GRE_Math_Bible_eBook

Geometry 171

69. In the figure, ABCD is a rectangle, and the area of quadrilateral AFCE is equal to the area of ∆ABC. What is the value of x ?

(A)5

(B)6

(C)7

(D)12

(E)15

The formula for the area of a triangle is 1/2 × base × height. Hence, the areas of triangles AEC, ACF, and

ABC are

The area of ∆AEC = (1/2)(x)(BC) = (1/2)(x)(4) = 2x

The area of ∆ACF = (1/2)(3)(AD)

The area of ∆ABC = (1/2)(AB)(AD)

=(1/2)(DF + FC)(BC)

=(1/2)(5 + 3)(4)

=(1/2)(8)(4)

=16

Now, the area of the quadrilateral AECF equals (the area of ∆AEC) + (the area of ∆ACF) = 2x + 6, and the

area of ∆ABC = 16. Since we are given that the area of the triangle ABC equals the area of the quadrilateral AFCE, 2x + 6 = 16. Solving for x in this equation yields x = 5. The answer is (A).

172GRE Math Bible

70.In the figure, ABCD is a rectangle. Points E and F cut the sides BC and CD of the rectangle respectively such that EC = 3, FC = 4, and AD = 12, and the areas of the crossed and the shaded regions in the figure are equal. Which one of the following equals the perimeter of rectangle ABCD ?

(A)42

(B)50

(C)56

(D)64

(E)72

A B

The formula for the area of a triangle is 1/2 × base × height. Hence, the area of ∆AFC (the crossed region) equals 1/2 × FC × AD, and the area of ∆AEC (the shaded region) equals 1/2 × EC × AB. We are given that the two areas are equal. Hence, we have the equation

1/2 × EC × AB = 1/2 × FC × AD

(3/2)AB = 24

AB = (2/3)(24) = 16

Now, the perimeter of a rectangle is two times the sum of the lengths of any two adjacent sides of the rectangle. Hence, the perimeter of ABCD is 2(AB + AD) = 2(16 + 12) = 56. The answer is (C).

Geometry 173

71.What is the perimeter of ∆ABC shown in the figure?

(A)2 + 42

(B)4 + 22

(C)8

(D)4 + 42

(E)4 + 43

Equating the vertical angles at points A and C in the figure yields A = 2z and y = z. Summing the angles of the triangle to 180° yields

So, A = 2z = 2(45) = 90, B = y = z = 45 and C = z = 45. Hence, ∆ABC is a right triangle. Also, since angles C and B are equal (equal to 45), the sides opposite these two angles, AB and AC, must be equal. Since AC equals 2 (from the figure), AB also equals 2. Now, applying The Pythagorean Theorem to the triangle yields

=4 + 4

=8

Now, the perimeter of ∆ABC = AB + BC + CA = 2 + 22 + 2 = 4 + 22 . The answer is (B).

174GRE Math Bible

72.In the figure, what is the value of x ?

(A)90

(B)95

(C)100

(D)105

(E)115

x°

4

1

5

1

Let’s name the vertices of the figure as shown below

Now, in ∆ABC, AB = 3, BC = 4, and AC = 5. Hence, A C2 equals 52 = 25, and A B2 + BC2 equals 32 + 42 = 9 + 16 = 25. Hence, AC2 = 25 = AB2 + BC2. From this, it is clear that the triangle satisfies The Pythagorean Theorem. Hence, by the theorem, ABC must be a right triangle, AC the longest side the hypotenuse, and the angle opposite the side, B must be the right angle. Hence, B = 90°.

Now, in ∆EGH, from the figure, we have each side (EG, GH, and EH) of the triangle measures 1 unit. Hence, all the sides of the triangle are equal, and the triangle is equilateral. Since in any equilateral triangle, each angle measures 60°, we have E = 60°.

Now, summing the angles of the quadrilateral to 360° yields B + E + D + F = 360°. Hence, we have 90 + 60 + x + x = 360. Solving for x yields 2x = 210 or x = 210/2 = 105. The answer is (D).

