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?
Answer & explanation
Correct answer: option 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
\(7\sqrt {2}-7-r=\sqrt {2}r\)
\(7(\sqrt {2}-1)=\sqrt {2}r+r\)
\(r(\sqrt {2}+1)=7(\sqrt {2}-1)\)
\(r=\frac{7(\sqrt {2}-1)}{\sqrt {2}+1}\)
\(r=7(\sqrt {2}-1)^2\)
\(r=7(2+1-2\sqrt {2}\)
= \(21-14\sqrt {2}cm\)