GRE_Math_Bible_eBook

Geometry 151

38.In the figure, the area of rectangle ABCD is 45. What is the area of the square EFGH ?

(A)20

(B)40

(C)50

(D)65

(E)70

The formula for the area of the rectangle is length width. Hence, the area of rectangle ABCD is AB AD = (x + 5)(x – 5) = x2 – 52. We are given that the area is 45, so x2 – 52 = 45. Solving the equation for x2 yields x2 = 45 + 25 = 70.

Now, the formula for the area of a square is side2. Hence, the area of square EFGH is EF2 = x2. Now, as shown earlier, x2 equals 70. The answer is (E).

39.In the figure, ABCD is a parallelogram, what is the value of b ?

(A)46

(B)48

(C)72

(D)84

(E)96

48°

D C

From the figure, A = 48 + 48 = 96. Since the sum of any two adjacent angles of a parallelogram equals 180°, we have A + b = 180 or 96 + b = 180. Solving for b yields b = 180 – 96 = 84. The answer is (D).

152GRE Math Bible

40.In the figure, AB and CD are the diameters of the circle. What is the value of x ?

(A)16°

(B)18°

(C)26°

(D)32°

(E)58°

x°

D

A

32°

O

B

C

OA and OC are radii of the circle and therefore equal. Hence, the angles opposite the two sides in ∆AOC are equal: C = A = 32° (from the figure). Now, summing the angles of the triangle to 180° yieldsA + C + AOC = 180 or 32 + 32 + AOC = 180. Solving the equation for AOC, we have AOC = 180 – (32 + 32) = 180 – 64 = 114.

Since BOD and AOC vertical angles, they are equal. Hence, we have BOD = AOC = 114.

Now, OD and OB are radii of the circle and therefore equal. Hence, the angles opposite the two sides in ∆BOD are equal: B = D. Now, summing the angles of the triangle to 180° yields B + D + BOD = 180 or D + D + BOD = 180 or 2D + 114 = 180. Solving the equation yields D = 32. Since D and angle x° are vertical angles, x also equals 32.

The answer is (D).

Geometry 153

41.In the figure, a, b, c, d, e, f, g, h, i, and j are chords of the circle. Which two chords are parallel to each other?

(A)a and f

(B)b and g

(C)c and h

(D)d and i

(E)e and j

The figure is not drawn to scale.

Two lines are parallel to each other when the angle swept between them is 180°. Use this property to determine the answer:

Choice (A): Angle swept between lines a and f equals Angle between a and b + Angle between b and c + Angle between c and d + Angle between d and e + Angle between e and f = 40 + 35 + 35 + 40 + 32 = 182 ≠ 180. Hence, a and f are not parallel. Reject.

Choice (B): Angle swept between lines b and g equals Angle between b and c + Angle between c and d + Angle between d and e + Angle between e and f + Angle between f and g = 35 + 35 + 40 + 32 + 36 = 178 ≠ 180. Hence, b and g are not parallel. Reject.

Choice (C): Angle swept between lines c and h equals Angle between c and d + Angle between d and e + Angle between e and f + Angle between f and g + Angle between g and h = 35 + 40 + 32 + 36 + 45 = 188 ≠ 180. Hence, c and h are not parallel. Reject.

Choice (D): Angle swept between lines d and i equals Angle between d and e + Angle between e and f + Angle between f and g + Angle between g and h + Angle between h and i = 40 + 32 + 36 + 45 + 32 = 185 ≠ 180. Hence, d and i are not parallel. Reject.

Choice (E): Angle swept between lines e and j equals Angle between e and f + Angle between f and g + Angle between g and h + Angle between h and i + Angle between i and j = 32 + 36 + 45 + 32 + 35 = 180. Hence, e and j are parallel. Correct.

The answer is (E).

154GRE Math Bible

42.A and B are centers of two circles that touch each other externally, as shown in the figure. What is the area of the circle whose diameter is AB ?

