sqrt {2}+1)=7(\sqrt {2}-1)\)

\(r=\frac{7(\sqrt {2}-1)}{\sqrt {2}+1}\)

\(r=7(\sqrt {2}-1)^2\)

\(r=7(2+1-2\sqrt {2}\)

= \(21-14\sqrt {2}cm\)