| In the given figure, O is the centre of the circle. |
| \[\angle AOD=120{}^\circ .\]If the radius of the circle be r, then find the sum of the areas (in sq units) of quadrilaterals \[AODP\] and \[OBQC.\] |
 |
A) \[\frac{\sqrt{3}}{2}{{r}^{2}}\]
B) \[3\sqrt{3}{{r}^{2}}\]
C) \[\sqrt{3}{{r}^{2}}\]
D) None of these
Correct Answer: C
Solution :
| \[OQ=OB=OC=r\,\,(say)\] |
| \[\angle AOD=\angle BOC=120{}^\circ \] |
| \[\therefore \] \[\angle BOQ=\angle COQ=60{}^\circ \] |
| \[\therefore \] \[\frac{SB}{OB}=\sin 60{}^\circ =\frac{\sqrt{3}}{2}\]\[\Rightarrow \]\[SB=\frac{r\sqrt{3}}{2}\] |
| \[\therefore \] \[BC=2\,\,SB=r\sqrt{3}\] |
| Area of quadrilateral \[BQCO=\frac{1}{2}\times BC\times OQ\] |
| \[=\frac{1}{2}\times r\sqrt{3}\times r=\frac{{{r}^{2}}\sqrt{3}}{2}\]sq units |
| \[\therefore \]Sum of the areas of both quadrilaterals |
| \[=2\times \frac{{{r}^{3}}\sqrt{3}}{2}={{r}^{2}}\sqrt{3}\]sq units |
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