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Equilateral triangle PQR is formed by joining the centers P, Q, and R
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19 Nov 2021, 06:27
19 Nov 2021, 06:27
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Equilateral triangle PQR is formed by joining the centers P, Q, and R of the three circles. Each pair of circles has exactly one point in common. If PQ = 12, what is the shaded area?
\(6\pi\)
\(12\pi\)
\(18\pi\)
\(24\pi\)
\(36\pi\)
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Equilateral triangle PQR is formed by joining the centers P, Q, and R
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22 Nov 2021, 09:44
22 Nov 2021, 09:44
1] \(PQR\) is an equilateral triangle
2] All circles are touching each other meaning they have single common point
Both of these 2 conditions obviously point out that all three circles have same radii = \(\frac{1}{2}\) x side length of \(\triangle PQR = \frac{1}{2}*12 = 6\)
Therefore, length of shaded region = \(\frac{60^o}{360^o}\pi r^2 = \frac{60^o}{360^o}\pi(6)^2 = \frac{1}{6}\pi(36) = 6\pi\)
Hence, Answer is A
2] All circles are touching each other meaning they have single common point
Both of these 2 conditions obviously point out that all three circles have same radii = \(\frac{1}{2}\) x side length of \(\triangle PQR = \frac{1}{2}*12 = 6\)
Therefore, length of shaded region = \(\frac{60^o}{360^o}\pi r^2 = \frac{60^o}{360^o}\pi(6)^2 = \frac{1}{6}\pi(36) = 6\pi\)
Hence, Answer is A
Re: Equilateral triangle PQR is formed by joining the centers P, Q, and R
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28 Jan 2025, 10:44
28 Jan 2025, 10:44
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Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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