GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
In the above figure, the areas of the circle and the triangle shown ar
[#permalink]
09 Feb 2025, 12:00
09 Feb 2025, 12:00
Expert Reply
00:00
Question Stats:
50% (01:15) correct
50% (00:47) wrong based on 2 sessions
Hide Show timer Statistics
Attachment:
GRe In the above figure, the areas of the circle and the triangle .png [ 36.5 KiB | Viewed 167 times ]
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
- Part of the project: GRE Quant & Verbal ADVANCED Daily Challenge 2025 NEW Edition - Gain 20 Kudos & Get FREE Access to GRE Prep Club TESTS
- Also replying to the unanswered questions
Re: In the above figure, the areas of the circle and the triangle shown ar
[#permalink]
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
Moderators:
