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yellowho: Master | Next Rank: 500 Posts; Posts: 233; Joined: Wed Aug 22, 2007 3:51 pmFollowed by: 2 members; Location: New York; Thanked: 7 times

by yellowho » Mon Jan 31, 2011 8:19 am

Triangle/Circle Problem:

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Source: Beat The GMAT — Problem Solving | Mon Jan 31, 2011 8:19 am

maihuna: Legendary Member; Posts: 1578; Joined: Sun Dec 28, 2008 1:49 amFollowed by: 9 members; Thanked: 82 times; GMAT Score:720

by maihuna » Mon Jan 31, 2011 8:35 am

connect C towards the line length 5 meeting circle: let it be D, we know CD will be perpendicular to BD, so RT anfle Triangle ADC with measurement : X, 2, 4, where 4 will be hyp so x = (4^2-2^2)^1/2 = 2(3)^1/2

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Mon Jan 31, 2011 9:42 am

Refer to the above figure.
N is the center of the semicircle.
As Ab is a tangent to the circle at point M, NM is perpendicular to AB.

Now, âˆ†ABC and âˆ†ANM are similar because,

Angle A is common in both
angle M = angle C = Right angle

Hence, NM/BC = AN/AB ................................................ (1)

Now, NM = CN = radius of the semicircle = r (say)
Hence, AN = AC - CN = (4 - r)
And, BC = 3, AB = 5

Replacing these values in (1),

.... r/3 = (4 - r)/5
=> 5r = 12 - 3r
=> 8r = 12
=> r = 12/8 = 3/2 = 1.5

Hence, radius of the semicircle is 1.5

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Mon Jan 31, 2011 9:43 am

maihuna wrote:connect C towards the line length 5 meeting circle: let it be D, we know CD will be perpendicular to BD

No, it won't.
C is not the center of the semicircle.

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