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by maihuna » Fri Sep 18, 2009 11:44 pm
A semi- circle is drawn with AB as its diameter. From C, a point on AB, a line perpendicular to AB is drawn meeting the circumference of the semi-circle at D. Given that AC = 2 cm and CD = 6 cm, the area of the semi-circle (in sq. cm.) will be:

a) 32Ï€

b) 50Ï€

c) 40.5Ï€

d) 81Ï€
e) 162 II
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by mruzeful » Sat Sep 19, 2009 2:05 am
is it A?

r=2+6

and area = (pi)r^2/2 = 32(pi)
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by real2008 » Sat Sep 19, 2009 8:13 am
it should be B
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Answer is B

by aa2kash » Sat Sep 19, 2009 9:03 am
draw a line from the centre of the circle to the point D. this is equal to the radius of the circle.

Now u have a right angle traingle with the sides
6, r and r-2
solve it and get r=10.
area of semicircle is pi*r*r/2 = 50pi
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by GMATQuantCoach » Sat Sep 19, 2009 3:47 pm
Connect AD and BD.

Triangle ADB must be a right triangle.

We can show that triangle ACD is similar to triangle DCB. Then
AC/CD = CD/CB

CD^2 = AC * CB
36 = 2 * CB
CB = 18

Radius = (AC + CB)/2 = 10

Area of Semi-Circle = 1/2 * pi * r^2 = 50pi

B is the answer.
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