Difficult triangles/circles problem

tonebeeze wrote:What is the quickest way to solve these types of problems. Please advise.

Quant Review PS #145

The figure shown (please see attached) consists of three identical circles that are tangent to each other. If the area of the shaded region (center space where the three circles do not touch) is (64)(square root 3) - 32 pi, what is the radius of each circle?

a. 4
b. 8
c. 16
d. 24
e. 32

OA: B

Assume that the radius of circle is r. The side of the equilateral triangle formed in the figure is 2r.

You may either remember (or quickly calculate) that area of an equilateral triangle of side a = a^2. sqrt(3)/4

The area of the equilateral triangle = sqrt(3). (side)^2/4 = sqrt(3).(2r)^2/4 = r^2.sqrt(3)

Now lets calculate the area of each of the three sectors inside the triangle. (i.e. the thtree symmetrical unshaded parts of the triangle)

Each sector is 60/360 = 1/6th of the area of the circle.

Thus area of three sectors = 3/6 = 1/2 area of circle = 1/2 pi . r^2

Area of shaded region = r^2. sqrt(3) -r^2.pi/2 = 64*sqrt(3) - 32.pi = 64*sqrt(3) - 64. pi/2

Thus r^2 = 64 => r = 8

tonebeeze wrote:What is the quickest way to solve these types of problems. Please advise.

Quant Review PS #145

The figure shown (please see attached) consists of three identical circles that are tangent to each other. If the area of the shaded region (center space where the three circles do not touch) is (64)(square root 3) - 32 pi, what is the radius of each circle?

a. 4
b. 8
c. 16
d. 24
e. 32

OA: B

Shaded region = triangle - 3 circle sectors = 64√3 - 32�

Looking at the equation above, we can see that the 3 circle sectors = 32�.
Since the triangle is equilateral, each of its angles is 60 degrees. Since 60/360 = 1/6, each circle sector is 1/6 the area of each circle.
Thus, the 3 sectors = 3*1/6 = 1/2 each circle area.
Since the 3 sectors = 32�, circle area = 64�.
Thus, �r² = 64�, and r=8.

The correct answer is B.

GMATGuruNY wrote:

tonebeeze wrote:What is the quickest way to solve these types of problems. Please advise.

Quant Review PS #145

The figure shown (please see attached) consists of three identical circles that are tangent to each other. If the area of the shaded region (center space where the three circles do not touch) is (64)(square root 3) - 32 pi, what is the radius of each circle?

a. 4
b. 8
c. 16
d. 24
e. 32

OA: B

Shaded region = triangle - 3 circle sectors = 64√3 - 32�

Looking at the equation above, we can see that the 3 circle sectors = 32�.
Since the triangle is equilateral, each of its angles is 60 degrees. Since 60/360 = 1/6, each circle sector is 1/6 the area of each circle.
Thus, the 3 sectors = 3*1/6 = 1/2 each circle area.
Since the 3 sectors = 32�, circle area = 64�.
Thus, �r² = 64�, and r=8.

The correct answer is B.

You really have a gift for explaining this very simply Thanks for your help!

Nice question. Good explanation stormier.