Geometry 175

73.In the figure, ABCD is a rectangle. Points P, Q, R, S, and T cut side AB of the rectangle such that AP = 3, PQ = QR = RS = ST = 1. E is a point on AD such that AE = 3. Which one of the following line segments is parallel to the diagonal BD of the rectangle?

(A)EP

(B)EQ

(C)ER

(D)ES

(E)ET

3

9 E

D C

The figure is not drawn to scale.

Let X represent the point on AB that makes EX parallel to the diagonal BD. Then ∆AXE and ∆ABD must be similar because

Hence, the sets of corresponding angles are equal and the triangles are similar.

Hence, the corresponding sides of the triangles must be in the same ratio. This yields the following equations:

AE/AD = AX/AB

3/9 = AX/15

AX = 15 3/9 = 15 1/3 = 5

Hence, the point X is 5 units away from the point A on the side AB. Now,

AP = 3 ≠ AX, which equals 5.

AQ = AP + PQ = 3 + 1 = 4 ≠ AX, which equals 5.

AR = AP + PQ + QR = 3 + 1 + 1 = 5 = AX, which equals 5.

Hence, point R coincides with point X and therefore just like EX, ER must be parallel to BD. Hence, the answer is (C).

176GRE Math Bible

74.In the, figure A, B and C are points on the circle. What is the value of x ?

(A)45

(B)55

(C)60

(D)65

(E)70

B

x°

C

A

75° 35°

O

OA and OB are radii of the circle. Hence, angles opposite them in ∆AOB are equal: OAB = ABO. Summing the angles of ∆AOB to 180° yields OAB + ABO + AOB = 180 or 2ABO + 75° = 180 [sinceOAB = OBA); OBA = (180 – 75)/2 = 105/2].

Similarly, OB equals OC (radii of a circle are equal) and angles opposite them in ∆BOC are equal: OBC =BCO. Summing angles of the triangle to 180° yields OBC + BCO + 35 = 180 or 2 OBC + 35 = 180 [since OBC = BCO; OBC = (180 – 35)/2 = 145/2].

Now, since an angle made by a line is 180°, we have

x + ABO + OBC = 180

x + 105/2 + 145/2 = 180

x + 250/2 = 180

x + 125 = 180

x = 180 – 125 = 55

The answer is (B).

178GRE Math Bible

76.In the figure, ABCD and PQRS are rectangles inscribed in the circle shown in the figure. If AB = 5, AD = 3, and QR = 4, then what is the value of l ?

(A)3

(B)4

(C)5

(D)15

(E)32

PQRS is a rectangle inscribed in the circle. Hence, its diagonal PR must pass through the center of the circle. So, PR is a diameter of the circle.

Similarly, AC is a diagonal of the rectangle ABCD, which is also inscribed in the same circle. Hence, AC must also be a diameter of the circle. Since the diameters of a circle are equal, PR = AC.

Applying The Pythagorean Theorem to ∆ABC yields AC2 = AB2 + BC2 = 52 + 32 = 25 + 9 = 34. Hence, PR2 also equals 34. Now, applying The Pythagorean Theorem to ∆PQR yields

PQ2 + QR2 = PR2 l2 + 42 = 34

l2 + 16 = 34 l2 = 18

l = 18 l = 32

The answer is (E)

180GRE Math Bible

78.Which one of the following relations is true regarding the angles of the quadrilateral shown in the figure?

(A)A = C

(B)B > D

(C)A < C

(D)B = D

(E)A = B

Joining the opposite vertices B and D on the quadrilateral yields the following figure:

Since the angle opposite the longer side in a triangle is greater, we have

AD (= 4) > AB(= 3) (from the figure). Hence, ABD > BDA and

CD (= 6) > BC(= 5) (from the figure). Hence, DBC > CDB.

Adding the two known inequalities ABD > BDA and DBC > CDB yields

ABD + DBC > BDA + CDB