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


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, ∆ABC and ∆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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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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