GRE_Math_Bible_eBook
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Geometry 161
54.In the figure, P =
(A)15°
(B)30°
(C)35°
(D)40°
(E)50°
B
A
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105° |
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115° |
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D |
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Since the angle in a line is 180°, we have PAD + DAB = 180, or PAD + 105 = 180 (from the figure,DAB = 105°). Solving for PAD yields PAD = 75.
Since the angle in a line is 180°, we have ADP + ADC = 180, or ADP + 115 = 180 (from the figure,ADC = 115°). Solving for ADP yields ADP = 65.
Now, summing the angles of the triangle PAD to 180° yields
P + PAD + ADP = 180
P + 75 + 65 = 180
P = 40
The answer is (D).
55. The length of a rectangular banner is 3 feet 2 inches, and the width is 2 feet 4 inches. Which one of the following equals the area of the banner?
(A)5 sq. feet
(B)5 1/2 sq. feet
(C)6 1/3 sq. feet
(D)7 sq. feet
(E)7 7/18 sq. feet
First, let’s convert all the measurements to feet. There are 12 inches in a foot, so 2 inches equals 2/12 = 1/6 feet, and 4 inches equals 4/12 = 1/3 feet.
Hence, 3 feet 2 inches equals 3 1/6 feet, and 2 feet 4 inches equals 2 1/3 feet.
Now, the area of a rectangle is length width. Hence,
Area = 3 1/6 2 1/3 = 19/6 7/3 = 133/18 = 7 7/18
The answer is (E).
162GRE Math Bible
56.In the figure, if A = C, then x =
(A)30
(B)60
(C)90
(D)150
(E)180
A
y°/2
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O |
C |
y° |
D
B
The dimensions in the figure may be different from what they appear to be.
From the figure, A = y/2. Since we are given that A = C, C also equals y/2. Also, in figure, O = x = y (vertical angles are equal). Summing the angles of ∆AOC to 180 yields
A + O + C = 180 |
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y/2 + x + y/2 = 180 |
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x/2 + x + x/2 = 180 |
(since x = y) |
2x = 180 |
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x = 180/2 = 90 |
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The answer is (C). |
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57.Column A
The area of the rectangle when its length is decreased by 4 inches and its width is not changed
The width and length of a rectangle are 6 inches and 12 inches, respectively.
Column B
The area of the rectangle when its width is decreased by 4 inches and its length is not changed
The formula for the area of a rectangle is length width. The original length and width of the rectangle are 12 inches and 6 inches, respectively.
Column A: After the length of the rectangle is decreased by 4 inches, the new length becomes 12 – 4 = 8 inches and the new area becomes 8 6 = 48.
Column B: After the width of the rectangle is decreased by 4 inches, the new width becomes 6 – 4 = 2 inches and the new area becomes 12 2 = 24.
Since Column A is greater than Column B, the answer is (A).
Geometry 163
58. |
Column A |
|a| is the distance point a is from |
Column B |
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the origin on the number line. x ≠ |
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|x| + |–2| |
0. |
|x – 2| |
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Suppose x = –2. Then Column A equals |x| + |–2| = |–2| + |–2| = 2 + 2 = 4, and Column B equals |x – 2| = |–2 – 2| = |–4| = 4. Here, Column A equals Column B.
Now, suppose x = 2. Then Column A equals |x| + |–2| = |2| + |–2| = 2 + 2 = 4, and Column B equals |x – 2| = |2 – 2| = |0| = 0. Here, Column A is greater than Column B.
This is a double case, and therefore the answer is (D).
59. |
Column A |
A regular polygon of 24 sides is |
Column B |
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The perimeter of the polygon |
inscribed in a circle. |
The circumference of the circle |
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Each side of the regular polygon represents a chord of the circle, and each chord subtends a unique arc on the circle. The length of a chord is always less than the length of the arc it subtends. Hence, the sum of the lengths of the 24 sides (which are the 24 chords of the circle) is less than the sum of the lengths of the arcs that each one of the cords subtends. Hence, the perimeter of the polygon is less than the circumference of the circle, and therefore Column A is less than Column B. The answer is (B).
Hard
60.In the figure, lines l and k are parallel. Which one of the following must be true?
(A)a < b
(B)a ≤ b
(C)a = b
(D)a ≥ b
(E)a > b
(b + 30)° |
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O a°
b° |
k |
Draw line m passing through O and parallel to both line l and line k.
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(b + 30)° |
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a° |
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x° |
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b° |
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Now, observe that angle x is only part of angle a, and x = b since they are alternate interior angles. Since x is only part of angle a, a > x and a > b. The answer is (E).
