GRE_Math_Bible_eBook

Geometry 131

Answers and Solutions to Problem Set I

Easy

1.If p is the circumference of the circle Q and the area of the circle is 25π, what is the value of p ?

(A)25

(B)10π

(C)35

(D)15π

(E)25π

The area of the circle Q = π radius2 = 25π. Solving the equation for the radius yields radius = 25ππ = 5. Now, the circumference of the circle Q = 2π radius = 2π 5 = 10π.

The answer is (B).

2.A rectangular field is 3.2 yards long. A fence marking the boundary is 11.2 yards in length. What is the area of the field in square yards?

(A)4.68

(B)7.68

(C)9.28

(D)11.28

(E)12.38

Let l be the length of the rectangle. Then l = 3.2 yards. We are given that the length of the fence required (perimeter) for the field is 11.2 yards. The formula for the perimeter of a rectangle is 2(length + width). Hence, the perimeter of the field is 2(l + w) = 11.2, or 2(3.2 + w) = 11.2. Solving for w yields w = 2.4. The formula for the area of a rectangle is length × width. Hence, the area of the rectangle is lw = 3.2 × 2.4 = 7.68. The answer is (B).

3.ABCD is a square and one of its sides AB is also a chord of the circle as shown in the figure. What is the area of the square?

(A)3

(B)9

(C)12

(D)122

(E)18

D A

3

O

3

C B

Side AB is the hypotenuse of the ∆AOB. Hence, by The Pythagorean Theorem, we have AB2 = AO2 + BO2 = 32 + 32 = 18. Hence, the area of the square ABCD equals side2 = AB2 = 18. The answer is (E).

132GRE Math Bible

4.In the figure, lines l and m are parallel. If y – z = 60, then what is the value of x ?

(A)60

(B)75

(C)90

(D)120

(E)135

z° s°

m

x°

Since the angle made by a line is 180°, z + y = 180. Also, we are given that y – z = 60. Adding the equations yields

z + y + y – z = 180 + 60 2y = 240

y = 120

Since the lines l and m are parallel, the alternate exterior angles x and y are equal. Hence, x equals 120. The answer is (D).

5.What is the value of y in the figure?

(A)20

(B)30

(C)35

(D)45

(E)50

3a°

Summing the angles of the triangle in the figure to 180° yields a + 3a + 5a = 180. Solving this equation for a yields a = 180/9 = 20. Angles y and a in the figure are vertical and therefore are equal. So, y = a = 20. The answer is (A).

Geometry 133

6.In the triangle, what is the value of x ?

(A)25

(B)55

(C)60

(D)77

(E)85

y° + x°

The angle sum of a triangle is 180°. Hence, (y + x) + (60 + x) + (70 – y) = 180. Simplifying the equation yields 2x + 130 = 180. Solving for x yields x = 25. The answer is (A).

7.In the figure, what is the value of a ?

(A)16

(B)18

(C)36

(D)54

(E)72

2a° 3a°

a° 4a°

l

The angle made by a line is 180°. Hence, from the figure, we have

a + 2a + 3a + 4a = 180 10a = 180

a = 18

The answer is (B).

Geometry 135

O B

C

Since the shortest distance between two points is a straight line, the length of a chord is always less than the length of the arc it makes (subtends) on the circle. Hence, from the figure, we have

AB < arc AB

BC < arc BC

CA < arc CA

Adding the three inequalities yields AB + BC + CA < arc AB + arc BC + arc CA. The left side of the inequality is the perimeter of triangle ABC (which Column A equals), and the right side is the circumference of the circle (which Column B equals). Hence, Column A is less than Column B, and the answer is (B).

O C

D

Since the shortest distance between two points is a straight line, the length of a chord is always shorter than the length of the arc that it makes on the circle. Hence, from the figure, we have

AB < arc AB

BC < arc BC

CD < arc CD

DA < arc DA

Adding the four inequalities yields AB + BC + CD + DA < arc AB + arc BC + arc CD + arc DA. The left side of the inequality is the perimeter of quadrilateral ABCD (which Column A equals), and the right side is the circumference of the circle (which Column B equals). Hence, Column A is less than Column B, and the answer is (B).

