Challenge question: In the above figure, the small circle...

[Brent@GMATPrepNow]

In the above figure, the small circle and big circle have diameters of 3 and 6 respectively. If AB||CD, and both circles share the same center, what is the area of the shaded region?

A) 9/2 + 2Ï€
B) 9/2 + 3Ï€
C) 3√3 + 3π
D) 6√3 + 2π
E) (9/2)√3 + 3π

Answer: E
Difficulty level: 700+
Source: www.gmatprepnow.com

[Brent@GMATPrepNow]

In the above figure, the small circle and big circle have diameters of 3 and 6 respectively. If AB||CD, and both circles share the same center, what is the area of the shaded region?

A) 9/2 + 2Ï€
B) 9/2 + 3Ï€
C) 3√3 + 3π
D) 6√3 + 2π
E) (9/2)√3 + 3π

If we recognize that the hypotenuse of the blue right triangle below is twice the length of one side, we can see that this is a 30-60-90 SPECIAL TRIANGLE

In the base triangle (on the right), the side opposite the 30° has length 1.
So, the magnification factor of the triangle in the question is 3/2 (i.e., the triangle in the diagram is 3/2 times the size of the base triangle)
So, the length of the 3rd side = (3/2)(√3) = (3√3)/2

Let's add this to our diagram...


From here, we can calculate the area of the shaded triangle.
The length of the base = (3√3)/2 + (3√3)/2 = 3√3

Area = (base)(height)/2
= (3√3)(3/2)/2
= (9√3)/4
Let's add this to our diagram...


Important: we earlier learned that the triangle in the first image is a 30-60-90 special triangle.
In fact, there are four such special triangles in the diagram.
So, let's add the 30° angles


Let's now find the area EACH sector.
Area of sector = (central angle/360°)(π)(radius²)
= (60°/360°)(π)(3²)
= (1/6)(Ï€)(9)
= 3Ï€/2


So, the area of the shaded region = (9√3)/4 + (9√3)/4 + 3π/2 + 3π/2
= (9√3)/2 + 3π

Answer: E

Cheers,
Brent

[swerve]

The area would be similar to that of a rectangle with sides 3*6 = 18.

Considering that pi = 3.14 and sqrt{3} = 1.7, the closest solution would be E.