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In the diagram (not drawn to scale), Sector PQ is a quarter-circle.

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In the diagram (not drawn to scale), Sector PQ is a quarter-circle.  [#permalink]

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In the diagram (not drawn to scale), Sector PQ is a quarter-circle. The distance from A to P is half the distance from P to B. The distance from C to Q is 2/7 of the distance from Q to B. If the length of AC is 100, what is the length of the radius of the circle with center B?


A. \(\frac{280\sqrt{85}}{51}\)

B. \(\frac{240\sqrt{70}}{61}\)

C. \(\frac{240\sqrt{67}}{43}\)

D. \(\frac{230\sqrt{51}}{43}\)

E. \(\frac{220\sqrt{43}}{51}\)

Attachment:
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Originally posted by venmic on 20 Sep 2012, 19:40.
Last edited by Bunuel on 04 Dec 2017, 09:11, edited 2 times in total.
Edited the question and added answer choices.
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Re: In the diagram (not drawn to scale), Sector PQ is a quarter-circle.  [#permalink]

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New post 21 Sep 2012, 02:36
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In the diagram (not drawn to scale), Sector PQ is a quarter-circle. The distance from A to P is half the distance from P to B. The distance from C to Q is 2/7 of the distance from Q to B. If the length of AC is 100, what is the length of the radius of the circle with center B?

A. \(\frac{280\sqrt{85}}{51}\)

B. \(\frac{240\sqrt{70}}{61}\)

C. \(\frac{240\sqrt{67}}{43}\)

D. \(\frac{230\sqrt{51}}{43}\)

E. \(\frac{220\sqrt{43}}{51}\)

Given: \(BP=BQ=radius\). Say \(BP=BQ=radius=14x\), for some positive \(x\). Then:
\(AP=\frac{1}{2}*BP=7x\) and \(CQ=\frac{2}{7}*BQ=4x\). Thus, \(AB=21x\) and \(CB=18x\).

Now, since hypotenuse \(AC=100\), then \((21x)^2+(18x)^2=100^2\) --> \(x^2=\frac{100^2}{765}\) --> \(x=\frac{100}{3\sqrt{85}}\).

Next, \(radius=14x=\frac{14*100}{3\sqrt{85}}\) --> rationalize by multiplying both numerator and denominator by \(\sqrt{85}\) to get: \(radius=\frac{14*100\sqrt{85}}{3*85}\) --> reduce by 5: \(radius=\frac{280\sqrt{85}}{51}\).

Answer: A.

Hope it's clear.

P.S. PLEASE ALWAYS POST ANSWER CHOICES WITH PS PROBLEMS.
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Re: In the diagram (not drawn to scale), Sector PQ is a quarter-circle.  [#permalink]

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New post 20 Sep 2012, 22:28
Please share the answer options. They are not visible.

Lets consider AP = x, than PB = 2x
Since PB = 2x, QB = 2x as well. Because both are radius of the same circle.
Since, QB = 2x we can calculate that CQ = 4x/7.
now use pythagoras theorem = CB^2+AB^2 = AC^2
(18x/7)^2+(3x)^2 = 100^2...Solve this equation to get the value of x and 2x will be the answer.

Hope it helps. Cheers!
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Re: In the diagram (not drawn to scale), Sector PQ is a quarter-circle.  [#permalink]

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New post 17 Jul 2016, 01:58
venmic wrote:
Attachment:
Untitled.png
In the diagram (not drawn to scale), Sector PQ is a quarter-circle. The distance from A to P is half the distance from P to B. The distance from C to Q is 2/7 of the distance from Q to B. If the length of AC is 100, what is the length of the radius of the circle with center B?

A. \(\frac{280\sqrt{85}}{51}\)

B. \(\frac{240\sqrt{70}}{61}\)

C. \(\frac{240\sqrt{67}}{43}\)

D. \(\frac{230\sqrt{51}}{43}\)

E. \(\frac{220\sqrt{43}}{51}\)

given CQ=2/7QB
CB=CQ+QB----->2/7QB+QB------>9/7QB------(1)
similarly given AP=1/2PB
AB=AP+PB--------->1/2PB+PB------>3/2PB-------(2)
As AB^2+CB^2=AC^2
(3/2PB)^2+(9/7QB)^2=100^2
as PB=QB=radius of circle
(9/4+81/49)PB^2=100^2
PB^2= (100^2*49*4)/765
as 765 can be written as 3*√85
we see √85 is only in option (A) hence
Ans A
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Re: In the diagram (not drawn to scale), Sector PQ is a quarter-circle.  [#permalink]

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Re: In the diagram (not drawn to scale), Sector PQ is a quarter-circle. &nbs [#permalink] 04 Dec 2017, 09:04
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