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# In the circle with center O above, chord QS is perpendicular

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Intern
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In the circle with center O above, chord QS is perpendicular [#permalink]

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27 Mar 2013, 04:05
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71% (03:41) correct 29% (03:33) wrong based on 76 sessions

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In the circle with center O above, chord QS is perpendicular to radius OR. If QS = 16 and PR = 4, what is the area of the circle ?

(A) 80 π
(B) 100 π
(C) 125 π
(D) 144 π
(E) 256 π

Please see the above mentioned circle in the attachment . The OA should be (B), this problem is part of a set of GMAT PS exercises given to me for training by a friend .
[Reveal] Spoiler: OA

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gmat q 22.jpg [ 96.4 KiB | Viewed 2182 times ]

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Re: In the circle with centre O above, chord QS is perpendicular [#permalink]

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27 Mar 2013, 04:37
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In the given circle,join OS ans OQ.
Let OR = OS = OQ = r. So, OP = r-4.

OR will bisect QS,Hence, QP = PS = 8.
Now solve for r in right angle triangle OPS.

OS ^2 = OP ^2 + PS ^ 2
=> r^2 = (r-4)^2 + 8^2
=> r = 10

So area = pi * 10 ^ 2 = 100 pi.
Hence option B.
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Re: In the circle with center O above, chord QS is perpendicular [#permalink]

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07 Mar 2016, 11:34
How do you solve the quadratic efficiently?
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Re: In the circle with center O above, chord QS is perpendicular [#permalink]

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26 Mar 2016, 23:20
subhendu009 wrote:
In the given circle,join OS ans OQ.
Let OR = OS = OQ = r. So, OP = r-4.

OR will bisect QS,Hence, QP = PS = 8.
Now solve for r in right angle triangle OPS.

OS ^2 = OP ^2 + PS ^ 2
=> r^2 = (r-4)^2 + 8^2
=> r = 10

So area = pi * 10 ^ 2 = 100 pi.
Hence option B.

Got the same answer. I took OR to be the bisector of QS, but i cannot remember the theorem for the same. Can you please suggest??
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Re: In the circle with center O above, chord QS is perpendicular [#permalink]

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27 Mar 2016, 07:14
STEMBusiness wrote:
How do you solve the quadratic efficiently?

the quadratic here is r^2=64+16+r^2+8r

The r^2 cancel out, so no need to solve the quadratic. you will just get r=10
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Re: In the circle with center O above, chord QS is perpendicular [#permalink]

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21 Oct 2016, 20:22
The following can be used to solve this problem:

8^2 + (x-4)^2 = x^2

Manipulate --> x = 10

Area = Pi(r^2) = (10)^2 pi = 100 pi
Re: In the circle with center O above, chord QS is perpendicular   [#permalink] 21 Oct 2016, 20:22
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# In the circle with center O above, chord QS is perpendicular

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