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Difficulty:
85%
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Question Stats:
61% (03:51) correct
39%
(03:31)
wrong
based on 249
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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. \(\frac{220\sqrt{43}}{51}\)
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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.
General Discussion
20 Sep 2012, 22:28
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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!
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!
