Bunuel wrote:
In the above diagram, a square is inscribed in a circle, which is inscribed in another square. Which is larger, the green region or the yellow region?
(1) The area of the larger square is 64.
(2) The area of the green region is 4.
Kudos for a correct solution.Attachment:
square-in-circle.JPG
GROCKIT OFFICIAL SOLUTION:Let’s start with
Statement 1.A(large square) = 64
s(large square) = 8
diameter = 8
r = 4
First, recognize that the figure is symmetrical. So while we may not explicitly be given an angle to find the sector area (not drawn) in which the yellow region resides, we do know its measure. The diagonals of a square intersect at a right angle, so we can deduce that the sector including the yellow region is 1/4 of the area of the circle. Since we know the radius…
A(large sector) = 1/4 * 16π = 4π
To find the yellow region itself, we must subtract the imaginary (not drawn) triangle from 4π. (Note that this imaginary triangle will be twice the green triangle.)
A(imaginary triangle) = 1/2 * r * r = 1/2 * 4 * 4 = 8
A(Yellow Region) = 4π – 8
The area of the green region can be found in two ways. Either we can see that it’s simply one-half of the 8 we just found, OR we can find both sides of the green triangle with the common 45-45-90 1:1:√2 formula. With a hypotenuse of 4, we derive 2√2 for each side, which yields an area of 4.
Which is greater, 4π – 8 or 4?
4π – 8. Sufficient. (We can save the calculations on the GMAT for DS questions, but it’s still good to go through it to practice for PS questions involving similar calculations.)
What about
Statement 2?
If we know the area of the green triangle equals 4, and that it is an isosceles right triangle, then we can set up a simple equation to find its sides, which can be denoted x:
1/2 * x² = 4
x² = 8
x = 2√2
If x = 2√2, then the hypotenuse (r) = 2√2 * √2 = 4. From here, we follow the same logic as we did for Statement 1, and determine Statement 2 is sufficient.
The answer choice is D. _________________