Question

In Fig. 6, 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.​

Answer 1

Answer:

$$\begin{matrix} By\, \, pythagorous\, \, theorem \\ A{ C^{ 2 } }=A{ B^{ 2 } }+B{ C^{ \begin{matrix} { l }2 \\ \end{matrix} } } \\ A{ C^{ 2 } }=\sqrt { { 3^{ 2 } }+{ 4^{ 2 } } } \\ AC=\sqrt { 25 } ,\, \, \, \, \\ Therefore,\, \, Area\; \, of\, \, right\, \, triangle=\frac { 1 }{ 2 } \times AB\times BC=\frac { 1 }{ 2 } \times 3\times 4=6c{ m^{ 2 } } \\ \Rightarrow AC=5\, \, cm, \\ \Rightarrow Area\, \, of\, semi-circle\, \, with\, \, diameter\, \, BC \\ \Rightarrow \frac { 1 }{ 2 } \pi { r^{ 2 } }\, \, =\, \, \frac { 1 }{ 2 } \pi Kt\frac { { B{ C^{ 2 } } } }{ 4 } \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \left[ { \therefore diameter\, =2r } \right] \\ =\frac { \pi }{ 2 } \times \frac { { { 4^{ 2 } } } }{ 4 } =2\pi c{ m^{ 2 } } \\ \Rightarrow Area\, \, of\, \, ler\, \, semi\, \, circle\, =\, \frac { \pi }{ 2 } \times \frac { { A{ C^{ 2 } } } }{ 4 } =\frac { \pi }{ 8 } \times { 5^{ 2 } }=\frac { { 25\pi } }{ 8 } c{ m^{ 2 } } \\ \Rightarrow Are\, \, of\, \, another\, \, swemi\, \, circle\, \, with\, \, diameter\, \, AB \\ \frac { 1 }{ 2 } \times \pi \times \frac { { A{ B^{ 2 } } } }{ 4 } =\frac { \pi }{ 8 } \times { 3^{ 2 } }=\frac { 9 }{ 8 } \pi c{ m^{ 2 } } \\ The\, \, area\, \, of\, \, shaded\, \, portion\, \, is: \\ =\, Area\, \, of\, \, triangle\, \, +Area\, \, of\, \, { { o } }th\, \, small\, \, semi\, \, circle-Area\, \, of\, \, bigger\, \, small\, \, semi-circle \\ =\left( { 6+2\pi +\frac { { 9\pi } }{ 8 } } \right) -\frac { { 25\pi } }{ 8 } =\frac { { 32 } }{ 8 } +\frac { { 25\pi } }{ 8 } -\frac { { 25\pi 8 } }{ { } } =4c{ m^{ 2 } } \\ e\, \, required\, \, shaded\, \, area\, \, is\, \, 4c{ m^{ 2 } }\, \, \, \, \, \, \, \, \, \, \, \, \\ \end{matrix}$$

Step-by-step explanation: