- B The length and width of the rectangle are the legs of the 30-60-90 triangle
ABC. Let the length of AB = x. The length of BC is thus x√3. Thus:
x × x√3 = 36√ 3
x^2 = 36
x = 6
The hypotenuse of a 30-60-90 triangle is double the shorter leg. Thus AC = 12. - D Since this question only provides relationships and no values, plugging in is
a good strategy. First, choose a value for the area of the square. If AB = 2, then
the area of the square = 2^2 = 4. If the area of the square is 4, then the area of
the rectangle must be 8. Since lw = AC × AE = 8, AE = 4. If AB = 2, then
BE = 2. The ratio of AC to BE is^22 =^11. - D To determine the area of the shaded region, subtract the area of the triangle
from the area of the square. You are told that the area of the square is 64, so
you must solve for the area of the triangle. Since the area of the square is 64,
the length of each side must be 8. If the length of each side is 8, then FC = 4
and EC = 4. Since ECF is a right triangle, its area is^12 × 4 × 4 = 8. The area of
the shaded region is thus 64 − 8 = 56. - 36 To maximize the area of a rectangle, make the length and width as close as
possible. You are told that
2(l + w) = 24
l + w = 12
Let l = w = 6. The resulting shape will be a square. The area of the square is
62 = 36. - C Identify shapes within the figure that you are familiar with. The shape can
be redrawn as a rectangle and right triangle, as indicated in the figure below:
AB
D C4 5553
To find the area of the figure, add the area of the resulting rectangle and
triangle. The area of the rectangle is l × w = 5 × 4 = 20. The area of the triangle
is^12 × b × h =^12 × 3 × 4 = 6. The area of the figure is 20 + 6 = 26. The correct
answer is C.- 35 To maximize the area of a triangle, make the given sides perpendicular.
When you do so, you will have a right triangle. Since 7 and 10 are the leg
lengths of this right triangle, the area of the triangle is^12 × 7 × 10 = 35. - 28 You are asked to solve for 2(l + w), where l = length of the rectangle and
w = width of the rectangle: area = lw = 48.
The formula for the perimeter of a rectangle is 2(l + w), where l = the
rectangle’s length and w = the rectangle’s width. Thus, our goal in this
question is to solve for l + w.
406 PART 4 ■ MATH REVIEW04-GRE-Test-2018_313-462.indd 406 12/05/17 12:05 pm