| Find the area of the shaded region in fig. If \[\text{PQ}=\text{24 cm},\] \[\text{PR}=\text{7 cm}\] and O is the centre of the circle. (NCERT EXERCISE; CBSE 2020) |
 |
A) \[161.54\,c{{m}^{2}}\]
B) \[165.63\,c{{m}^{2}}\]
C) \[167.24\,c{{m}^{2}}\]
D) \[169.43\,c{{m}^{2}}\]
Correct Answer: A
Solution :
| [a] Given, \[PQ=24cm\] and \[PR=7cm\] |
| Since RQ is the diameter of the circle. |
| Therefore, \[\angle RPQ\]will be \[90{}^\circ \]. |
| By applying Pythagoras theorem in \[\Delta RPQ\]. |
| \[R{{P}^{2}}+P{{Q}^{2}}=R{{Q}^{2}}\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{(7)}^{2}}+{{(24)}^{2}}=R{{Q}^{2}}\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,RQ=\sqrt{625}=25\] |
| \[\therefore \] Radius of circle \[OR=\frac{RQ}{2}=\frac{25}{2}\] |
| Since RQ is the diameter of the circle, it divides the circle in two equal parts. |
| \[\therefore \] Area of semicircle \[=\frac{1}{2}\pi {{r}^{2}}=\frac{1}{2}\pi {{\left( \frac{25}{2} \right)}^{2}}\] |
| \[=\frac{1}{2}\times \frac{22}{7}\times \frac{625}{4}\] \[\left[ r=OR=\frac{25}{2}cm \right]\] |
| \[=\frac{6875}{28}c{{m}^{2}}\] |
| and area of \[\Delta PQR=\frac{1}{2}\times PQ\times PR\] |
| \[=\frac{1}{2}\times 24\times 7=84c{{m}^{2}}\] |
| So. area of shaded region |
| \[=\text{Area of semicircle}-\text{Area of }\Delta \text{PQR}\] |
| \[=\frac{6875}{28}-84=\frac{6875-2392}{28}=\frac{4523}{28}c{{m}^{2}}\] |
| \[=161.54\,c{{m}^{2}}\] |
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