18.
The figure above shows a square that is tangent to one circle at four
points, and inscribed in another. If the diameter of the large circle is 10,
what is the diameter of the smaller circle?
(A)
5
=
3
2
(B) 5
(C) 2p
(D)
5
=
2
(E) 7.5
The correct answer is (D). The square’s diagonal is equal in length to the large
circle’s diameter, which is 10. This diagonal is the hypotenuse of a triangle whose
legs are two sides of the square. The triangle is right isosceles, with sides in the
ratio 1:1:
=
2. The length of each side of the square 5
10
=
2
,or5
=
2. This length
is also the diameter of the small circle.
Tangents and Inscribed Circles
A circle is tangent to a line (or line segment) if they intersect at one and only one point
(called the point of tangency). Here’s the key rule to remember about tangents: A line
that is tangent to a circle is always perpendicular to the line passing through the
circle’s center at the point of tangency.
The next figure shows a circle with center O inscribed in a square. Point P is one of
four points of tangency. By definition,
OP ⊥ AB.
Chapter 11: Math Review: Geometry 283
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A
P
B
O
Also, notice the following relationships between the circle in the preceding figure and
the square in which it is inscribed (r 5 radius):
• Each side of the square is 2r in length.
• The square’s area is (2r)
2
,or4r
2
.
• The ratio of the square’s area to that of the inscribed circle is
4
p
.
• The difference between the two areas—the total shaded area—is 4r
2
2pr
2
.
• The area of each separate (smaller) shaded area is
1
4
(4r
2
2pr
2
).
For any regular polygon (including squares) that inscribes a circle:
• The point of tangency between each line segment and the circle bisects the
segment.
• Connecting each vertex to the circle’s center creates an array of congruent angles,
arcs, and triangles.
For example, the left-hand figure below shows a regular pentagon, and the right-hand
figure shows a regular hexagon. Each polygon inscribes a circle. In each figure, the
shaded region is one of five (or six) identical ones.
PART IV: Quantitative Reasoning284
TIP
A line that is tangent to a
circle is always perpendicular
to a line connecting the point
of tangency and the circle’s
center.
www.petersons.com
19.
A
D
B
O
C
40º
xº
In the figure above, a circle with center O is tangent to AB at point D and
tangent to
AC at point C. If m∠A 5 40°, then x 5
(A) 140
(B) 145
(C) 150
(D) 155
(E) 160
The correct answer is (A). Since
AC is tangent to the circle, AC ⊥ BC.
Accordingly, DABC is a right triangle, and m∠B 5 50°. Similarly,
AB ⊥ DO,
DDBO is a right triangle, and ∠DOB 5 40°. ∠DOC (the angle in question) is
supplementary to ∠DOB. Thus, m∠DOC 5 140° (x 5 140).
20. One side of a rectangle is the diameter of a circle.
The opposite side of the rectangle is tangent to the circle.
Column A
Column B
The perimeter of the
rectangle
The circumference of the circle
(A) The quantity in Column A is greater.
(B) The quantity in Column B is greater.
(C) The quantities are equal.
(D) The relationship cannot be determined from the information given.
The correct answer is (B). The centered information describes the following
figure:
Chapter 11: Math Review: Geometry 285
TIP
When tackling a GRE
geometry question, don’t
hesitate to draw your own
figure to help you visualize the
problem and answer the
question.
www.petersons.com
Calling the radius r, the rectangle’s perimeter is 2(2r)+2(r)=6r. The circle’s
circumference is 2pr. Since p.3, 2p.6, and the circle’s circumference is larger.
Comparing Circles
On the GRE, questions asking you to compare circles come in two varieties. You will
be required to do one of the following:
Calculate the difference between radii, circumferences, and areas.
Determine ratios involving the two circles and their radii,circumferences,andareas.
To calculate a difference between the radii, circumferences, or areas, just calculate
each area or circumference, then subtract.
To handle questions involving ratios, you need to understand that the relationship
between a circle’s radius (circumference) and its area is exponential, not linear
(because A 5pr
2
). For example, if one circle’s radius is twice that of another, the ratio
of the circles’ areas is 1:4[pr
2
:p(2r)
2
]. If the larger circle’s radius is three times that of
the smaller circle, the ratio is 1:9[pr
2
:p(3r)
2
]. A 1:4 ratio between radii results in a
1:16 area ratio, and so forth.
The same proportions apply if you compare circumferences and areas. If the circum-
ference ratio is 2:1, then the area ratio is 4:1. If the circumference ratio is 3:1, then the
area ratio is 9:1, and so forth.
21.
In the figure above, point O lies at the center of both circles. If the length
of
OP is 6 and the length of PQ is 2, what is the ratio of the area of the
smaller circle to the area of the larger circle?
(A) 3:8
(B) 7:16
(C) 1:2
(D) 9:16
(E) 5:8
The correct answer is (D). The ratio of the small circle’s radius to that of the
large circle is 6:8, or 3:4. Since Area 5pr
2
, the area ratio is p(3)
2
:p(4)
2
,or9:16.
