The area not covered by six pointed star

Question

In a circle with radius $r$, two equi triangles overlapping each other in the form of a six pointed star touching the circumference is inscribed! What is the area that is not covered by the star?

Progress

By subtracting area of the star from area of circle , the area of the surface can be found! But how to calculate the area of the star?

Answer 1

The length of a side of an equilateral triangle is $$\sqrt{r^2+r^2-2\cdot r\cdot r\cdot \cos (120^\circ)}=\sqrt 3r.$$ The distance between the center of the circle and each side of an equilateral triangle is $$\sqrt 3r\cdot \frac{\sqrt 3}{2}\cdot \frac{1}{3}=\frac 12r.$$

Hence, the length of a side of the smaller equilateral triangle, which is the 'corner' of the star, is $$\frac 12r\cdot \frac{2}{\sqrt 3}=\frac{1}{\sqrt 3}r.$$

Hence, the area of the star is $$\frac{\sqrt 3}{4}\cdot (\sqrt 3r)^2+3\times \frac{\sqrt 3}{4}\left(\frac{1}{\sqrt 3}r\right)^2=\sqrt 3r^2.$$

So, the answer is $$\pi r^2-\sqrt 3r^2=(\pi-\sqrt 3)r^2.$$

Answer 2

Answer:

The required Area = Area of the Circle - (2*Area of the Equilateral Triangle - Area of the Hexagon that is formed by the superimposition of two equilateral trianle)

Area of the Circle $=\pi r^2$

Length of the equilateral triangle $\sqrt{3}r$

Area of the equilateral triangle then is = $\frac{3\sqrt{3}r^2}{4}$

Length of the hexagon (l) $= \frac{r}{\sqrt{3}}$

Area of the hexagon = $\frac{3\sqrt{3}l^2}{2}$

Thus Area of the hexagon = $\frac{3\sqrt{3}r^2}{6}$

The required area of the part of the circle uncovered by the six sided star = $(\pi r^2 - (2*\frac{3\sqrt{3}r^2}{4} - \frac{3\sqrt{3}r^2}{6}))$

which reduces to $$ = r^{2}(\pi -\sqrt{3})$$

Answer 3

If the six-pointed star is regular, then the answer is $r^2(\pi-\sqrt{3})$. If it is not, then the answer can be larger, up to a limit of $r^2\big(\pi-\frac{3}{4}\sqrt{3}\big)$.

Proof

The required area is the area of the circle ($\pi r^2$) minus the area of the star. The area of the star is the area of a large equilateral triangle (A) plus the area of three small ones. Each small one has a side-length $\frac{1}{3}$ of the large triangle's and therefore an area $\frac{1}{9}$ of its area. So the area we want is $\pi r^2 - A(1+\frac{3}{9}) = \pi r^2 - \frac{4A}{3}$.

A is equal to 6 times the area of the right-angled triangle shown. Stand it on its short side. Its area is $B=\frac{\mathrm{base * height}}{2} = \frac{r^2\sin{30^\circ}\cos{30^\circ}}{2}=r^2(\frac{1}{2})\frac{\sqrt{3}}{2}\frac{1}{2} = r^2(\frac{\sqrt{3}}{8})$. So $A=6B=\frac{3r^2\sqrt{3}}{4}$

So the required area is $\pi r^2 - \big(\frac{3r^2\sqrt{3}}{4}\big)\frac{4}{3} = r^2(\pi-\sqrt{3})$

However, all you say is that the triangles overlap to make a star. You do not define a star. If we allow the overlap such that the points on the circle are unequally spaced, then we might get a shape like this:

At the limit, the area of the star is the area of a large triangle, A, giving required area $\pi r^2 - A = r^2\big(\pi-\frac{3}{4}\sqrt{3}\big)$