The two identical squares shown above are the largest squares

[Brent@GMATPrepNow]

The two identical squares shown above are the largest squares that can be drawn inside the semicircle. If the length of AB = √12 - √6, what is the area of rectangle BCDE?
A) 6
B) 2√6
C) 6√2
D) 12
E) 12√2

Answer: D
Source: www.gmatprepnow.com

[Brent@GMATPrepNow]

The two identical squares shown above are the largest squares that can be drawn inside the semicircle. If the length of AB = √12 - √6, what is the area of rectangle BCDE?
A) 6
B) 2√6
C) 6√2
D) 12
E) 12√2

Answer: D
Source: www.gmatprepnow.com

Draw a diagonal in one of the squares
Let x = the length of each side of the square
Since we have a right triangle we can apply the Pythagorean Theorem to write: x² + x² = hypotenuse²
Simplify: 2x² = hypotenuse²
Solve: hypotenuse = (√2)x

IMPORTANT: This tells us that the radius of the circle has length (√2)x

Now that we have one way to express the circle's radius, notice that we have another way to find the circle's radius.
Since x = the length of one side of a square, and since AB = √12 - √6, we can see that the SUM of those two lengths = the circle's radius

So, we can write: The circle's radius = x + √12 - √6

Now that we have two different ways to express the circle's radius, we can write: (√2)x = x + √12 - √6
Subtract x from both sides of the equation to get: (√2)x - x = √12 - √6
Factor both sides: x(√2 - 1) = √6(√2 - 1)
Divide both sides by (√2 - 1) to get: x = √6

Since x is the length of one side of a square, the area of ONE square = (√6)(√6) = 6
So the area of BOTH squares = 6 + 6 = 12

Answer: D

Cheers,
Brent