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JubtaGubar
Joined: 23 Oct 2011
Last visit: 06 Jun 2012
Posts: 34
Given Kudos: 7
Location: Ukraine
Schools: LBS '14 (M)
GMAT 1: 650 Q44 V35
WE:Corporate Finance (Mutual Funds and Brokerage)
21 Apr 2012, 10:28
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
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Question Stats:
54% (02:35) correct
46%
(02:24)
wrong
based on 174
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Consider a quarter of a circle of radius 16. Let r be the radius of the circle inscribed in this quarter of a circle. Find r.
A. \(16*(\sqrt{2} -1)\)
B. \(8*(\sqrt{3} -1)\)
C. \(4*(\sqrt{7} - 1)\)
D. \(12* (\sqrt{7} -1)\)
E. None of these
A question from MBA Strategy course
A. \(16*(\sqrt{2} -1)\)
B. \(8*(\sqrt{3} -1)\)
C. \(4*(\sqrt{7} - 1)\)
D. \(12* (\sqrt{7} -1)\)
E. None of these
A question from MBA Strategy course
21 Apr 2012, 12:09
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JubtaGubar
Look at the diagram below:
Attachment:
Quarter of circle.png [ 5.62 KiB | Viewed 11540 times ]
Now, since the sides of a square equal to \(r\), then its diagonal equals to \(r\sqrt{2}\), hence \(r\sqrt{2}+r=16\) --> \(r=\frac{16}{\sqrt{2}+1}\).
Rationalise by multiplying both numerator and denominator by \(\sqrt{2}-1\): \(r=\frac{16(\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2}-1)}\) --> apply \((a+b)(a-b)=a^2-b^2\) to the expression in the denominator: \(r=\frac{16(\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2}-1)}=\frac{16(\sqrt{2}-1)}{2-1}=16(\sqrt{2}-1)\).
Answer: A.
Hope it's clear.
General Discussion
JubtaGubar
Joined: 23 Oct 2011
Last visit: 06 Jun 2012
Posts: 34
Given Kudos: 7
Location: Ukraine
Schools: LBS '14 (M)
GMAT 1: 650 Q44 V35
WE:Corporate Finance (Mutual Funds and Brokerage)
23 Apr 2012, 08:13
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Thank you Bunuel!
I got 16/(sqr2 +1) and just forgot to multiply by (sqr2 -1).
I got 16/(sqr2 +1) and just forgot to multiply by (sqr2 -1).
