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line segments QS and RT

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Manager
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Joined: 06 Feb 2010
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line segments QS and RT [#permalink] New post 12 Nov 2010, 21:44
00:00
A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

35% (01:45) correct 65% (01:01) wrong based on 31 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: line segments QS and RT [#permalink] New post 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.

Answer: C.
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Re: line segments QS and RT [#permalink] New post 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

Answer:- C
Re: line segments QS and RT   [#permalink] 14 Nov 2010, 15:12
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