PART IV: Quantitative Reasoning286
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POLYGONS
Polygons include all plane figures formed only by straight segments. Up to this point,
we’ve focused on only three-sided polygons (triangles) and four-sided polygons (quad-
rilaterals). Now take a quick look at the key characteristics of all polygons.
You can use the following formula to determine the sum of the measures of all interior
angles of any polygon whose angles each measure less than 180° (n 5 number of
sides):
(n 2 2)(180°) 5 sum of interior angles
For regular polygons, the average angle measure is also the measure of every angle.
But for any polygon (except for those with an angle exceeding 180°), you can find the
average angle measure by dividing the sum of the measures of the angles by the
number of sides. One way to shortcut the math is to memorize the angle sums and
averages for polygons with 3–8 sides:
3 sides: (3 2 2)(180°) 5 180° 4 3 5 60°
4 sides: (4 2 2)(180°) 5 360° 4 4 5 90°
5 sides: (5 2 2)(180°) 5 540° 4 5 5 108°
6 sides: (6 2 2)(180°) 5 720° 4 6 5 120°
7 sides: (7 2 2)(180°) 5 900° 4 7 ' 129°
8 sides: (8 2 2)(180°) 5 1,080° 4 8 5 135°
A GRE question might simply ask for the measure of any interior angle of a certain
regular polygon; to answer it, just apply the preceding formula. If the polygon is not
regular, you can add up known angle measures to find unknown angle measures.
22. Exactly two of the angles of the polygon shown below are congruent if x <
180, what is the LEAST possible sum of the degree measures of two of the
polygon’s interior angles?
degrees
Enter a number in the box.
The correct answer is (138). The figure shows a hexagon. The sum of the
measures of all six angles is 720°. Subtracting the measures of the three
known angles from 720° leaves 420°, which is the sum of the measures of the
three unknown angles. Set up an equation, then solve for x:
Chapter 11: Math Review: Geometry 287
ALERT!
A polygon in which all sides
are congruent and all angles
are congruent is called a
regular polygon. But for the
GRE, you only need to know
the principle, not the
terminology.
www.petersons.com
x 1 x 1
1
2
x 5 420
5
2
x 5 420
x 5
~420!~2!
5
x 5~84!~2! = 168
Of the three unknown angles, two are 168° each. The other is
1
2
of 168°, or
84°. The polygon’s two smallest angles measure 54° and 84°. Their sum is
138°.
Another, more difficult, type of problem requires you to determine the area
of a polygon, which might be either regular or irregular. To do so, you need
to partition the polygon into an assemblage of smaller geometric figures.
23.
A
B
C
D
E
What is the area of polygon ABCDE shown above?
(A)
4 1 2
=
3
(B)
3 1 3
=
2
(C)
6
=
3
(D)
2 1 6
=
2
(E)
8
=
2
The correct answer is (A). Divide the polygon into three triangles as shown
below. The area of each of the two outer triangles 5
1
2
bh 5
1
2
~2!~2!52. (Their
combined area is 4.) Since the two outer triangles are both 1:1:
=
2 right tri-
angles,
BE ≅ BD, and both line segments are 2
=
2 units long. Accordingly, the
central triangle is equilateral. Calculate its area.
s
2
=
3
4
5
~2
=
2!
2
=
3
4
5
8
=
3
4
5 2
=
3
Thus, the area of the polygon is 4 1 2
=
3.
PART IV: Quantitative Reasoning288
ALERT!
GRE geometry figures are not
necessarily drawn to scale
(except when a f igure i s
accompanied by a note
stating otherwise). So don’t rely
on visual proportions when
analyzing GRE geometry
questions.
www.petersons.com
A
B
C
D
E
CUBES AND OTHER RECTANGULAR SOLIDS
GRE questions about rectangular solids always involve one or both of two basic
formulas (l 5 length, w 5 width, h 5 height):
Volume 5 lwh
Surface Area 5 2lw 1 2wh 1 2lh 5 2(lw 1 wh 1 lh)
For cubes, the volume and surface-area formulas are even simpler than for other
rectangular solids (let s 5 any edge):
Volume 5 s
3
or s 5
=
3
Volume
Surface Area 5 6s
2
s
s
s
A GRE question might require you to apply any one of the preceding formulas. If so,
plug what you know into the formula, then solve for whatever characteristic the
question asks for.
Chapter 11: Math Review: Geometry 289
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24. s . 1
Column A
Column B
The volume of a cube with
side s
The volume of a rectangular solid with
sides of s, s +1,ands 2 1
(A) The quantity in Column A is greater.
(B) The quantity in Column B is greater.
(C) The quantities are equal.
(D) The relationship cannot be determined from the information given.
The correct answer is (A). Quantity A = s
3
. Find Quantity B by multiplying
together the three expressions given as the lengths of the sides:
~s!~s +1!~s 2 1! = ~s
2
+ s!~s 2 1!
