# 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 cm^{2}

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 cm

^{2}

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

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