GRE_Math_Bible_eBook

Geometry 141

Now, when two lines (here p and q) cut by a transversal (here l) are parallel, we have

(I)Corresponding angles are equal: No corresponding angles are listed in the figure.

(II)Alternate interior angles are equal: b = c. In choice (E).

(III)Alternate exterior angles are equal: a = d. Not listed in any answer-choice.

(IV) Interior angles are supplementary.

(V)Exterior angles are supplementary. The answer is (E).

20.A, B, C, and D are points on a line such that point B bisects line AC and point A bisects line CD. What is the ratio of AB to CD ?

(A)1/4

(B)1/3

(C)1/2

(D)2/3

(E)3/4

Drawing the figure given in the question, we get

D A B C

AB = BC

CA = DA

Suppose AB equal 1 unit. Since point B bisects line segment AC, AB equals half AC. Hence, AC equals twice AB = 2(AB) = 2(1 unit) = 2 units. Again, since point A bisects line segment DC, DC equals twice AC = 2(AC) = 2(2 units) = 4 units. Hence, AB/DC = 1 unit/4 units = 1/4. The answer is (A).

142GRE Math Bible

21.If A, B, C, D, and E are points in a plane such that line CD bisects ACB and line CB bisects right angle ACE, then DCE =

(A)22.5°

(B)45°

(C)57.5°

(D)67.5°

(E)72.5°

Drawing the figure given in the question yields

E

B

D

A

We are given that CB bisects the right-angle ACE. Hence, ACB = BCE = ACE/2 = 90°/2 = 45°. Also, since CD bisects ACB, ACD = DCB = ACB/2 = 45°/2 = 22.5°. Now, DCE =DCB + BCE = 22.5° + 45° = 67.5°. The answer is (D).

22. From the figure, which of the following must be true?

(I)x + y = 90

(II)x is 35 units greater than y

(III)x is 35 units less than y

(A)I only

(B)II only

(C)III only

(D)I and II only

(E)I and III only

l

35°

x° y°

m

Angle x is an exterior angle of the triangle and therefore equals the sum of the remote interior angles, 35 and y. That is, x = y + 35. This equation says that x is 35 units greater than y. So, (II) is true and (III) is false. Now, if x is an obtuse angle (x > 90), then x + y is greater than 90. Hence, x + y need not equal 90. So,

(I) is not necessarily true. The answer is (B).

Geometry 143

23.In the figure, triangles ABC and ABD are right triangles. What is the value of x ?

(A)20

(B)30

(C)50

(D)70

(E)90

B

In triangle ABC, A is 20°. Hence, the right angle in ∆ABC is either ABC or BCA. If ABC is the right angle, then ABD, of which ABC is a part, would be greater than the right angle (90°) and ABD would be

an obtuse angle, not a right triangle. So, ABC is not 90° and therefore BCA must be a right angle. Since AD is a line, ACB + BCD = 180°. Solving for BCD yields BCD = 180 – 90 = 90. Hence, ∆BCD is

also a right triangle, with right angle at C. Since there can be only one right angle in a triangle, D is not a right angle. But, we are given that ∆ABD is right angled, and from the figure A equals 20°, which is not a right angle. Hence, the remaining angle ABD is right angled. Now, since the sum of the angles in a triangle is 180°, in ∆ABC, we have 20 + 90 + x = 180. Solving for x yields x = 70. The answer is (D).

24.In the figure, ABC is a right triangle. What is the value of y ?

(A)20

(B)30

(C)50

(D)70

(E)90

B

y°

In the given right triangle, ∆ABC, A is 20°. Hence, A is not the right angle in the triangle. Hence, either of the other two angles, C or B, must be right angled.

144GRE Math Bible

Now, BDA (= ABC = y°, from the figure) is an exterior angle to ∆BCD and therefore equals the sum of the remote interior angles C and DBC. Clearly, the sum is larger than C and therefore if C is a right angle, BDA (= ABC) must be larger than a right angle, so BDA, hence, ABC must be obtuse. But a

triangle cannot accommodate a right angle and an obtuse angle simultaneously because the angle sum of the triangle would be greater than 180°. So, C is not a right angle and therefore the other angle, B, is a right angle. Hence, y° = B = 90°. The answer is (E).

25.The following are the measures of the sides of five different triangles. Which one of them represents a right triangle?

