The figure shown above consists of three identical circles that are tangent to each other. If the area of the shaded region is \(64\sqrt{3}-32\pi\), what is the radius of each circle?

(A) 4
(B) 8
(C) 16
(D) 24
(E) 32


Let the radius of the circle be \(r\), then the side of equilateral triangle will be \(2r\).

Area of the shaded region equals to area of the equilateral triangle minus area of three 60 degrees sectors.

Area of a 60 degree sector is 1/6 of the are of whole circle (as whole circle is 360 degrees and 60 is 1/6 of it), hence the area of 3 such sectors will be 3/6=1/2 of the area of whole circle, so \(area_{sectors}=\frac{\pi{r^2}}{2}\) (here if you could spot that \(\frac{\pi{r^2}}{2}\) should correspond to \(32\pi\) then you can write \(\frac{\pi{r^2}}{2}=32\pi\) --> \(r=8\));

Area of equilateral triangle equals to \(a^2\frac{\sqrt{3}}{4}\), where \(a\) is the length of a side. So in our case \(area_{equilateral}=(2r)^2*{\frac{\sqrt{3}}{4}}=r^2\sqrt{3}\);

Area of the shaded region equals to \(64\sqrt{3}-32\pi\), so \(area_{equilateral}-area_{sectors}=r^2\sqrt{3}-\frac{\pi{r^2}}{2}=64\sqrt{3}-32\pi\) --> \(r^2=\frac{2(64\sqrt{3}-32\pi)}{2\sqrt{3}-\pi}=\frac{64(2\sqrt{3}-\pi)}{(2\sqrt{3}-\pi)}=64\) --> \(r=8\).

Answer: B.

GEOMETRY: Shaded Region Problems: https://gmatclub.com/forum/geometry-sha ... 75005.html
_________________

Find the are of equilateral triangle with side as 2r
Find the are of the three sectors (eq. trianlge has \(\angle\) = \(60^{\circ}\) )
Subtract both to get the area of shaded region. You will get the equation as

\(\sqrt{3}r^{2} - \frac{ \pi r^{2}}{2} = 64\sqrt{3} - 32 \pi\)

Don't even need to solve the equation you can see \(r^{2} = 64\) OR \(\frac{r^{2}}{2} = 32\)
Hence, r = 8


Answer - B
Time Taken - 2:00
Difficulty level - 700

Let the radius of each circle = r
The triangle formed is a equilateral triangle with each side = 2r

So Area of the Triangle = \(\frac{\sqrt{3}}{4} * (2r)^2 = \sqrt{3} r^2\) ....................... (1)

This triangle consists of 3 sectors each of 60 Degrees i.e total 180 Deg

Area of these sectors = \(\frac{\pi r^2}{2}\) .................. (2)

Area of shaded region = (1) - (2)

\(\sqrt{3}r^2 - \frac{\pi r^2}{2} = 64\sqrt{3} - 32\pi\)

Solving the above, radius r = 8 = Answer = B

Option B.
The triangle formed by the circles will be an equilateral triangle.
Area of triangle=\(\sqrt{3}/4*(side)^2\)
And side =2*radius
Now area of triangle=area shaded+area of 3 sectors of the circles.
Area shaded=given
Area of 3 sectors=\(3*[60/360*pi*r^2]\)
We'll equate the 2 sides[Area triangle=Area shaded+area of 3 sectors]
and simplify to get r=8

Is this a valid way of approaching the question, or just a lucky coincidence:

Area of equilateral triangle is \((S^2\sqrt{3})/4\)

So \((S^2\sqrt{3})/4\) = \(64\sqrt{3}\)

Simplify this to\(S = 16\sqrt{3}/\sqrt{3}\) -> = 16. One side is 16, divide by 2 to get 8.

I simply ignored the "-32" - is this a valid way of doing it?

Hi erikvm,

The best approach would be to write down the equations for the area of shaded region and then equate both sides. Your method worked because the final expression was written in the form where you could equate two expressions from LHS and RHS. This may not be the case always and hence is not the best practice to follow.

