In the diagram, each of the three identical circles touch the other two. The circumference of each circle is 36. What is the area of the shaded region?

Guest Apr 21, 2021

#1**+2 **

https://gyazo.com/0bb2e632441b774f3d412d086fa55957

I made an equilateral triangle using the centers of the triangle.

Since an equilateral triangle has angles of 60^{o}, each sector part is just \(\frac{1}{6}\) of the circle.

Circumference of each circle = 36 units

Radius = \(\frac{18}{π}\)

Sector areas = \(3 \cdot \frac{1}{6} \cdot (\frac{18}{π})^2 π\)

= \(\frac{1}{2} \cdot \frac{324}{π^2}π\)

= \(\frac{1}{2} \cdot \frac{324}{π}\)

= \(\frac{162}{π}\)

Now the area of the equilateral triangle is:

\(\frac{\sqrt{3}}{4}\cdot (\frac{36}{π})^2\)

= \(\frac{\sqrt{3}}{4} \cdot \frac{1296}{π^2}\)

= \(\frac{1296\sqrt{3}}{4π^2}\)

Which means the area of the shaded region is:

\(\frac{1296\sqrt{3}}{4π^2} - \frac{162}{π}\)

= \(\frac{1296\sqrt{3}}{4π^2} - \frac{162π}{π^2}\)

= \(\frac{1296\sqrt{3}}{4π^2} - \frac{648π}{4π^2}\)

= \(\frac{1296\sqrt{3} - 648π}{4π^2}\)

= \(\frac{4(324\sqrt{3} - 162π)}{4π^2}\)

= \(\frac{324\sqrt{3} - 162π}{π^2}\)

please tell me if I did anything wrong...

Logarhythm Apr 21, 2021

#2**0 **

LOL

I got something complicated as well... I thought I got it incorrect. i actually got the same answer as you, so nothing's wrong... I think.

SparklingWater2 Apr 21, 2021

#3**0 **

lol I was debating whether or not to post that because there was a similar question, however this was asking for area of the shaded region instead of the perimeter of it...

Logarhythm
Apr 21, 2021

#4**+2 **

See image below:

Triangle area - cicle sector area * 3 = area in question

sqrt(3)/4 ( 36/ pi)^{2} - 3 * 1/6 * pi ( 18/pi)^{2 }= shaded (we use 1/6 because the sector of the circle is 60/360 of the circle)

324 sqrt(3) / pi^{2} - 162 / pi ** <============ Same answer as LR Way to go Logarhythm!! **

ElectricPavlov Apr 21, 2021