164GRE Math Bible
61.In the figure, lines l and k are parallel. If a is an acute angle, then which one of the following must be true?
(A)b > 10
(B)b > 15
(C)b < 20
(D)b < 30
(E)b > 45
(b + 30)° |
l |
O a°
b° |
k |
Draw a line parallel to both of the lines l and k and passing through O.
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(b + 30)° |
l |
O |
p° |
m |
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q° |
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b° |
k |
We are given that a is an acute angle. Hence, a < 90. Since angles p and b + 30 are alternate interior angles, they are equal. Hence, p = b + 30. Similarly, angles q and b are alternate interior angles. Hence, q = b. Since angle a is the sum of its sub-angles p and q, a = p + q = (b + 30) + b = 2b + 30. Solving this equation for b yields b = (a – 30)/2 = a/2 – 15. Now, dividing both sides of the inequality a < 90 by 2 yields a/2 < 45. Also, subtracting 15 from both sides of the inequality yields a/2 – 15 < 30. Since a/2 – 15 = b, we have b < 30. The answer is (D).
62. The diagonal length of a square is 14.1 sq. units. What is the area of the square, rounded to the nearest integer? (
2 is approximately 1.41.)
(A)96
(B)97
(C)98
(D)99
(E)100
If a is the length of a side of the square, then a diagonal divides the square into two congruent (equal) right triangles. Applying The Pythagorean Theorem to either triangle yields diagonal2 = side2 + side2 = a2 + a2 =
2a2. Taking the square root of both sides of this equation yields diagonal = |
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diagonal length is 14.1. Hence, a 2 |
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= 198.81 = 99.4. The number 99.4 is nearest to 99. Hence, the answer is (D). |
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Geometry 165
14.1
Note: If you had approximated 2 with 10, you would have mistakenly gotten 100 and would have answered (E). Approximation is the culprit. Prefer doing it in the last.
63. In the figure, ABC and ADC are right triangles. Which of the following could be the lengths of AD and DC, respectively?
(I)
3 and 
4
(II)4 and 6
(III) 1 and 
24 (IV) 1 and 
26
(A)I and II only
(B)II and III only
(C)III and IV only
(D)IV and I only
(E)I, II and III only
A
4
B 
D
3
C
Figure not drawn to scale
From the figure, we have that B is a right angle in triangle yields AC2 = AB2 + BC2 = 42 + 32 = 25. Hence,
∆ABC. Applying The Pythagorean Theorem to the AC = 
25 = 5.
Now, we are given that ∆ADC is a right-angled triangle. But, we are not given which one of the three angles of the triangle is right-angled. We have two possibilities: either the common side of the two triangles, AC, is the hypotenuse of the triangle, or it is not.
In the case AC is the hypotenuse of the triangle, we have by The Pythagorean Theorem,
AC2 = AD2 + DC2 52 = AD2 + DC2
This equation is satisfied by III since 52 = 12 + (
24)2 . Hence, III is possible.
In the case AC is not the hypotenuse of the triangle and, say, DC is the hypotenuse, then by applying The Pythagorean Theorem to the triangle, we have
AD2 + AC2 = DC2
AD2 + 52 = DC2
This equation is satisfied by IV: 52 +12 = (
26)2.
Hence, we conclude that III and IV are possible. The two are available in choice (C). Hence, the answer is
(C).
166GRE Math Bible
64.In the figure, what is the value of x ?
(A)15
(B)30
(C)45
(D)60
(E)75
x°
x°
x°
x°
Let’s name the vertices as shown in the figure
x°
A
x°
G
x° C
B
E
F
D
x°
Let’s start evaluating the unknown angles of the triangles in the figure from left most located triangle through the right most located triangle. Then the value of x can be derived by summing the angles of right most located triangle to 180°. This is done as follows:
In the first triangle from the left: ∆ABC, we have from the figure that B = x° and A = x° (vertical angles are equal). Summing the angles of the triangle to 180° yields x + x + C = 180. Solving the equation forC yields C = 180 – 2x.
Geometry 167
In the second triangle from equal), and we have C [in Now, summing the angles ofE yields E = x.
the left: ∆CED, we have from the figure that D = x° (vertical angles are ∆CED] = C in ∆ACB (They are vertical angles) = 180 – 2x (Known result). the triangle to 180° yields (180 – 2x) + x + E = 180. Solving the equation for
In the third triangle from the left: ∆GFE, we have from the figure that G = x° (vertical angles are equal) and E [in ∆GFE] = E in ∆CED (Vertical angles are equal) = x° (Known result). Now, we also haveF = 90° (From the figure). Now, summing the three angles of the triangle to 180° yields x + x + 90 = 180. Solving for x yields 2x = 90 or x = 45. Hence, the answer is (C).
