In terms of $R$ which is the radius of all four circles, what is the area of the intersection region of these four equal circles and the height of the marked arrow in the figure? The marked arrow is along the line CD, also the midpoint of all the circles are points A, B, C and D. Looking for a very short intuitive solution. I have checked similar questions on this site for example this and this.
Best Answer
Notice that the area of the equilateral triangle with edge $R$ plus $\frac12$ the area of the shaded region is $\frac16$ the area of the circle. The height of the triangle is $\frac{R\sqrt3}{2}$ and the area is $\frac12\cdot R\cdot\frac{R\sqrt3}{2}=\frac{R^2\sqrt3}{4}$. The area of the shaded region is: $$2\left(\frac{\pi R^2}{6}-\frac{R^2\sqrt3}{4}\right)=\frac{2\pi R^2}{6}-\frac{3R^2\sqrt3}{6}=\frac{R^2(2\pi-3\sqrt3)}{6}$$ Also notice that the height of the triangle plus $\frac12$ the height of the shaded area is equal to the radius of the circle. The height of the shaded area is: $$2\left(R-\frac{R\sqrt3}{2}\right)=2R-R\sqrt3=R(2-\sqrt3)$$