question_answer

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 \[POQ=\frac{150}{360}\times \pi \times 36=15\pi \,c{{m}^{2}}\]