# Common area between Circle and Equilateral triangle [closed]

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.

• Welcome to MSE! Can you edit your question to include your thoughts and efforts on this problem? What have you tried, and where are you having difficulty? This will help people write an appropriate answer the addresses your problem. Questions that include this information tend to have a much better response.
– user296602
Commented Jan 20, 2016 at 19:32
• the difference of Areas is given by $\Delta A = r^2(\pi - \sqrt 3)$ Are you sure the question doesn't say $\pi - \sqrt 3$ instead of $\pi - 3$ so that we have $r=1$ instead of $r=\sqrt \frac{\pi - 3}{\pi - \sqrt 3}$
– WW1
Commented Jan 20, 2016 at 20:18
• @WW1 Yes I am sure. Commented Jan 21, 2016 at 12:29

Hint: Divide the triangle into 4 triangles.

• I have to find the area without using a calculator. Commented Jan 20, 2016 at 20:02
• @RezwanArefin The area of the smaller triangles is 1/4 area of larger triangle. The area of one small triangle plus the are of one of circular segment is 1/6 area of circle. Add these to get the area of the shaded region: 1/2 area of triangle plus 1/6 area of circle. Commented Jan 20, 2016 at 20:22

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?

• But, I am not allowed to use a calculator. Commented Jan 20, 2016 at 19:52
• None of this requires a calculator. You should know the formulas for the area of a circle in terms of its diameter and the area of an equilateral triangle in terms of its side (or be able to derive the second). This will give you an algebraic expression for $d$ Commented Jan 20, 2016 at 19:54
• How to find find the area of the clear part of the triangle near A? Commented Jan 20, 2016 at 19:57

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 $A=A_1+2A_2$