Question

the points c1 and c2 are centers of the two equal circles shown in the figure. if the distance between the center is 3 cm, find the area of the shaded region?

Answer 1

`3pi-(9/4)sqrt(3)~=5.528`

Explanation:

First let's call "`D`" the midpoint between `C_1` and `C_2` or between `A` and `B`.

Since `C_1C_2=r` => `C_1D=r/2`

If we consider the `triangle_(BC_1D)`
`cos B hat C_1 D=(C_1D)/(BC_1)=(cancel(r)/2)/cancel(r)=1/2` => `B hat C_1 D=60^@`
`-> sin B hat C_1 D=(BD)/(BC_1)` => `BD=rsin B hat C_1 D=rsin 60^@` => `BD=(rsqrt(3))/2`

Since the area of the circle1 is `pi*r^2`, the area of circular sector defined by the arc `BC_2` is
`S_1=(60^@/360^@)*pi*r^2=(pi*r^2)/6`

The area defined by the arc `BC_2` and the chord `AB` is
`S_(BC_2D)=S_1-S_(triangle_(BC_1D))`

`S_(triangle_(BC_1D))=(b*h)/2=(BD*C_1D)/2=((rsqrt(3))/2*r/2)/2=r^2sqrt(3)/8`
Then
`S_(BC_2D)=(pi*r^2)/6-r^2sqrt(3)/8=r^2/24(4pi-3sqrt(3))`

Since the two circles are equal
`S_(C_1BC_2)=2*S_(BC_2D)=r^2/12(4pi-3sqrt(3))`
`S_(C_1BC_2)=3^2/12(4pi-3sqrt(3))=(3/4)(4pi-3sqrt(3))=3pi-9/4*sqrt(3)~=5.528`