| Find the area of the shaded region in figure, if AC = 20 cm, AB = 15 cm and O is the centre of the circle. |
| \[\left[ take\,\,\pi =\frac{22}{7} \right]\] |
 |
A) \[95.54\text{ }c{{m}^{2}}\]
B) \[95.45\,c{{m}^{2}}\]
C) \[59.54\text{ }c{{m}^{2}}\]
D) None of the above
Correct Answer: A
Solution :
| It is clear from the figure, BC is diameter of the circle |
| \[\therefore \,\,\,\,\,\,\,\,\,\,\angle BAC=90{}^\circ \] |
| [\[\because \]angle in a semi-circle is a right angle] |
| Now, in right angle \[\Delta BAC\], |
| \[B{{C}^{2}}=A{{B}^{2}}+A{{C}^{2}}\] |
| [using Pythagoras theorem] |
| \[\therefore \,\,\,\,BC=\sqrt{{{\left( 15 \right)}^{2}}+{{\left( 20 \right)}^{2}}}\] |
| \[\left[ \because \,\,\,AC=20\,cm,\,AB=15\,cm \right]\] |
| \[\Rightarrow \,\,\,BC=\sqrt{625}\Rightarrow BC=25\,\,cm\] |
| \[\therefore \] Radius of the circle, \[r=\frac{25}{2}cm\] |
| Now, area of shaded region |
| = Area of semi-circle - Area of \[\Delta ABC\] |
| \[=\frac{\pi {{r}^{2}}}{2}-\frac{1}{2}\times AB\times AC\] |
| \[=\frac{1}{2}\times \frac{22}{7}\times \frac{25}{2}-\frac{1}{2}\times 20\times 50\] |
| \[=\frac{6875}{28}-150=245.54-150=95.54\,\,c{{m}^{2}}\] |
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