A cool-aboration Problem

The figure shows 3 lines connecting the centres of the 3 circles to form a triangle.

The answer to this question is the area of the $\color{#ff8000}{orange}$ region in the middle.

The angles of the triangle are 60 degrees each because it is an equilateral triangle and each side is 2 units.

So the $\color{#efef00}{yellow}$ areas are 3 sectors with angle 60 degrees and radius 1.

$Area=\frac { \theta }{ 360 } \pi { r }^{ 2 }=\frac { 60 }{ 360 } \pi =\frac { \pi }{ 6 }$

So area of 3 sectors is $3\times \frac { \pi }{ 6 } =\frac { \pi }{ 2 }$

Now we know the area of $\color{#ffff00}{yellow}$ region. We can subtract it from the area of the equilateral triangle of side length 2.

$Area\quad of\quad eq.\quad triangle=\frac { \sqrt { 3 } }{ 4 } { a }^{ 2 }=\frac { \sqrt { 3 } }{ 4 } \left( { 2 }^{ 2 } \right) =\sqrt { 3 }$

So our answer is $\sqrt { 3 } -\frac { \pi }{ 2 } \approx \boxed { 0.161 }$

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Please upvote if you liked the solution.

Since Mehul has not elaborated properly.Let me post a nice solution :

Construct a triangle by joining the centres of the 3 circles.Indeed this triangle will have all its sides of length $2r=2$ and thus is an equilateral triangle.

Thus area of an equilateral triangle $=\dfrac{\sqrt{3}}{4}(side)^2 = \dfrac{\sqrt{3}}{4}(2)^2 = \sqrt{3}$

We can see that 3 congruent sectors are formed.Since there is an equilateral triangle , each angle of these sectors : $\theta=60^0$

Area of 3 sectors $=3 \times\dfrac{\theta}{360}\times \pi r^2 \\ = 3\times\dfrac{60}{360}\times \left(\dfrac{22}{7}\right) (1^2) =\dfrac{11}{7}$ .

Hence required area = $\sqrt{3}-\dfrac{11}{7} \approx \boxed{0.161}$

Note: This question assumes that $\pi=\dfrac{22}{7}$ .

Construct a Equilateral triangle by Joining The Centers of the circles.The triangle is equilateral because all the sides of the triangle are of the same length.The triangle will have a side of two units.

The area of the triangle is $\dfrac{\sqrt3}{4}*2*2$ = $\sqrt 3$ square units.

Now, We subtract The area of the Three sectors of the three circles.

= $3*\dfrac {22}{7}*1*1*60*\dfrac{1}{360}=\dfrac{11}{7}\ units.$

Because Area of a sector Is $Pi*{r}^{2}*\theta*1/360$ . Where theta is the angle of the sector at the center. r is the radius of the circle. And since we have 3 sectors, we multiply by three.

Hence, The area enclosed is $\sqrt 3-\dfrac{11}{7}$ which is Approximately 0.161 square units.

And, We are done.

Cheers!

the figure formed is a curvalinear triangle with a formula generated by FOX is 0.161r^2 since r = 1 then area = 0.161