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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?

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

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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 $\pi/2$.

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