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

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Manager
Joined: 02 Nov 2009
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In the diagram (not drawn to scale), Sector PQ is a quarter-circle.  [#permalink]

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Updated on: 04 Dec 2017, 08:11
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Question Stats:

65% (03:44) correct 35% (03:17) wrong based on 158 sessions

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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:

Untitled.png [ 15.12 KiB | Viewed 10001 times ]

Originally posted by venmic on 20 Sep 2012, 18:40.
Last edited by Bunuel on 04 Dec 2017, 08:11, edited 2 times in total.
Edited the question and added answer choices.
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Joined: 02 Sep 2009
Posts: 50621
Re: In the diagram (not drawn to scale), Sector PQ is a quarter-circle.  [#permalink]

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21 Sep 2012, 01:36
6
6

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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Joined: 11 May 2011
Posts: 331
Re: In the diagram (not drawn to scale), Sector PQ is a quarter-circle.  [#permalink]

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20 Sep 2012, 21:28
1
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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17 Jul 2016, 00:58
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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30 Sep 2018, 23:41
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Re: In the diagram (not drawn to scale), Sector PQ is a quarter-circle. &nbs [#permalink] 30 Sep 2018, 23:41
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# In the diagram (not drawn to scale), Sector PQ is a quarter-circle.

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