Common area between Circle and Equilateral triangle

Question

A circle is drawn with diameter BC of a equilateral triangle ABC. Area of triangle is $\pi - 3$ less than the area of the circle. What is the area of the common region between circle and the triangle? I have to do this without using a calculator.

Answer 1

Hint: Divide the triangle into 4 triangles.

Answer 2

Hint: 1) let $|BC|=d$. Compute the area of the circle and the area of the triangle. The difference you are given will let you find $d$ 2) now find the length of the segment from A to the circle. 3) find the area of the clear part of the triangle near A 4) subtract this from the triangle area. Where are you stuck?

Answer 3

Hint:

Let $OB=r$. Note that , since the triangle is equilateral, its height is $h=AO=r\sqrt{3}$.

1) As a first step find $r$ from the condition about the areas of the circle and the triangle: $\pi r^2\sqrt{3}=\pi-3$

2) prove that all the triangle $COM$, $BON$, $MNO$, $ANM$ are equilateral, so $\angle MON = 60°=\frac{\pi}{3}$.

3) Find the area of the sector of circle $OMN$: $A_1=r^2\frac{\pi}{3}$

4) Find the area of the triangle $COM$ : $A_2=\left(\frac{r}{2}\right)^2\sqrt{3}$ , that, by symmetry, is the same as the area of the triangle $BON$.

5) the searched area is $A=A_1+2A_2$