7. If a square’s sides are each increased by 50%, by what percent does the
square’s area increase?
(A) 75%
(B) 100%
(C) 125%
(D) 150%
(E) 200%
The correct answer is (C). Letting s 5 the length of each side before the
increase, area 5 s
2
.If
3
2
s 5 the length of each side after the increase, the new
area 5
S
3
2
s
D
2
5
9
4
s
2
. The increase from s
2
to
9
4
s
2
is
5
4
, or 125%.
GRE questions involving non-square rectangles also come in many possible flavors.
For example, a question might ask you to determine area based on perimeter, or vice
versa.
8. The length of a rectangle with area 12 is three times the rectangle’s width.
What is the perimeter of the rectangle?
inches
Enter a number in the box.
The correct answer is (16). The ratio of length to width is 3:1. The ratio 6:2 is
equivalent, and 6 3 2 5 12 (the area). Thus, the perimeter 5 (2)(6) 1 (2)(2) 5 16.
A question might involve the properties of a square or rectangle as well as those of
another geometric figure, such as a right, isosceles, or equilateral triangle.
9. A rectangular block of wood is to be cut into sections to form
either two triangular enclosures or three rectangular
enclosures. All wood sections must be equal in length on all sides.
Column A
Column B
The total area of the two
proposed triangular
enclosures
The total area of the three proposed
rectangular enclosures
(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). Since the sections must all be equal in length, the
triangles must each be equilateral and the rectangles must each be square. The
easiest way to make the comparison is to assign to the length of the original block
a number that is the least common multiple of 6 and 12 (the total number of sides
among the triangles and squares, respectively). That multiple, of course, is 12.
Each triangle side = 2. Applying the area formula for equilateral triangles, the
Chapter 11: Math Review: Geometry 273
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area of each triangle is
23
4
3
2
=
, and the total area for both triangles is
23
, or about 3.4 (using 1.7 as an approximate value for
3
). Given a block
length of 12, and 12 sides for three squares, the length of each side of each square
=
12
12
1=
. Accordingly, the area of each square is 1, and the total area of all three
squares is 3.
Or, a question might require you to determine a combined perimeter or area of
adjoining rectangles.
10.
In the figure above, all intersecting line segments are perpendicular.
What is the area of the shaded region, in square units?
(A) 84
(B) 118
(C) 128
(D) 139
(E) 238
The correct answer is (C). The figure provides the perimeters you need to
calculate the area. One way to find the area of the shaded region is to consider it
as what remains when a rectangular shape is cut out of a larger rectangle. The
area of the entire figure without the “cut-out” is 14 3 17 5 238. The “cut-out”
rectangle has a length of 11, and its width is equal to 17 2 4 2 3 5 10. Thus, the
area of the cut-out is 11 3 10 5 110. Accordingly, the area of the shaded region is
238 2 110 5 128.
Another way to solve the problem is to partition the shaded region into three
smaller rectangles, as shown in the next figure, and sum up the area of each.
A GRE question about a non-rectangular parallelogram might focus on angle mea-
sures. These questions are easy to answer. In any parallelogram, opposite angles are
congruent, and adjacent angles are supplementary. (Their measures total 180°.) So if
PART IV: Quantitative Reasoning274
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one of a parallelogram’s angles measures 65°, then the opposite angle must also
measure 65°, while the two other angles each measure 115°.
11.
w
x
z
y
WXZY is a parallelogram.
r = 180 2 s
Column A
Column B
pq
(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 (D). In the figure, r° and s° are opposite angles, and
therefore r = s. Given that r = 180 2 s, r and s must both equal 90. Since the
diagonals of WXZY form right angles, WXZY must be a rhombus (a parallelogram
in which all four sides are equal in length). However, p = q if and only if WXZY is
a square.
A more difficult question about a non-rectangular parallelogram might focus on area.
To determine the parallelogram’s altitude, you might need to apply the Pythagorean
theorem (or one of the side or angle triplets).
12.
In the figure above, AB i CD and AD i BC. If BC is 4 units long and CD is
2 units long, what is the area of quadrilateral ABCD?
(A) 4
(B)
4
=
2
(C) 6
(D) 8
(E)
6
=
2
The correct answer is (B). Since ABCD is a parallelogram, its area 5 base (4)
3 altitude. To determine altitude (a), draw a vertical line segment connecting
point A to
BC, which creates a 45°-45°-90° triangle. The ratio of the triangle’s
Chapter 11: Math Review: Geometry 275
ALERT!
A GRE geometry figure will not
necessarily be drawn to
scale—unless the figure is
accompanied by a note
stating that it is.
TIP
A non-rectangular
parallelogram in which all four
sides are congruent (called a
rhombus) has the following in
common with a square:
Perimeter 5 4s;
Area 5 one half the product
of the diagonals.
www.petersons.com
hypotenuse to each leg is
=
2:1. The hypotenuse AB 5 2. Thus, the altitude (a)of
ABCD is
2
=
2
,or
=
2. Accordingly, the area of ABCD 5 4 3
=
2, or 4
=
2.
Trapezoids
A trapezoid is a special type of quadrilateral. The next figure shows a trapezoid.
