question_answer

In the given figure, \[\Delta \,ABC\] is a right-angled triangle in which \[\angle A\] is \[90{}^\circ \]. Semicircles are drawn on AB, AC and BC as diameters. Find the area of the shaded region.

Answer:

In right \[\Delta \text{ }BAC\], by Pythagoras theorem,

\[B{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}\]

\[={{(3)}^{2}}+{{(4)}^{2}}\]

\[=9+16=25\]

\[BC=\sqrt{25}=5\,\,cm\]

Area of semi-circle with diameter \[BC=\frac{1}{2}\pi {{r}^{2}}\]

\[=\frac{1}{2}\times \pi {{\left( \frac{5}{2} \right)}^{2}}=\frac{25}{8}\pi \,c{{m}^{2}}\]

Area of semi-circle with diameter \[AB=\frac{1}{2}\pi {{r}^{2}}\]

\[=\frac{1}{2}\pi {{\left( \frac{3}{2} \right)}^{2}}=\frac{9}{8}\pi \,c{{m}^{2}}\]

Area of semi-circle with diameter\[AC=\frac{1}{2}\pi {{r}^{2}}\].

\[=\frac{1}{2}\pi {{\left( \frac{4}{2} \right)}^{2}}=\frac{16}{8}\pi \,\,c{{m}^{2}}\]

Area of rt \[\Delta \,BAC=\frac{1}{2}\times AB\times AC\]

\[=\frac{1}{2}\times 3\times 4=6\,\,c{{m}^{2}}\]

Area of dotted region \[=\left( \frac{25}{8}\pi -6 \right)c{{m}^{2}}\]

Area of shaded region \[=\frac{16}{8}\pi +\frac{9}{8}\pi -\left( \frac{25}{8}\pi -6 \right)\]

\[=\frac{16}{8}\pi +\frac{9}{8}\pi -\frac{25}{8}\pi +6\]

\[=6\,\,c{{m}^{2}}\]