Area of 3 overlapping circles

Three equal circles with radius r are drawn as shown,
each with its centre on the circumference of the other two circles.
A, B and C are the centres of the three circles.
Prove that an expression for the area of the shaded region is:

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

$\begin{array}{|rcll|} \hline A_{\triangle ABC} &=& \frac12 \cdot r \cdot h_{\triangle ABC} \\\\ h_{\triangle ABC}^2 + (\frac{r}{2})^2 &=& r^2 \\ h_{\triangle ABC}^2 + \frac{r^2}{4} &=& r^2 \\ h_{\triangle ABC}^2 = r^2 - \frac{r^2}{4} \\ h_{\triangle ABC}^2 = \frac34 \cdot r^2 \\ h_{\triangle ABC} = \frac{\sqrt{3}}{2} \cdot r \\\\ A_{\triangle ABC} &=& \frac12 \cdot r \cdot h_{\triangle ABC} \\ A_{\triangle ABC} &=& \frac12 \cdot r \cdot \frac{\sqrt{3}}{2} \cdot r \\ \mathbf{A_{\triangle ABC} }&\mathbf{=}& \mathbf{\frac{r^2}{4} \cdot \sqrt{3} }\\ \hline \end{array}$

$\begin{array}{|rcll|} \hline A_{Arc} &=& \pi r^2 \cdot \frac{60^{\circ}} {360^{\circ}} \\ A_{Arc} &=& \frac{\pi r^2} {6} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline A_{\text{shaded region}} &=& 3\cdot (A_{Arc}-A_{\triangle ABC})+A_{\triangle ABC} \\ A_{\text{shaded region}} &=& 3\cdot A_{Arc}-3\cdot A_{\triangle ABC}+A_{\triangle ABC} \\ A_{\text{shaded region}} &=& 3\cdot A_{Arc}-2\cdot A_{\triangle ABC} \\ A_{\text{shaded region}} &=& 3\cdot \frac{\pi r^2} {6}-2\cdot \frac{r^2}{4} \cdot \sqrt{3} \\ A_{\text{shaded region}} &=& \frac{\pi r^2} {2}- \frac{r^2}{2} \cdot \sqrt{3} \\ \mathbf{A_{\text{shaded region}} }&\mathbf{=}& \mathbf{\frac{r^2}{2} \cdot ( \pi- \sqrt{3} )} \\ \hline \end{array}$

Hi Kreyn,

Three equal circles with radius r are drawn as shown, each with its centre on the circumference of the other two circles. A, B and C are the centres of the three circles. Prove that an expression for the area of the shaded region is:

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

Now AB, AC and BC are all radii, so they are all equal, so ABC is an equilateral triangles and all the angles are 60 degrees.

$\boxed{\text{Area of triangle ABC=}\frac{1}{2}absinC }\\ \\~\\ Area\;\;ABC=\frac{1}{2}r^2sin60^\circ\\ Area\;\;ABC=\frac{1}{2}r^2*\frac{\sqrt3}{2}=\frac{\sqrt3\;r^2}{4}$

Now I want to know what the area of minor segment AB is on the circle centred at C

$Segment\; area = \frac{60}{360}\pi r^2\;-\;\frac{\sqrt3\;r^2}{4}\\ Segment\; area = \frac{2}{12}\pi r^2\;-\;\frac{3\sqrt3\;r^2}{12}\\ Segment\; area = \frac{2\pi r^2-3\sqrt3\;r^2}{12}\\ so\\ Shaded \;area = 3* \frac{2\pi r^2-3\sqrt3\;r^2}{12} + \;\frac{\sqrt3\;r^2}{4}\\ Shaded\; area = \frac{2\pi r^2-3\sqrt3\;r^2+\sqrt3\;r^2}{4} \\ Shaded\; area = \frac{2\pi r^2-2\sqrt3\;r^2}{4} \\ Shaded\; area = \frac{\pi r^2-\sqrt3\;r^2}{2} \\ Shaded\; area = \frac{ r^2}{2}(\pi -\sqrt3\;) \\$

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