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 \[=\frac{16}{8}\pi +\frac{9}{8}\pi -\left( \frac{25}{8}\pi -6 \right)\] \[=\frac{16}{8}\pi +\frac{9}{8}\pi -\frac{25}{8}\pi +6\] \[=6\,\,c{{m}^{2}}\]
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