The Official Guide For GMAT® Quantitative Review, 2ND Edition


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






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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

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
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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.

Area of shaded = area of triangle - the sum of the sectors
area of shaded = 64\sqrt{3} - 32pi
area of triangle = 1/2(2r)(r\sqrt{3}) = r^2 \sqrt{3}
sum of the sectors = (3(60/360 x pi r^2))

64\sqrt{3} - 32pi = r^2\sqrt{3} - (3(1/6 x pi r^2))
1/2 x 3r^2 = r^2 - 64 + 32
3r^2 = 2r^2 - 128 + 64
r^2 = -64

square each side
r=8
thus, the answer is (B) 8

hope this helps

Bunuel - i clicked on the link provided but link doesnt exist any other options ?

64sqrt{3}-32П, what is the radius of each circle?

S of shaded area = S of triangle - 1/6S of circle*3. SO in our formula only 32П looks like S of a circle.

32П = 1/6 ПR^2*3 (as we have 60 degree angle) ==> R=8 (B)

It took 40 sec for me.