# Calculating the area between three tangent circles of unit radius, and using the result to show $\pi^2<12$

Diagram shows three circles, each of radius 1cm, centres A, B, and C. Each circle touches the other two.

From here you can get the area of triangle ABC:

Height of equilateral triangle ABC: $$h = \sqrt{2^2 - 1^2} = \sqrt3$$

Area of equilateral triangle ABC: $$A = \frac{1}{2}\cdot 2\cdot\sqrt3 = \sqrt3$$

Area of single circle: $$C = \pi\cdot 1^2 = \pi$$

Circle sector has angle 60deg (angles in equilateral triangle)

Therefore area of single circle sector $$S = \frac{\pi}{6}$$

Area of shaded region: SW $$= S = \sqrt3 - \frac{\pi}{6} \cdot 3 = \sqrt3 - \frac{\pi}{2}$$

From the area of the shaded region, prove that $$\pi^2 < 12$$.

Possible i've miscalculated the shaded region area, but i'm not sure how one relates to the other, any pointers?

• $\sqrt3-\frac{\pi}{2}>0$, $\sqrt3>\frac{\pi}{2}$,$3>(\frac{\pi}{2})^2$,$\pi^2<12$ Sep 20, 2021 at 20:48
• Clear and concise, thank you. Sep 20, 2021 at 21:23

I would not even bother with the shaded area except to note it's strictly positive. Thus the area of the full triangle $$\sqrt3$$ is greater than the area of the three sectors $$\pi/2$$.