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02 Aug 2017, 01:01
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
86% (01:20) correct
14%
(01:25)
wrong
based on 57
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There are 2 circles, 1 big circle and 1 small circle, and they meet at one point. The diameter of the small circle is equal to the radius of the big circle, and the area of the big circle is \(14π\). What is the area of the small circle?
\(A. 3π\)
\(B. 3.5π\)
\(C. 4π\)
\(D. 4.5π\)
\(E. 7π\)
sashiim20
Joined: 04 Dec 2015
Last visit: 05 Jun 2024
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Location: India
Concentration: Technology, Strategy
Schools: ISB '19 IIMB IIMA XLRI HEC Sept19 intake Rotman '21 NUS '21 NTU '20 Trinity MBA '20 Bocconi '21 Smurfit "21
WE:Information Technology (Consulting)
Schools: ISB '19 IIMB IIMA XLRI HEC Sept19 intake Rotman '21 NUS '21 NTU '20 Trinity MBA '20 Bocconi '21 Smurfit "21
Posts: 608
02 Aug 2017, 01:29
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MathRevolution
Given Area of big circle \(= 14π\)
Let the radius of big circle be \(= R\)
\(\pi\)\(R^2 = 14π\)
\(R^2 = 14\)
\(R = \sqrt{14}\)
Given diameter of small circle is equal to radius of big circle.
Let the radius of small circle be \(= r\)
Therefore \(R = 2r\)
\(r = \frac{R}{2}\) \(= \frac{\sqrt{14}}{ 2}\)
Area of small circle \(= \pi r^2 = \pi * \frac{\sqrt{14}}{ 2} * \frac{\sqrt{14}}{ 2} = \pi * \frac{14}{4} = \frac{7}{2} \pi = 3.5\pi\)
Answer (B)...
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02 Aug 2017, 02:08
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MathRevolution
Let the radius of bigger circle be R & radius of smaller circle be r.
So, r = R/2
\(πR^2 = 14π\)
\(R = \sqrt{14}\)
\(r = \sqrt{14}/2\)[/m]
Area of smaller circle
\(= πr^2 = π(R/2)^2= πR^2 /4\)
\(=14π/4 = 3.5π\)
