McGraw-Hill Education GRE 2019

Since the area of the rectangle is 48, we know that lw = 48. Next, note that

the diagonal of a rectangle is equivalent to the hypotenuse of a right triangle

whose legs are the length and width of the rectangle. We can thus create the

formula: l^2 + w^2 = 10^2 = 100.

Now, let’s combine these two equations to solve for l + w.

Note that (l + w)^2 = l^2 + 2lw + w^2. Substitute 100 for l^2 + w^2 , and 48 for lw to

arrive at:

(l + w)^2 = 100 + 2(48) = 196.

Square-root both sides:

l + w = 14

Finally, substitute 14 for (l + w) in the perimeter formula and arrive at:

2(l + w) = 2(14) = 28.

Quantitative Comparison Questions

1. B Plug in numbers. Let the area of square A = 16. In this case, the area of
   square B = 8. Each side of square A has a length of √ 16 = 4. Each side of square
   B has a length of √ 8 = 2√ 2. The ratio of 4 to 2√ 2 is less than 2.


2. A Interior angles of a parallelogram must add up to 180. Thus the sum of the
   measures of angles ACD and CAB is 180. If the measure of angle ACD is less
   than 60, then the measure of angle CAB must be greater than 120.


3. A Recall that the sum of the interior angles of a regular polygon can be
   arrived at with the following formula: (n – 2)180, where n = number of sides.
   Since an octagon has 8 sides, the sum of its angles is (8 − 2)180 = 1,080. The
   measure of each angle is 1,080 8 = 135. Quantity A is greater.


4. B The area of a parallelogram can be determined by using the formula
   base × height. The base of the parallelogram is 8. To determine the relationship
   between the quantities, you must determine the relationship between the
   height and 6. Note that when you drop the height, a right triangle is formed
   in which the height of the parallelogram is one of the legs, and side AB is the
   hypotenuse.

A D

C

6

8

B

Since each leg of a right triangle must be smaller than the hypotenuse, the

height of the parallelogram must be less than 6. The area of the parallelogram

is thus less than 48.

5. B Instead of calculating, recognize that if BC were extended to have a length
   of 10, the area of the resulting rectangle would be 40. Since BC is less than 10,
   the area of the trapezoid must be less than 40.

CHAPTER 13 ■ GEOMETRY 407

04-GRE-Test-2018_313-462.indd 407 12/05/17 12:05 pm