Thus 5 + 3 + x + √x^2 + 16 = 20. Solve for x:
8 + x + √x^2 + 16 = 20
x + √x^2 + 16 = 12
√x^2 + 16 = 12 – x
Square both sides:
√x^2 + 162
= (12 – x)^2
x^2 + 16 = 144 – 24x + x^2
− x^2 − x^2
16 = 144 – 24x
↓
−128 = −24x
163 = x- C Since c is the hypotenuse of the right triangle, solve for a and b, and use the
Pythagorean theorem to solve for c. Since the area is 24,
a × b ×^12 = 24
ab= 48
Substitute (b + 2) for a:
(b + 2)b = 48
b^2 + 2b = 48
b^2 + 2b – 48 = 0
(b + 8)(b – 6) = 0
b = −8 or b = 6
Since b represents the length of a side, it must be positive. Thus b = 6. If b = 6,
then a = 8. The triangle is thus a 6-8-10 triangle. c = 10. - C Plug in values: Let XZ = 2√ 2. In this case, XY and XZ will each equal 2.
The area of triangle XYZ will be 2 × 2 ×^12 = 2. If XZ = 2√ 2 , then AC = √ 2.
If AC = √ 2 , then AB and BC = 1. The area of triangle ABC is thus
1 × 1 ×^12 =^12. 2 is 4 times as great as^12. - C Since AB and ED are parallel lines cut by a transversal, the measure of angle
BAD = the measure of angle EDC. Solve for EDC. Since ED = DC, angle ECD
= angle DEC. Thus angle DEC = 50. Thus 50 + 50 + EDC = 180 → EDC = 80.
Since EDC = 80, BAD = 80. - A Use the unknown multiplier:
3 x = smallest angle
4 x = middle angle
5 x = largest angle
CHAPTER 13 ■ GEOMETRY 39504-GRE-Test-2018_313-462.indd 395 12/05/17 12:04 pm