In figure ABC is a right angled triangle right-angled at A. Semicircles are drawn on AB, AC and BC as diameters. Find the area of the shaded region.

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Solution

$\begin{matrix} B y p y t h a g o r o u s t h e o r e m A C^{2} = A B^{2} + B C^{\begin{matrix} l 2 \end{matrix}} A C^{2} = \sqrt{3^{2} + 4^{2}} A C = \sqrt{25}, T h e r e f o r e, A r e a o f r i g h t t r i a n g l e = \frac{1}{2} \times A B \times B C = \frac{1}{2} \times 3 \times 4 = 6 c m^{2} \Rightarrow A C = 5 c m, \Rightarrow A r e a o f s e m i - c i r c l e w i t h d i a m e t e r B C \Rightarrow \frac{1}{2} π r^{2} = \frac{1}{2} π K t \frac{B C^{2}}{4} [∴ d i a m e t e r = 2 r] = \frac{π}{2} \times \frac{4^{2}}{4} = 2 π c m^{2} \Rightarrow A r e a o f l e r s e m i c i r c l e = \frac{π}{2} \times \frac{A C^{2}}{4} = \frac{π}{8} \times 5^{2} = \frac{25 π}{8} c m^{2} \Rightarrow A r e o f a n o t h e r s w e m i c i r c l e w i t h d i a m e t e r A B \frac{1}{2} \times π \times \frac{A B^{2}}{4} = \frac{π}{8} \times 3^{2} = \frac{9}{8} π c m^{2} T h e a r e a o f s h a d e d p o r t i o n i s : = A r e a o f t r i a n g l e + A r e a o f o t h s m a l l s e m i c i r c l e - A r e a o f b i g g e r s m a l l s e m i - c i r c l e = (6 + 2 π + \frac{9 π}{8}) - \frac{25 π}{8} = \frac{32}{8} + \frac{25 π}{8} - \frac{25 π 8}{} = 4 c m^{2} e r e q u i r e d s h a d e d a r e a i s 4 c m^{2} \end{matrix}$

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