| In the following figure, O is the centre of the circle with \[AC=24\,cm,\] \[\text{AB}=\text{7 cm}\] and \[\angle BOD=90{}^\circ \]. The area of the shaded region, (Use \[\pi =3.14\]) is: (CBSE 2012,17) |
 |
A) \[248\,c{{m}^{2}}\]
B) \[284\,c{{m}^{2}}\]
C) \[\text{298 c}{{\text{m}}^{\text{2}}}\]
D) \[\text{318 c}{{\text{m}}^{\text{2}}}\]
Correct Answer: B
Solution :
| [b] Since, angle in a semicircle is always a right angle. |
| \[\therefore \,\,\,\,\,\,\,\,\,\angle BAC=90{}^\circ \] |
| In right-angled \[\Delta BAC,\] |
| \[B{{C}^{2}}=A{{C}^{2}}+A{{B}^{2}}\] (By Pythagoras theorem) |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,B{{C}^{2}}={{(24)}^{2}}+{{(7)}^{2}}\] |
| [\[AC=24\]and \[AB=7\,cm\](given)] |
| \[=576+49=625={{(25)}^{2}}\] |
| \[BC=25\,cm\] |
| \[\therefore \] Radius \[=OB=OC=\frac{25}{2}cm\] |
| \[\therefore \] Area to the shaded region |
| \[=\text{ Area of circle}-\text{Area of }\Delta \text{BAC}-\text{Area of COD}\] |
| \[=\pi {{\left( \frac{25}{2} \right)}^{2}}-\frac{1}{2}\times AB\times AC-\frac{1}{4}\times \pi {{\left( \frac{25}{2} \right)}^{2}}\] |
| \[=3.14\times \frac{25}{2}\times \frac{25}{2}-\frac{1}{2}\times 7\times 24-\frac{1}{4}\times 3.14\times \frac{25}{2}\times \frac{25}{2}\]\[=490.625-84-122.656\] |
| \[=490.625-206.656\] |
| \[=283.969\approx 284c{{m}^{2}}\] |
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