A
4
D^10
C B
QUANTITY A QUANTITY B
- The area of trapezoid ABCD 40 A B C D
Exercise Answers
Discrete Quantitative Questions
- D ADC is a right triangle with sides 30, 40, and 50 (this is a multiple of the
3-4-5 triangle). Thus AD = 30. If AD = 30, the area of ABCD = lw = 30(40) =
1,200. Since the area of MNOP = 1,200:
0 × (MN) = 1,200
MN = 60 - 60 Recall that the sum of the angles of a polygon = (n – 2)180, where n =
number of sides. Thus the sum of the angles in a six-sided polygon is (6 − 2)180
= 4(180) = 720. Use the unknown multiplier to represent the quantities of the
angles as 2x, 3 x, 4 x, 4x, 5x, and 6x. The sum of these quantities = 720:
2 x + 3 x + 4x + 4x + 5x + 6x = 720
24 x = 720
x = 30
The smallest angle = 2x = 2(30) = 60. - C Let l = length of the rectangle and w = width of the rectangle:
Area = lw = 36
Perimeter = 2(l + w) = 26
l + w = 13
Use substitution to solve for l: w = 13 – l. Substitute (13 – l) for w in the first
equation:
l(13 – l) = 36
13 l – l^2 = 36
l^2 – 13l + 36 = 0
(l – 4)(l – 9) = 0
l =4 or l = 9
If l = 4, then w = 9. If l = 9, then w = 4. Since you are told that l > w, l = 9.
The correct answer is C.
CHAPTER 13 ■ GEOMETRY 405
04-GRE-Test-2018_313-462.indd 405 12/05/17 12:05 pm