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# In the figure shown, line segments QS and RT are diameters of the circ

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In the figure shown, line segments QS and RT are diameters of the circ [#permalink]  12 Nov 2010, 21:44
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Difficulty:

65% (hard)

Question Stats:

45% (01:51) correct 55% (01:00) wrong based on 40 sessions
In the figure shown, line segments QS and RT are diameters of the circle. If the distance between Q and R is 8/root2, what is the area of the circle?

(A) 4 pie
(B) 8 Pie
(C) 16 pie
(D) 32 Pie
(E) 64 Pie
[Reveal] Spoiler: OA

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Re: In the figure shown, line segments QS and RT are diameters of the circ [#permalink]  12 Nov 2010, 22:19
Expert's post
monirjewel wrote:
In the figure shown, line segments QS and RT are diameters of the circle. If the distance between Q and R is 8/root2, what is the area of the circle?
(A) 4 pie
(B) 8 Pie
(C) 16 pie
(D) 32 Pie
(E) 64 Pie

Let the center of the circle be O. Then OQ and OR will be the radii. Now as triangle OQR is right angle then QR will be its hypotenuse hence $$QR^2=(\frac{8}{\sqrt{2}})^2=r^2+r^2$$ --> $$2r^2=32$$ --> $$r^2=16$$ --> $$area=\pi{r^2}=16\pi$$.

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Re: In the figure shown, line segments QS and RT are diameters of the circ [#permalink]  14 Nov 2010, 15:12
Line joining Q n R makes right isoscelleous(45, 45, 90) triangle where hypotenus = 8/sqrt2

Thus radius = 8/(sqrt2 * sqrt2) = 4

Area = pie*r^2 = 16 pie

Re: In the figure shown, line segments QS and RT are diameters of the circ   [#permalink] 14 Nov 2010, 15:12
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