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?
Best Answer
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 $\pi/2$.