(A)4π

(B)25π/4

(C)9π

(D)16π

(E)25π

B A

2 3

Since the two circles touch each other, the distance between their centers, AB, equals the sum of the radii of the two circles, which is 2 + 3 = 5. Hence, the area of a circle with diameter AB (or radius = AB/2) is

π × radius2 =

π(AB/2)2 =

π(5/2)2 =

25π/4

The answer is (B).

Geometry 155

43.In the figure, which one of the following angles is the greatest?

(A)A

(B)B

(C)C

(D)D

(E)CDB

The figure is not drawn to scale.

In ∆ABD, AD = 4, AB = 3, and BD = 2 (from the figure). Forming the inequality relation for the side lengths yields AD > AB > BD. Since in a triangle, the angle opposite the longer side is greater, we have a similar inequality for the angles opposite the corresponding sides: ABD > BDA > A.

Similarly, in ∆BCD, DC = 6, BC = 5, and BD = 2. Forming the inequality for the side lengths yields

D C > B C > B D . Also the angles opposite the corresponding sides follow the relationDBC > CDB > C.

Now, summing the two known inequalities ABD > BDA > A and DBC > CDB > C yieldsABD + DBC > BDA + CDB > A + C; B > D > A + C. From this inequality, clearly B is

the greatest angle. Hence, the answer is (B).

A B C D

44.In the figure, A, B, C, and D are points on a line in that order. If AC = 5, BD = 10, and AD = 13, then what is the length of BC ?

(A)2

(B)8

(C)15

(D)18

(E)28

From the number line, we have the equations

AC = AB + BC

BD = BC + CD

AD = AB + BC + CD

Adding the first two equations yields

AC + BD = (AB + BC) + (BC + CD) = AB + 2BC + CD

Subtracting the third equation from this equation yields

156 GRE Math Bible

AC + BD – AD = AB + 2BC + CD – (AB + BC + CD) = BC

Hence, BC = AC + BD – AD = 5 + 10 – 13 = 2 (given AC = 5, BD = 10, and AD = 13). The answer is (A).

45.The side length of a square inscribed in a circle is 2. What is the area of the circle?

(A)π

(B)2π

(C)2π

(D)22π

(E)π2

diameter

The diagonal of a square inscribed in a circle is a diameter of the circle. The formula for the diagonal of a square is 2 side. Hence, the diameter of the circle inscribing the square of side length 2 is 2 2 =

radius2 = π(2)2 = 2π. The answer is (C).

46.In the figure shown, if A = 60°, B = C, and BC = 20, then AB =

(A)20

(B)102

(C)103

(D)202

(E)203

A

60°

Hence, A = B = C = 60°. Since the three angles of ∆ABC are equal, the three sides of the triangle must also be equal. Hence, AB = BC = 20. The answer is (A).

Geometry 157

47. Which of the following could be the four angles of a parallelogram?

(I)50°, 130°, 50°, 130°

(II)125°, 50°, 125°, 60°

(III)60°, 110°, 60°, 110°

(A)I only

(B)II only

(C)I and II only

(D)I and III only

(E)I, II and III

A quadrilateral is a parallelogram if it satisfies two conditions:

1)The opposite angles are equal.

2)The angles sum to 360°.

Now, in (I), opposite angles are equal (one pair of opposite angles equals 50°, and the other pair of opposite angles equals 130°). Also, all the angles sum to 360° (= 50° + 130° + 50° + 130° = 360°). Hence, (I) is true.

In (II), not all opposite angles are equal (50˚ ≠ 60˚). Hence, (II) is not a parallelogram.

In (III), the angle sum is not equal to 360° (60° + 110° + 60° + 110° = 340° ≠ 360°). Hence, (III) does not represent a quadrilateral.

Hence, only (I) is true, and the answer is (A).

48. In the figure, ABCD is a rectangle and E is a point on the side AB. If AB = 10 and AD = 5, what is the area of the shaded region in the figure?

(A)25

(B)30

(C)35

(D)40

(E)45

A E B

D C

The shaded region contains two triangles ∆AED and ∆EBC. Hence, the area of the shaded region equals the sum of the areas of these two triangles.

