A CRESCENT IS FORMED BY TWO CIRCLES WHICH TOUCH AT A. C IS THE CENTRE OF THE LARGER CIRCLE. THE WIDTH OF THE CRESCENT AT BD IS 9 cm AND AT EF IS 5 cm. FIND THE AREA OF THE SHADED REGION.

Asked by  | 16th Dec, 2012, 11:19: PM

Expert Answer:

Let x = DC, C ' be the center of the smaller circle, r be the radius of the smaller circle, and R be the radius of the larger circle. We have
BC = BD+DC = 9+x
and
AC = AD-CD = 2r-x.
Since
R = BC = AC,
(1) R = 9+x = 2r-x,
(2) r = (2x+9)/2.
In the right triangle C 'CE,
CE = CF-EF = R-5 = x+9-5 = x+4,
C 'C = C 'D-CD = r-x,
and
C 'E = r.
By the Pythagorean theorem,
(x+4)^2+(r-x)^2 = r^2,
r^2-(r-x)^2 = (x+4)^2,
(r-r+x)(r+r-x) = (x+4)^2,
x(2r-x) = (x+4)^2.
Using (1) in the last equation, we have
x(9+x) = (x+4)^2,
x = 16.
It follows that, by (2), r = (2x+9)/2 = 41/2 and by (1), R = 9+x = 25. Hence the area of the crescent is pi*(R^2-r^2) = pi*(R-r)(R+r) = pi*(9/2)*(91/2) = 819*pi/4 = 643.5 cm2

Answered by  | 17th Dec, 2012, 09:47: AM

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