(A)3 , 4 , 5

(B)1, 5, 4

(C)7, 3, 4

(D)3 , 7 , 4

(E)4, 8, 10

A right triangle must satisfy The Pythagorean Theorem: the square of the longest side of the triangle is equal to the sum of the squares of the other two sides. Hence, we look for the answer-choice that satisfies this theorem:

Choice (A): (5)2 ≠ (3)2 + (4)2 . Reject.

Choice (B): 52 ≠ 12 + 42. Reject.

Choice (C): 72 ≠ 32 + 42. Reject.

Choice (D): (7)2 = (3)2 + (4)2. Correct.

Choice (E): 102 ≠ 82 + 42. Reject.

The answer is (D).

26. ∆ABC is a right-angled isosceles triangle, and B is the right angle in the triangle. If AC measures 72 , then which one of the following would equal the lengths of AB and BC, respectively?

(A)7, 7

(B)9, 9

(C)10, 10

(D)11, 12

(E)7, 12

A

B C

In a right-angled isosceles triangle, the sides of the right angle are equal. Now, in the given right-angled isosceles triangle ∆ABC, B is given to be the right angle. Hence, the sides of the angle, AB and BC, are equal. Applying The Pythagorean Theorem to the triangle yields

Hence, AB = BC = 7. The answer is (A).

Geometry 145

27.In the figure, if AB = 8, BC = 6, AC = 10 and CD = 9, then AD =

(A)12

(B)13

(C)15

(D)17

(E)24

A

8

The lengths of the three sides of ∆ABC are AB = 8, BC = 6, and AC = 10. The three sides satisfy The Pythagorean Theorem: AC2 = BC2 + AB2 (102 = 62 + 82). Hence, triangle ABC is right angled and B, the angle opposite the longest side AC (hypotenuse), is a right angle. Now, from the figure, this angle is part of ∆ADB, so ∆ADB is also right angled. Applying The Pythagorean Theorem to the triangle yields

=82 + (6 + 9)2

=82 + 152

=289

=172

The answer is (D).

28.The average length of the sides of ∆ABC is 12. What is the perimeter of ∆ABC ?

(A)4

(B)6

(C)12

(D)24

(E)36

A

60°

B C

The perimeter of a triangle equals the sum of the lengths of the sides of the triangle.

The average length of the sides of the triangle equals (the sum of the lengths of the three sides)/3.

Hence, the perimeter of a triangle equals three times the average length of the sides of the triangle.

Now, we are given that the average length of the triangle is 12. Hence, the perimeter of the triangle equals 3 12 = 36. The answer is (E).

146GRE Math Bible

29.Which one of the following is true regarding the triangle shown in the figure?

(A)x > y > z

(B)x < y < z

(C)x = y = z

(D)2x = 3y/2 = z

(E)x/2 = 2y/3 = z

From the figure, we have the following inequality between the sides of the triangle: BC (= 2) > AB (= 1 1/2) > AC (= 1). In a triangle, the longer the side, the bigger the angle opposite it. Hence, we have the following inequality between the angles of the triangle: A > C > B. Replacing the angles in the inequality with their respective vertical angles in the figure yields x > y > z. The answer is (A).

30.In the figure, ABCD is a rectangle, and the area of ∆ACE is 10. What is the area of the rectangle?

(A)18

(B)22.5

(C)36

(D)44

(E)45

A B

4

E

5

D C

The formula for the area of a triangle is 1/2 × base × height. Hence, the area of ∆ACE (which is given to equal 10) is 1/2 × CE × AB. Hence, we have

Now, the formula for the area of a rectangle is length × width. Hence, the area of the rectangle ABCD =

BC × AB

The answer is (C).

Geometry 147

31.In the figure, what is the area of ∆ABC ?

(A)2

(B)2

(C)1

(D)

1

2

(E) 1/2

2z°

Equating vertical angles at point B in the figure yields y = z and A = 2 z. So, the triangle is isosceles (B =C = y). Now, the angle sum of a triangle is 180°, so 2z + z + z = 180. Solving for z yields z = 180/4 = 45. Hence, we have 2z = 2 45 = 90. So, ∆ABC is a right-angled isosceles triangle with right angle at vertex A and equal angles at B and C (both equaling 45°). Since sides opposite equal angles in a triangle are equal, sides AB and AC are equal and each is 1 unit (given that AB is 1 unit). Now, the area of the right triangle ABC is 1/2 AB AC = 1/2 1 1 = 1/2. The answer is (E).

32. In the figure, O is the center of the circle. Which one of the following must be true about the perimeter of the triangle shown?