Now, the formula for the area of a triangle is 1/2 × base × height. Hence, the area of ∆AED = 1/2 × AE × AD, and the area of ∆EBC = 1/2 × EB × BC = 1/2 × EB × AD (since BC = AD).

Hence, the area of the shaded region equals 1/2 × AE × AD + 1/2 × EB × AD = 1/2 × AD(AE + EB) = 1/2 × AD × AB [since AB = AE + EB] = 1/2 × 5 × 10 = 25.

The answer is (A).

158GRE Math Bible

49.What is the area of the equilateral triangle if the base BC = 6?

(A)93

(B)183

(C)263

(D)303

(E)363

A

C D B

Since the side BC of the equilateral triangle measures 6 (given), the other side AC also measures 6. Since he altitude AD bisects the base (this is true in all equilateral or isosceles triangles), CD = BD = (1/2)BC =

1/2 × 6 = 3. Applying The Pythagorean Theorem to ∆ADC yields AD2 = AC2 – CD2 = 62 – 32 = 36 – 9 = 27, or AD = 33 . Hence, the area of the triangle is 1/2 × base × height = 1/2 × BC × AD = 1/2 × 6 × 33 =

93 . The answer is (A).

50. In the figure, ABCD is a square, and OB is a radius of the circle. If BC is a tangent to the circle and PC = 2, then what is the area of the square?

(A)16

(B)20

(C)25

(D)36

(E)42

P

D C

Side BC in the square is also a tangent to the circle shown. Since, at the point of tangency, a tangent is perpendicular to the radius of a circle, CBO is a right angle. Hence, ∆CBO is right angled; and by The Pythagorean Theorem, we have OC2 = OB2 + BC2.

Now, we have PC = 2. Hence, OC = OP + PC = Radius + PC = 3 + 2 = 5. Putting the results in the known equation OB2 + BC2 = OC2 yields 32 + BC2 = 52, or BC2 = 52 – 32 = 25 – 9 = 16.

Now, the area of the square is side2 = BC2 = 16. The answer is (A).

Geometry 159

51. In the figure shown, AB is a diameter of the circle and O is the center of the circle. If A = (3, 4), then what is the circumference of the circle?

(A)3

(B)4

(C)4π

(D)5π

(E)10π

y-axis

A

B

Since AB is a diameter of the circle, the point A(3, 4) is on the circle. The center of the circle is O(0, 0). Since the radius of a circle equals the distance between a point on the circle and the center, AO is a radius.

By the formula for the distance between two points, AO = (3− 0)2 + (4 − 0)2 = 32 + 42 = 25 = 5.

The formula for the circumference of a circle is 2π × radius. Hence, the circumference is 2π × 5 = 10π. The answer is (E).

52. A, B, and C are three unequal faces of a rectangular tank. The tank contains a certain amount of water. When the tank is based on the face A, the height of the water is half the height of the tank. The dimensions of the side B are 3 ft × 4 ft and the dimensions of side C are 4 ft × 5 ft. What is the measure of the height of the water in the tank in feet?

(A)2

(B)2.5

(C)3

(D)4

(E)5

Draw a rectangular tank as given (based on face A).

Mapping the corresponding given values to the sides of the faces yields

160 GRE Math Bible

The height of the tank is 4 ft, and the height of the water is half the height of the tank, which is 4/2 = 2 ft. The answer is (A).

53.In the figure, if line CE bisects ACB, then x =

(A)45

(B)50

(C)55

(D)65

(E)70

A

40°

D

E

90°

x°

B C

Summing the angles of ∆ABC to 180° yields

A + B + C = 180

40 + 90 + C = 180 by substituting known valuesC = 180 – (40 + 90) = 50

We are given that CE bisects ACB. Hence, ECB = ACE = one half of the full angle ACB, which equals 1/2 50 = 25. Now, summing the angles of ∆ECB to 180° yields

BEC + ECB + CBE = 180 x + 25 + 90 = 180

x = 180 – 25 – 90 = 65

The answer is (D).