GRE_Math_Bible_eBook

Geometry 191

Now, substituting this value of x in equation (1) yields

52 + 122 = r2

25 + 144 = r2 r = 169 = 13

Hence, the radius is 13, and the answer is (C).

The ordering ABCD indicates that AB and BC are adjacent sides of a rectangle with common vertex B (see figure below). Remember that the perimeter of a rectangle is equal to twice its length plus twice its width. Hence,

P = 2AB + 2BC

We are given that the perimeter is 5/2 times the length of side AB. Replacing the left-hand side of the

5

equation with 2 AB yields

25 AB = 2AB + 2BC

Thus, the length of side AB is four times as long as side BC. Hence, side AB is greater than side BC. The answer is (A).

92. A closed rectangular tank contains a certain amount of water. When the tank is placed on its 3 ft by 4 ft side, the height of the water in the tank is 5 ft. When the tank is placed on another side of dimensions 4 ft by 5 ft, what is the height, in feet, of the surface of the water above the ground?

(A)2

(B)3

(C)4

(D)5

(E)6

When based on the 3 ft 4 ft side, the height of water inside the rectangular tank is 5 ft. Hence, the volume of the water inside tank is length × width × height = 3 4 5 cu. ft.

When based on 4 ft 5 ft side, let the height of water inside the rectangular tank be h ft. Then the volume of the water inside tank would be length × width × height = 4 5 h cu. ft.

Equating the results for the volume of water, we have 3 4 5 = v = 4 5 h. Solving for h yields h = (3 4 5)/(4 5) = 3 ft.

The answer is (B).

Geometry 193

95.In the figure, ∆ABC is inscribed in the circle. The triangle does not contain the center of the circle O. Which one of the following could be the value of x in degrees?

(A)35

(B)70

(C)85

(D)90

(E)105

A

x°

C B

.O

A chord makes an acute angle on the circle to the side containing the center of the circle and makes an obtuse angle to the other side. In the figure, BC is a chord and does not have a center to the side of point A. Hence, BC makes an obtuse angle at point A on the circle. Hence, A, which equals x°, is obtuse and therefore is greater than 90°. Since 105 is the only obtuse angle offered, the answer is (E).

96.In the figure, ABCD is a square, and BC is tangent to the circle with radius 3. P is the point of intersection of the line OC and the circle. If PC = 2, then what is the area of square ABCD ?

(A)9

(B)13

(C)16

(D)18

(E)25

A B O

P

2

D C

In the figure, since OB and OP are radii of the circle, both equal 3 units. Also, the length of line segment OC is OP + PC = 3 + 2 = 5. Now, since BC is tangent to the circle, OBC = 90. Hence, triangle OBC is a

right triangle. Applying The Pythagorean Theorem yields

196GRE Math Bible

100.In the figure, lines l1, l2, and l3 are parallel to one another. Line-segments AC and DF cut the three lines. If AB = 3, BC = 4, and DE = 5, then which one of the following equals DF ?

(A)3/30

(B)15/7

(C)20/3

(D)6

(E)35/3

l3

C F

In the figure, AC and DF are transversals cutting the parallel lines l1, l2, and l3. Let’s move the line-segment DF horizontally until point D touches point A. The new figure looks like this:

l3

C F F

Now, in triangles ABE and ACF, B equals C and E equals F because corresponding angles of parallel lines (here l2 and l3) are equal. Also, A is a common angle of the two triangles. Hence, the three angles of triangle ABE equal the three corresponding angles of the triangle ACF. Hence, the two triangles are similar. Since the ratios of the corresponding sides of two similar triangles are equal, we have

The answer is (E).

Geometry 197

101.In the figure, AB is parallel to CD. What is the value of x ?

(A)36

(B)45

(C)60

(D)75

(E)Cannot be determined

A E B

x° y°

Lines AB and CD are parallel (given) and cut by transversal ED. Hence, the alternate interior angles x and y are equal. Since x = y, ∆ECD is isosceles (C = D). Hence, angles x and y in ∆ECD could each range between 0° and 90°. No unique value for x is derivable. Hence, the answer is (E).

A

B

Q P

Since AB is tangent to both circles, BAQ = 90° and ABP = 90°. Hence, AQ is parallel to BP. Now, draw a line through point P and parallel to AB as shown in the figure.

198 GRE Math Bible

Then ABPD must be a rectangle. Hence, AB = DP. Also, PDQ equals the corresponding angle, BAD. So, both equal 90°. Since a right angle is the greatest angle in a triangle, the side opposite the angle is the longest. PQ is the side opposite the right angle PDQ in ∆PQD. Hence, PQ is greater than the other side DP. Hence, PQ is also greater than AB, which equals DP (we know). Hence, Column B is greater than Column A, and the answer is (B).

x°

C B

.O

A chord makes an acute angle on the circle to the side containing the center of the circle and makes an obtuse angle to the other side. In the figure, BC is a chord and does not have a center to the side of point A. Hence, BC makes an obtuse angle at point A on the circle. Hence, A, which equals x°, is obtuse and therefore is greater than 90°. Hence, Column A is greater than Column B, and the answer is (A).

Since no two sides of the triangle in Column A are equal, no two angles of the triangle are equal. Since the sum of the three angles of the triangle is 180 degrees, at least one angle must be greater than 60 degrees. Otherwise, the angles would not sum to 180 degrees. Hence, Column A is greater than 60 (= Column B). The answer is (A).

Geometry 199

Very Hard

105.In the figure, if y = 60, then what is the value of z ?

(A)20

(B)30

(C)55

(D)75

(E)90

x°

x°

Let’s name the vertices in the figure as shown below.

B F

x°

G

x°

D

In the figure, FGE equals y° (vertical angles). Summing the angles of ∆GEF to 180° yields

GEF + EFG + FGE = 180GEF + 90 + y = 180

GEF = 180 – (90 + y)

=90 – y = 90 – 60 (since y = 60, given)

=30°

Now, since ABD = BDF (both equal x, from the figure), lines AB and DF are parallel (alternate interior angles). Hence, CAB = GEF (corresponding angles) = 30° (we know GEF = 30˚). Now, z = CAB (vertical angles), and therefore z = 30. The answer is (B).

200GRE Math Bible

106.In the figure, ABCD is a rectangle and AF is parallel to BE. If x = 5, and y = 10, then what is the area of ∆AFD ?

(A)2.5

(B)5

(C)12.5

(D)50

(E)50 + 5y

x

Since opposite sides of a rectangle are equal and parallel, AD is parallel to BC and AD = BC.

We are also given that AF is parallel to BE.

Further, points D, C, and E must be on same line because the angle made by points D and E at C is 180° (DCE = DCB + BCE = angle in a rectangle + a right angle [given] = 90° + 90° = 180°). So, DC and CE can be considered parallel.

Any three pairs of parallel lines (here AF and BE, DF and CE, and AD and BC) make two similar triangles (AFD and BEC). And if one pair of corresponding sides of two similar triangles are equal (here, AD = BC), then the triangles are congruent (equal).

The areas of congruent triangles are equal. So, area of ∆AFD = area of ∆BEC = 1/2 base height = 1/2 BC CE = 1/2 5 5 = 1/2 25 = 12.5. The answer is (C).