= s
3
2 s
2
+ s
2
2 s
= s
3
2 s
Subtracting s
3
from both columns leaves the comparison between zero (0)
(Column A) and 2s (Column B). You can now see that QuantityAmust be greater
than Quantity B.
Or a question might require you to deal with the formulas for both surface area and
volume.
25. A closed rectangular box with a square base is 5 inches in height. If the
volume of the box is 45 cubic inches, what is the box’s surface area in
square inches?
(A) 45
(B) 66
(C) 78
(D) 81
(E) 90
The correct answer is (C). First, determine the dimensions of the square base.
The box’s height is given as 5. Accordingly, the box’s volume (45) 5 5lw, and lw 5
9. Since the base is square, the base is 3 inches long on each side. Now you can
calculate the total surface area: 2lw 1 2wh 1 2lh 5 (2)(9) 1 (2)(15) 1 (2)(15) 5
78.
A variation on the preceding question might ask the number of smaller boxes you
could fit, or “pack,” into the box that the question describes: for instance, the number
of cube-shaped boxes, each one 1.5 inches on a side, that you could pack into the 3 3
3 3 5 box is 12 (3 levels of 4 cubes, with a half-inch space left at the top of the box).
PART IV: Quantitative Reasoning290
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A test question involving a cube might focus on the ratios among the cube’s linear,
square, and cubic measurements.
26. If the volume of one cube is 8 times greater than that of another, what is
the ratio of the area of one square face of the larger cube to that of the
smaller cube?
(A) 16:1
(B) 12:1
(C) 8:1
(D) 4:1
(E) 2:1
The correct answer is (D). The ratio of the two volumes is 8:1. Thus, the linear
ratio of the cubes’ edges is the cube root of this ratio:
=
3
8:
=
3
1 5 2:1. The area
ratio is the square of the linear ratio, or 4:1.
CYLINDERS
The only kind of cylinder theGRE covers is a “right” circular cylinder (a tube sliced at
90° angles). The surface area of a right cylinder is the sum of the following areas:
• The circular base
• The circular top
• The rectangular surface around the cylinder’s vertical face (visualize a rectan-
gular label wrapped around a can)
The area of the vertical face is the product of the circular base’s circumference (i.e.,
the rectangle’s width) and the cylinder’s height. Thus, given a radius r and height h of
a cylinder:
Surface Area (SA) 5 2pr
2
1 (2pr)(h)
Volume 5pr
2
h
On the GRE, a cylinder problem might require little more than a straightforward
application of formula either for surface area or volume. As with rectangular-solid
Chapter 11: Math Review: Geometry 291
TIP
The relationships among a
linear edge, a square face,
and the volume of a cube are
exponential. For the GRE,
make sure you know how to
express each one in terms of
the others.
www.petersons.com
questions, just plug what you know into the formula, then solve for what the question
asks.
27. Column A
Column B
The volume of a right
cylinder whose circular
base has a radius of 3 and
whose height is 6.
The volume of a right
cylinder whose circular
base has a radius of 6 and
whose height is 3.
(A) The quantity in Column A is greater.
(B) The quantity in Column B is greater.
(C) The quantities are equal.
(D) The relationship cannot be determined from the information given.
The correct answer is (B). Use the formula for the volume of a right cylinder
(V = pr
2
h) to compare volumes:
Quantity A = p(3
2
)(6) = p(3)(3)(6)
Quantity B = p(6
2
)(3) = p(6)(6)(3)
You can easily see that Quantity B is greater than Quantity A.
A tougher cylinder problem might require you to apply other math concepts as well, or
it might call for you to convert one unit of measure into another.
28. A cylindrical pail whose diameter is 14 inches and whose height is 10
inches is filled to one fourth its capacity with water. Which of the following
most closely approximates the volume of water in the pail, in gallons? [231
cubic inches 5 1 gallon]
(A) 6.7
(B) 4.2
(C) 2.9
(D) 1.7
(E) 0.8
The correct answer is (D). The volume of the cylindrical pail is equal to the
area of its circular base multiplied by its height:
V 5pr
2
h '
S
22
7
D
~49!~10! = (22)(7)(10) = 1,540 cubic inches
The gallon capacity of the pail ' 1,540 4 231, or about 6.7. One fourth of
this capacity is approximately 1.7 gallons.
COORDINATE GEOMETRY—THE XY-PLANE
GRE coordinate geometry questions involve the rectangular coordinate plane (or
xy-plane) defined by two axes—a horizontal x-axis and a vertical y-axis. You can define
any point on the coordinate plane by using two coordinates: an x-coordinate and a
y-coordinate. A point’s x-coordinate is its horizontal position on the plane, and its
PART IV: Quantitative Reasoning292
TIP
Rounding your calculations will
suffice to answer GRE
questions that ask for an
appoximation. Just don’t
round up or down too far.
www.petersons.com
. opposite side of the rectangle is tangent to the circle.
Column A
Column B
The perimeter of the
rectangle
The circumference of the circle
(A) The quantity. greater.
(B) The quantity in Column B is greater.
(C) The quantities are equal.
(D) The relationship cannot be determined from the information given.
The