136 GRE Math Bible

Let ABC be the triangle, and let P, Q, and R be the points at which the inscribed circle touches the sides of the triangle.

A

P

Q

C B

R

From the figure, it is clear that to go from one point on the circle, say, point P to another point, say, point Q, the shortest available path is the arc PQ. Hence, arc PQ < PA + AQ. Similarly, arc QR < QB + BR, and arc RP < RC + CP. Summing the three inequalities yields arc PQ + arc QR + arc RP < (PA + AQ) + (QB + BR) + (RC + CP). The right side of the inequality is the perimeter of the triangle ABC (which Column A equals), and the left side is the circumference of the circle (which Column B equals). Hence, Column A is greater than Column B, and the answer is (A).

Summing the angles of ∆ABC to 180° yields A + B + C = 180. Substituting the value of B (= 72°) into this equation yields A + 72 + C = 180. Solving the equation for A yields A = 180 – 72 – C = 108 – C = Column A.

Now, Suppose C = 1°. Then A = 108° – 1° = 107°. In this case, A is greater than 72°. Now, SupposeC = 107°. Then A = 108° – 107° = 1°. In this case, A is less than 72°.

Hence, we have a double case, and the answer is (D).

Geometry 137

14.In the figure, AD and BC are lines intersecting at O. What is the value of a ?

(A)15

(B)30

(C)45

(D)60

(E)135

B

Equating vertical angles AOB and COD in the figure yields y = 2x + 30. Also, equating vertical anglesAOC and BOD yields 5y/2 = 5x + 5a. Multiplying this equation by 2/5 yields y = 2x + 2a. Subtracting this equation from the equation y = 2x + 30 yields 2a = 30. Hence, a = 30/2 = 15, and the answer is (A).

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

(A)y = z

(B)y < z

(C)y ≤z

(D)y > z

(E)y ≥ z

yz

x 3x

Equating the two pairs of vertical angles in the figure yields y = 3x and x = z. Replacing x in first equation with z yields y = 3z. This equation says that y is 3 times as large as z. Hence, y > z. The answer is (D).

138GRE Math Bible

16.From the figure, which one of the following could be the value of b ?

(A)20

(B)30

(C)60

(D)75

(E)90

2a° – b° O a°/2 + 2b°

2a° – 2b°

In the figure, angles b° – a°/2 and 2a° – 2b° must be positive. Hence, we have the inequalities, b – a/2 > 0 and 2a – 2b > 0.

Adding a/2 to both sides of first inequality and 2b to both sides of second inequality yields the following two inequalities:

b > a/2 2a > 2b

Dividing the second inequality by 2 yields a > b.

Now, summing angles around point O to 360° yields a + (b – a/2) + (a/2 + 2b) + (2a – 2b) + (2a – b) = 360. Simplifying this yields 5a = 360, and solving yields a = 360/5 = 72.

Substituting this value in the inequalities b > a/2 and a > b yields

b > a/2 = 72/2 = 36, and 72 > b

Combining the inequalities b > 36 and 72 > b yields 36 < b < 72. The only choice in this range is (C), so the answer is (C).

140GRE Math Bible

18.In the figure, what is the value of a ?

(A)30

(B)45

(C)60

(D)72

(E)90

6c° 3c° 2b°

a° b°

n

Equating vertical angles in the figure yields a = 2b and b = 3c. From the first equation, we have b = a/2. Plugging this into the second equation yields a/2 = 3c, from which we can derive c = a/6. Since the angle made by a line is 180˚, we have for line l that b + a + 6c = 180. Replacing b with a/2 and c with a/6 in this equation yields a/2 + a + 6(a/6) = a/2 + a + a = 180. Summing the left-hand side yields 5a/2 = 180, and multiplying both sides by 2/5 yields a = 180(2/5) = 72. The answer is (D).

19. In the figure, lines l and m are parallel. Which one of the following, if true, makes lines p and q parallel?

(A)a = b

(B)a = c

(C)c = d

(D)d = b

(E)b = c

p q

c° d°

m

Superimposing parallel line m on line l yields a figure like this: