Question: 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
Options:
A) $ 24,\pi cm^{2} $
B) $ 21,\pi cm^{2} $
C) $ 15,\pi cm^{2} $
D) $ 12,\pi cm^{2} $
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Answer:
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 ,cm^{2} $