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There are 2 circles, 1 big circle and 1 small circle, and they meet at

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MathRevolution
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There are 2 circles, 1 big circle and 1 small circle, and they meet at [#permalink] New post   02 Aug 2017, 01:01
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A
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15% (low)

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84% (01:19) correct 16% (01:25) wrong based on 60 sessions
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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?

\(A. 3π\)
\(B. 3.5π\)
\(C. 4π\)
\(D. 4.5π\)
\(E. 7π\)
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B
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sashiim20
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Re: There are 2 circles, 1 big circle and 1 small circle, and they meet at [#permalink] New post   02 Aug 2017, 01:29
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MathRevolution wrote:
Attachment:
8.2.png


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?

\(A. 3π\)
\(B. 3.5π\)
\(C. 4π\)
\(D. 4.5π\)
\(E. 7π\)


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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Re: There are 2 circles, 1 big circle and 1 small circle, and they meet at [#permalink] New post   02 Aug 2017, 02:08
MathRevolution wrote:
Attachment:
8.2.png


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?

\(A. 3π\)
\(B. 3.5π\)
\(C. 4π\)
\(D. 4.5π\)
\(E. 7π\)


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
\(= πr^2 = π(R/2)^2= πR^2 /4\)
\(=14π/4 = 3.5π\)
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There are 2 circles, 1 big circle and 1 small circle, and they meet at [#permalink] New post   02 Aug 2017, 02:14
MathRevolution wrote:
Attachment:
8.2.png


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?

\(A. 3π\)
\(B. 3.5π\)
\(C. 4π\)
\(D. 4.5π\)
\(E. 7π\)


The ratio of areas of two circles is the same as the ratio of the squares of their respective Radii / Diameters. Hence the correct answer should be 1/4 of the area of the bigger circle.
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Re: There are 2 circles, 1 big circle and 1 small circle, and they meet at [#permalink] New post   04 Aug 2017, 02:23
Expert Reply
==> Ratio of length^2=Ratio of area, and since the ratio of the length is 2, the ratio of the area becomes 4, so you get \(\frac{14π}{4}=3.5π\).

The answer is B.
Answer: B
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Re: There are 2 circles, 1 big circle and 1 small circle, and they meet at [#permalink]
04 Aug 2017, 02:23
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