Question

Consider 3 equal circles of radius r within a given circle of radius R each to touch the other two and the given circle as shown in figure, then the area of shaded region is equal to ?

Answer 1

We can form an expression for the area of the shaded region like so:

`A_"shaded" = piR^2 - 3(pir^2)-A_"centre"`

where `A_"centre"` is the area of the small section between the three smaller circles.

To find the area of this, we can draw a triangle by connecting the centres of the three smaller white circles. Since each circle has a radius of `r`, the length of each side of the triangle is `2r` and the triangle is equilateral so have angles of `60^o` each.

We can thus say that the angle of the central region is the area of this triangle minus the three sectors of the circle. The height of the triangle is simply `sqrt((2r)^2-r^2) = sqrt(3)r^`, so the area of the triangle is `1/2 * base * height = 1/2 * 2r * sqrt(3)r = sqrt(3)r^2`.

The area of the three circle segments within this triangle are essentially the same area as half of one of the circles (due to having angles of `60^o` each, or `1/6` a circle, so we can deduce the total area of these sectors to be `1/2 pir^2`.

Finally, we can work out the area of the centre region to be `sqrt(3)r^2-1/2pir^2 = r^2(sqrt(3)-pi/2)`

Thus going back to our original expression, the area of the shaded region is

`piR^2-3pir^2-r^2(sqrt(3)-pi/2)`

Answer 2

`A = r^2( 1 /6 (8 sqrt(3) - 1) pi - sqrt(3) )`

Explanation:

Let's give the white circles a radius of `r=1`. The centers form an equilateral triangle of side `2`. Each median/altitude is `sqrt{3}` so the distance from a vertex to the centroid is `2/3 sqrt{3}`.

The centroid is the center of the big circle so that's the distance between the center of the big circle and the center of the little circle. We add a little radius of `r=1` to get

`R = 1 + 2/3 sqrt{3}`

The area we seek is the area of the big circle less the equilateral triangle and the remaining `5/6` of each little circle.

`A = pi R^2 - 3( 5/6 pi r^2 ) - \sqrt{3}/4 (2r)^2`

`A = pi( 1 + 2/3 sqrt{3})^2 - 3( 5/6 pi) - sqrt{3}`

`A = 1/6 (8 sqrt(3) - 1) pi - sqrt(3)`

We scale by `r^2` in general.