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31 Mar 2012, 06:52
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D
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Difficulty:
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79% (01:49) correct
21%
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The shaded region in the gure above represents a circular tire. If the distance from point O to point T is 18 inches and the area of the shaded region is equal to the area of the unshaded region, what is the radius of the unshaded region, in inches?
(A) \(\frac{9}{2}\sqrt{2}\)
(B) \(\frac{9}{2}\sqrt{2}\)
(C) 9
(D) \(9\sqrt{2}\)
(E) \(9\sqrt{3}\)
Attachment:
Circle.GIF [ 5.99 KiB | Viewed 8545 times ]
31 Mar 2012, 07:19
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enigma123
The area of the big circle is \(\pi{R^2}=324\pi\);
The area of the small circle (the unshaded region) is \(\pi{r^2}\);
The area of the shaded region is \(324\pi-\pi{r^2}\);
We are told that the area of the shaded region is equal to the area of the unshaded region, so \(\pi{r^2}=324\pi-\pi{r^2}\) --> \(2\pi{r^2}=324\pi\) --> \(r^2=162\) --> \(r=9\sqrt{2}\).
Answer: D.
31 Mar 2012, 14:17
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Outer ring's area = Bigger circle area - inner circle area = pi*R_out ^2 - pi*R_in ^2
it also equals to the inner circle area, by the question stem: pi*R_out ^2 - pi*R_in ^2 = pi*R_in ^2
So:
2*pi*R_in^2 = pi R_out^2
R_in^2 = (R_out^2)/2 => (18^2)/2
R_in = sqrt[ (18^2)/2 ] = 18/sqrt(2) = 9*sqrt(2) ===> answer D.
it also equals to the inner circle area, by the question stem: pi*R_out ^2 - pi*R_in ^2 = pi*R_in ^2
So:
2*pi*R_in^2 = pi R_out^2
R_in^2 = (R_out^2)/2 => (18^2)/2
R_in = sqrt[ (18^2)/2 ] = 18/sqrt(2) = 9*sqrt(2) ===> answer D.