(A)Always less than 10

(B)Always greater than 40

(C)Always greater than 30

(D)Always less than 30

(E)Less than 40 and greater than 20

O

10

B

A

In ∆AOB, OA and OB are radii of the circle. Hence, both equal 10 (since OA = 10 in the figure).

Now, the perimeter of a triangle equals the sum of the lengths of the sides of the triangle. Hence, perimeter of ∆AOB = OA + OB + AB = 10 + 10 + AB = 20 + AB.

148GRE Math Bible

In a triangle, the length of any side is less than the sum of the lengths of the other two sides and is greater than their difference. So, AB is less than AO + OB (= 10 + 10 = 20) and is greater than 10 – 10 = 0. So, we have 0 < AB < 20. Adding 20 to each part of this inequality yields 20 < 20 + AB < 20 + 20, or 20 < Perimeter < 40. Hence, the answer is (E).

33. Which of the following must be true?

(I)The area of triangle P.

(II)The area of triangle Q.

(III)The area of triangle R.

(A)I = II = III

(B)I < II < III

(C)I > II > III

(D)III < I < II

(E)III > I > II

In the figure, triangle P is equilateral, with each side measuring 10 units. So, as in any equilateral triangle, the altitude (x here) is shorter than any of the other sides of the triangle. Hence, x is less than 10. Now,

I = Area of Triangle P

=1/2 × base × height

=1/2 × 10 × x

=5x and this is less than 50, since x is less than 10

Triangle Q has both base and altitude measuring 10 units, and the area of the triangle is 1/2 × base × height = 1/2 × 10 × 10 = 50. So, II = 50. We have one more detail to pick up: y, being the hypotenuse in the right triangle in figure Q, is greater than any other side of the triangle. Hence, y is greater than 10, the measure of one leg of the right triangle.

Triangle R has both the base and the altitude measuring y units. Hence,

II= Area of the Triangle R

=1/2 × base × height

=1/2 × y × y

=(1/2)y2, and this is greater than 1/2 × 102 (since y > 10), which equals 50

Summarizing, the three results I < 50, II = 50, and III > 50 into a single inequality yields I < II < III. The answer is (B).

Geometry 149

34.In the figure, ABCD is a parallelogram. Which one of the following is true?

(A)x < y

(B)x > q

(C)x > p

(D)y > p

(E)y > q

Since ABCD is a parallelogram, opposite sides are equal. So, x = q and y = p. Now, line BD is a transversal cutting opposite sides AB and DC in the parallelogram. So, the alternate interior angles ABD and BDC both equal 31°. Hence, in ∆ABD, B (which equals 31°) is greater than D (which equals 30°, from the figure). Since the sides opposite greater angles in a triangle are greater, we have x > y. But, y = p (we know). Hence, x > p, and the answer is (C).

35. In the figure, the areas of parallelograms EBFD and AECF are 3 and 2, respectively. What is the area of rectangle ABCD ?

(A)3

(B)4

(C)5

(D)43

(E)7

The area of the rectangle ABCD is length × width = DC × AD.

The area of the parallelogram AECF is FC × AD (base × height). Also, the area of parallelogram EBFD is D F × A D . Now, summing the areas of the two parallelograms AECF and EBFD yields FC × AD + DF × AD = (FC + DF)(AD) = DC × AD = the area of the rectangle ABCD. Hence, the area of rectangle ABCD equals the sum of areas of the two parallelograms, which is 2 + 3 = 5. The answer is (C).

150GRE Math Bible

36.In the figure, the area of rectangle ABCD is 100. What is the area of the square EFGH ?

(A)256

(B)275

(C)309

(D)399

(E)401

The area of the rectangle ABCD, length × width, is (x + 3)(x + 2) = x2 + 5x + 6 = 100 (given the area of the rectangle ABCD equals 100). Now, subtracting 6 from both sides yields x2 + 5x = 94.

The area of square EFGH is (2x + 5)2 = 4x 2 + 20x + 25 = 4(x2 + 5x) + 25 = 4(94) + 25 = 401. The answer is

(E).

37. In the figure, the area of rectangle EFGH is 3 units greater than the area of rectangle ABCD. What is the value of ab if a + b = 8?

(A)9

(B)12

(C)15

(D)18

(E)21

(x + 6)(x + 2) = x2 + 8x + 12, and the area of the rectangle EFGH is (x + a)(x + b) = x2 + (a + b)x + ab = x2 + 8x + ab (given that a + b = 8). Now, we are given that the area of the rectangle EFGH is 3 units greater than the area of the rectangle ABCD. Hence, we have