Also to answer bgpower's question, the final equation can be written as

\(\sqrt{3}r^2 - \frac{πr^2}{2} = 64\sqrt{3} - 32π\)

There are two easy ways to solve the equation for r.

Equating the expression
You can see the similarity in expression on both sides of the equation and can equate \(\frac{πr^2}{2} = 32π\) or \(\sqrt{3}r^2 = 64\sqrt{3}\) . This would give you the value of \(r = 8\).

Solving the equation
Alternatively the equation can also be solved very quickly. You just need to be aware of the possibilities of the terms canceling out.

The equation can be written as \(r^2 =\) (\(64\sqrt{3} - 32π)/(\sqrt{3}- \frac{π}{2})\)

If we take \(64\) common from the numerator we will have the expression as \(r^2 =\) \(64 *( \sqrt{3} - \frac{π}{2})/(\sqrt{3}- \frac{π}{2})\) thus resulting in \(r^2 = 64\) and \(r = 8\)

So with the above approach, it is possible to solve the question in less than 2 minutes.

Hope this helps

Regards
Harsh
_________________

I used guesstimation which works for people who suck at math like me.

All you need to know is √3 = 1.7
pi=3.14

64*1.7 - 32*3.14

Is more or less 8

The shaded region is about (1/4) of the triangle so
a^2(√3/4)=4*8
a^2 = 32*(4/√3)
a^2 = 128/1.7 is less than 260 but more than 192
16*16=256
so r = 16/2 = 8

Took me 5 minutes though

Attached is a visual that should help.

_________________

We can let each radius = r, and so the side of each triangle = 2r.

Notice that the area of the equilateral triangle consists of the central shaded region and three identical circular sectors, each of which is a 60-degree sector from its circle. Using this information, we can create the following equation:

(Area of equilateral triangle) - (3 x area of 1/6 of each circle) = area of shaded region

(2r)^2√3/4 - 3(1/6 x π r^2) = 64√3 − 32π

[(4r^2)√3]/4 - (πr^2)/2 = 64√3 − 32π

(r^2)√3 - (πr^2)/2 = 64√3 − 32π

Multiplying both sides by 2, we have:

2(r^2)√3 - πr^2 = 128√3 − 64π

r^2(2√3 - π) = 128√3 − 64π

Dividing both sides by (2√3 - π), we have:

r^2 = 64

r = 8

Answer: B
_________________

Area Equilateral Triangle - Shaded region = Half area of one circle

\((2r)^2 * \frac{\sqrt{3}}{4} - (64\sqrt{3} - 32\pi) = \frac{1}{2}\pi r^2\)

\(r^2\sqrt{3} - 128\sqrt{3} + 32\pi = \frac{1}{2}\pi r^2\)

\(2r^2\sqrt{3} - 128\sqrt{3} + 64\pi - \pi r^2 = 0\)

\((r^2 - 64)(2\sqrt{3} - \pi) = 0\)

\(r^2 = 64\)

\(r = 8 \)

There is a very simple of way of solving this.

Basically the equation is nothing but the Area of the Triangle - The 3 Sectors = 64\sqrt{3}-32π

So area of the equilateral triangle is 64\sqrt{3}

Area of an equilateral triangle = side^2*\sqrt{3}/4

side^2=64*4
side=16

Radius is 16/2=8.

Answer is B.

Hello,

can you help me understand how did we arrive at pi*r^2/2? Usually the area of the sector formula is angle/360* pi*r^2

We have three 60-degree sectors. Together they make 180-dgeree sector, or half the circle, which is 360-degree. The area of half a circle is pi*r^2/2.
_________________

Area of equilateral triangle is given by \(\frac{s^2\sqrt{3}}{4}\), where s is the length of a side.

In this case, the length of the side is 2r. S = 2r -->

\(\frac{(2r)^2\sqrt{3}}{4}\) = \(r^2\sqrt{3}\)

Now we know that the irrational terms in the expression for the area is not likely to be affected by each other: \(64\sqrt{3} - 32Pi\)

So we can conclude that r^2 = 64.