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In the above figure, the areas of the circle and the triangle shown ar
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09 Feb 2025, 12:00
09 Feb 2025, 12:00
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85% (00:58) correct
14% (00:47) wrong based on 7 sessions
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In the above figure, the areas of the circle and the triangle shown are $\(4 \pi\)$ and 3 respectively. What is the length of the line segment $\(P Q\)$ ?
(A) 3
(B) $\(\sqrt{12}\)$
(C) $\(\sqrt{13}\)$
(D) $\(\sqrt{15}\)$
(E) 4
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Re: In the above figure, the areas of the circle and the triangle shown ar
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06 Mar 2025, 11:33
06 Mar 2025, 11:33
Expert Reply
In the figure above $\(\mathrm{OP}=2\)$ is the radius of the circle and the area of triangle POQ is given as 3.
Area of right triangle $\(\mathrm{POQ}=\frac{1}{2} \times$ Base $\times\)$ Height $\(=\frac{1}{2} \times \mathrm{OP} \times \mathrm{OQ}=\frac{1}{2} \times 2 \times \mathrm{OQ}=\mathrm{OQ}\)$ which is given as 3 , so we get $\(\mathrm{OQ}=3\)$
Now, applying Pythagoras theorem i.e. Hypotenuse $\({ }^2=\)$ Perpendicular $\(^2+\)$ Base $\(^2\)$ in the triangle $\(P O Q\)$ we get $\(\mathrm{PQ}^2=\mathrm{OQ}^2+\mathrm{OP}^2\)$ i.e. $\(P Q=\sqrt{O Q^2+\mathrm{OP}^2}=\sqrt{3^2+2^2}=\sqrt{13}\)$
Hence the answer is (C).
Area of right triangle $\(\mathrm{POQ}=\frac{1}{2} \times$ Base $\times\)$ Height $\(=\frac{1}{2} \times \mathrm{OP} \times \mathrm{OQ}=\frac{1}{2} \times 2 \times \mathrm{OQ}=\mathrm{OQ}\)$ which is given as 3 , so we get $\(\mathrm{OQ}=3\)$
Now, applying Pythagoras theorem i.e. Hypotenuse $\({ }^2=\)$ Perpendicular $\(^2+\)$ Base $\(^2\)$ in the triangle $\(P O Q\)$ we get $\(\mathrm{PQ}^2=\mathrm{OQ}^2+\mathrm{OP}^2\)$ i.e. $\(P Q=\sqrt{O Q^2+\mathrm{OP}^2}=\sqrt{3^2+2^2}=\sqrt{13}\)$
Hence the answer is (C).
gmatclubot
Re: In the above figure, the areas of the circle and the triangle shown ar [#permalink]
06 Mar 2025, 11:33
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