SATHEE: Quantitative Aptitude Ques 1434

Quantitative Aptitude Ques 1434

Question: 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. $

Options:

A) $ \frac{\sqrt{3}}{2}r^{2} $

B) $ 3\sqrt{3}r^{2} $

C) $ \sqrt{3}r^{2} $

D) None of these

Show Answer

Answer:

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=2SB=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

Quantitative Aptitude Ques 1433

Quantitative Aptitude Ques 1435

Quantitative Aptitude Ques 1434

Question: In the given figure, O is the centre of the circle.

Options:

Answer:

Solution:

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