# 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 ?

## May 30, 2018

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

${A}_{\text{shaded" = piR^2 - 3(pir^2)-A_"centre}}$

where ${A}_{\text{centre}}$ 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 $2 r$ 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 $\frac{1}{2} \cdot b a s e \cdot h e i g h t = \frac{1}{2} \cdot 2 r \cdot \sqrt{3} r = \sqrt{3} {r}^{2}$.

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 $\frac{1}{6}$ a circle, so we can deduce the total area of these sectors to be $\frac{1}{2} \pi {r}^{2}$.

Finally, we can work out the area of the centre region to be $\sqrt{3} {r}^{2} - \frac{1}{2} \pi {r}^{2} = {r}^{2} \left(\sqrt{3} - \frac{\pi}{2}\right)$

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

$\pi {R}^{2} - 3 \pi {r}^{2} - {r}^{2} \left(\sqrt{3} - \frac{\pi}{2}\right)$

May 30, 2018

$A = {r}^{2} \left(\frac{1}{6} \left(8 \sqrt{3} - 1\right) \pi - \sqrt{3}\right)$

#### 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 $\frac{2}{3} \sqrt{3}$.

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

$R = 1 + \frac{2}{3} \sqrt{3}$

The area we seek is the area of the big circle less the equilateral triangle and the remaining $\frac{5}{6}$ of each little circle.

$A = \pi {R}^{2} - 3 \left(\frac{5}{6} \pi {r}^{2}\right) - \setminus \frac{\sqrt{3}}{4} {\left(2 r\right)}^{2}$

$A = \pi {\left(1 + \frac{2}{3} \sqrt{3}\right)}^{2} - 3 \left(\frac{5}{6} \pi\right) - \sqrt{3}$

$A = \frac{1}{6} \left(8 \sqrt{3} - 1\right) \pi - \sqrt{3}$

We scale by ${r}^{2}$ in general.