| If O is the centre of a circle of radius 5 cm. At a distance of 13 cm from O, a point P is taken. From this point, two tangents PQ and PR are drawn to the circle. Then, the area of quadrilateral PQOR is |
 |
A) \[60\,c{{m}^{2}}\]
B) \[32.5\,c{{m}^{2}}\]
C) \[65\,c{{m}^{2}}\]
D) \[30\,c{{m}^{2}}\]
Correct Answer: A
Solution :
| Clearly, \[OQ=OR=5\,cm,\] |
| \[\angle OQP=\angle ORP=90{}^\circ \] |
| and \[OP=13\,cm\] |
| \[\therefore \] \[P{{Q}^{2}}={{(O{{P}^{2}}-OQ)}^{2}}={{(13)}^{2}}-{{(5)}^{2}}\] |
| \[=(169-25)=144\] |
| \[\Rightarrow \] \[PQ=\sqrt{144}=12\,cm\] |
| \[\therefore \] \[ar(\Delta OQP)=\frac{1}{2}\times PQ\times OQ\] |
| \[=\left( \frac{1}{2}\times 12\times 5 \right)=30\,c{{m}^{2}}\] |
| Similarly, \[ar(\Delta ORP)=30\,c{{m}^{2}}\] |
| \[\therefore \] ar(quadrilateral PQOR) \[=(30+30)=60\,c{{m}^{2}}\] |
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