If the chords AP and AQ of a circle of radius 6 cm are at distance 3 cm and \[3\sqrt{2}\,cm\] respectively from the centre O of the circle, then the area of the smaller sector POQ is |
A) \[24\,\pi \text{ }c{{m}^{2}}\]
B) \[21\,\pi \text{ }c{{m}^{2}}\]
C) \[15\,\pi \text{ }c{{m}^{2}}\]
D) \[12\,\pi \text{ }c{{m}^{2}}\]
Correct Answer: C
Solution :
In \[\Delta AMO,\] \[\sin \theta =\frac{3}{6}\]\[\Rightarrow \]\[\sin \theta =\frac{1}{2}\] |
\[\sin \theta =\sin 30{}^\circ \] |
\[\theta =30{}^\circ \] (i) |
Similarly, in \[\Delta ANO,\] |
\[\sin \alpha =\frac{3\sqrt{2}}{6}\]\[\Rightarrow \]\[\sin \alpha =\frac{1}{\sqrt{2}}\] |
\[\Rightarrow \] \[\alpha =45{}^\circ \] |
\[\therefore \] \[(\theta +\alpha )=75{}^\circ \] |
Again \[\angle POQ=2\,\angle PAQ\] \[[\because \angle PAQ=75{}^\circ ]\] |
\[=2\times 75{}^\circ =150{}^\circ \] |
Area of sector \[POQ=\frac{150}{360}\times \pi \times 36=15\pi \,c{{m}^{2}}\] |
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