Brent@GMATPrepNow wrote: ↑Wed May 13, 2020 8:05 am
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