65.In the figure, what is the value of x ?
(A)10°
(B)30°
(C)45°
(D)60°
(E)75°
5
5
5√2
5
5
5
x°
Let’s name the vertices of the figure as shown
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5√2 |
5 |
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D
x°
In ∆ABC, all the sides are equal (each equals 5). Hence, the triangle is equilateral, and A = B = C = 60°.
168GRE Math Bible
Also, ∆EFG is a right-angled isosceles triangle (since EG = FG = 5), and The Pythagorean Theorem is satisfied (EG2 + FG2 = 52 + 52 = 50 = EF2 = (5
2)2. Hence, E = F = 45° (Angles opposite equal sides
of an isosceles right triangle measure 45° each). Now, in ∆CED, we have:
D = x° |
vertical angles, from the figure |
C = C in ∆ABC |
vertical angles, from the figure |
= 60° |
we know |
E in ∆CED = E in ∆EFG |
vertical angles |
= 45° |
we know |
Now, summing these three angles of ∆CED to 180° yields 60 + 45 + x = 180. Solving this equation for x yields x = 75. The answer is (E).
66.If ABCD is a square and the area of ∆AFG is 10, then what is the area of ∆AEC ?
(A)5
(B)
10
2
(C)
10
3
(D)10
(E)20
A B
F
4
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The formula for the area of a triangle is 1/2 × base × height. By the formula, the area of ∆AFG (which is given to be 10) is 1/2 × FG × AB. Hence, we have
1/2 × 4 × AB = 10 |
given that the area of ∆AFG = 10 |
AB = 5 |
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Also, by the same formula, the area of ∆AEC is |
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1/2 × EC × DA |
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= 1/2 × 2 × DA |
from the figure, EC = 2 units |
= 1/2 × 2 × AB |
ABCD is a square. Hence, side DA = side AB |
= 1/2 × 2 × 5 |
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= 5 |
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The answer is (A). |
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Geometry 169
67. In the figure, ABCD is a rectangle, and F and E are points on AB and BC, respectively. The area of ∆DFB is 9 and the area of ∆BED is 24. What is the perimeter of the rectangle?
(A)18
(B)23
(C)30
(D)42
(E)48
A |
F |
2 |
B |
4
E
D |
C |
The formula for the area of a triangle is 1/2 × base × height. Hence, the area of ∆DFB is 1/2 × FB × AD. We are given that the area of ∆DFB is 9. Hence, we have
1/2 × FB × AD = 9
1/2 × 2 × AD = 9
AD = 9
Similarly, the area of ∆BED is 1/2 × BE × DC, and we are given that the area of the triangle is 24. Hence, we have
1/2 × BE × DC = 24
1/2 × 4 × DC = 24 |
from the figure, BE = 4 |
DC = 12
Now, the formula for the perimeter of a rectangle is
2 × (the sum of the lengths of any two adjacent sides of the rectangle)
Hence, the perimeter of the rectangle ABCD = 2(AD + DC) = 2(9 + 12) = 2 × 21 = 42. The answer is (D).
170GRE Math Bible
68.In the figure, ABCD is a rectangle. The area of quadrilateral EBFD is one-half the area of the rectangle ABCD. Which one of the following is the value of AD ?
(A)5
(B)6
(C)7
(D)12
(E)15
A B
4
F
E
3
D C
From the figure, it is clear that the area of quadrilateral EBFD equals the sum of the areas of the triangles ∆EBD and ∆DBF. Hence, the area of the quadrilateral EBFD
= area of ∆EBD + area of ∆DBF |
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= 1/2 × ED × AB + 1/2 × BF × DC |
area of a triangle equals 1/2 × base × height |
= 1/2 × 3 × AB + 1/2 × 4 × DC |
from the figure, ED = 3 and BF = 4 |
= (3/2)AB + 2DC |
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= (3/2)AB + 2AB |
opposite sides AB and DC must be equal |
= (7/2)AB |
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Now, the formula for the area of a rectangle is length × width. Hence, the area of the rectangle ABCD equals AD × AB. Since we are given that the area of quadrilateral EBFD is half the area of the rectangle ABCD, we have
12 (AD × AB) = 72 AB
AD × AB = 7AB
AD = 7 |
(by canceling AB from both sides) |
The answer is (C).