Referring to this figure, all trapezoids share these four properties:
Only one pair of opposite sides are parallel (BC i AD).
The sum of the measures of all four angles is 360°.
Perimeter 5 AB 1 BC 1 CD 1 AD
Area 5
BC 1 AD
2
3 altitude (that is, one half the sum of the two parallel sides
multiplied by the altitude).
B
C
A
D
On the GRE, a trapezoid problem might require you to determine the altitude, the
area, or both.
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13.
To cover the floor of an entry hall, a 1' 3 12' strip of carpet is cut into two
pieces, shown as the shaded strips in the figure above, and each piece is
connected to a third carpet piece as shown. If the 1' strips run parallel to
each other, what is the total area of the carpeted floor, in square feet?
(A) 46
(B) 48
(C) 52.5
(D) 56
(E) 60
The correct answer is (E). The altitude of the trapezoidal piece is 8. The sum
of the two parallel sides of this piece is 12' (the length of the 1' 3 12' strip before
it was cut). You can apply the trapezoid formula to determine the area of this
piece:
A 5 8 3
12
2
5 48
The total area of the two shaded strips is 12 square feet, so the total area of
the floor is 60 square feet.
A GRE trapezoid problem might require you to find the trapezoid’s altitude by the
Pythagorean theorem.
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14.
B
C
AD
3
5
4
120
°
In the figure above, BC i AD. What is the area of quadrilateral ABCD?
(A)
5
=
2
(B)
9
=
3
2
(C)
27
=
3
4
(D)
27
2
(E) 16
The correct answer is (C). The figure shows a trapezoid. To find its area, first
determine its altitude by creating a right triangle:
BC
AD
3
5
120
60
33
2
4
√
°
This right triangle conforms to the 30°-60°-90° Pythagorean angle triplet.
Thus, the ratio of the three sides is 1:
=
3:2. The hypotenuse is given as 3,
so the trapezoid’s altitude is
3
=
3
2
. Now you can calculate the area of the
trapezoid:
S
1
2
D
~4 1 5!
S
3
=
3
2
D
5
S
9
2
DS
3
=
3
2
D
5
27
=
3
4
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CIRCLES
For the GRE, you’ll need to know the following basic terminology involving circles:
Circumference: The distance around the circle (its “perimeter”).
Radius: The distance from a circle’s center to any point on the circle’s cir-
cumference.
Diameter: The greatest distance from one point to another on the circle’s
circumference (twice the length of the radius).
Chord: A line segment connecting two points on the circle’s circumference
(a circle’s longest possible chord is its diameter, passing through the circle’s
center).
Finding a Circle’s Circumference or Area
In solving most GRE circle problems, you apply one or both of the following two
formulas (r 5 radius, d 5 diameter):
Circumference 5 2pr,orpd
Area 5pr
2
Note that the value of p is approximately 3.14, or
22
7
. For the GRE, you won’t need to
work with a value for p any more precise. In fact, in most circle problems, the solution
is expressed in terms of p rather than numerically.
With the two formulas, all you need is one value—area, circumference, diameter, or
radius—and you can determine all the others. For example:
Given a circle with a diameter of 6:
Radius 5 3
Circumference 5 (2)(3)p56p
Area 5p(3)
2
5 9p
15. If a circle’s circumference is 10p centimeters long, what is the area of the
circle, in square centimeters?
(A) 12.5
(B) 5p
(C) 22.5
(D) 25p
(E) 10p
The correct answer is (D). First, determine the circle’s radius. Applying the
circumference formula C 5 2pr, solve for r:
Chapter 11: Math Review: Geometry 279
TIP
If answering a GRE question
requires you to use a number
value for p, rest assured that a
rough approximation—e.g., a
little greater than 3—will
suffice.
www.petersons.com
10p52pr
5 5 r
Then, apply the area formula, with 5 as the value of r:
A 5p~5!
2
5 25p
Arcs and Degree Measures of a Circle
An arc is a segment of a circle’s circumference. A minor arc is the shortest arc
connecting two points on a circle’s circumference. For example, in the next figure,
minor arc AB is the one formed by the 60° angle from the circle’s center (O).
A
B
A circle, by definition, contains a total of 360°. The length of an arc relative to the
circle’s circumference is directly proportionate to the arc’s degree measure as a
fraction of the circle’s total degree measure of 360°. For example, in the preceding
figure, minor arc AB accounts for
60
360
,or
1
6
, of the circle’s circumference.
16.
Circle O, as shown in the figure above, has diameters of DB and AC and has
a circumference of 9. What is the length of minor arc BC?
(A) 4
(B)
11
3
(C)
7
2
(D)
13
4
(E) 3
PART IV: Quantitative Reasoning280
TIP
An arc of a circle can be
defined either as a length (a
portion of the circle’s
circumference) or as a degree
measure of the central angle
that determines the arc.
www.petersons.com
The correct answer is (C). Since AO and OB are both radii, DAOB is an
isosceles triangle, and therefore m∠BAO 5 70°. It follows that m∠AOB 5
40°. ∠BOC is supplementary to ∠AOB, therefore m∠BOC 5 140°.
(Remember: angles from a circle’s center are proportionate to the arcs they
create.) Since m∠BOC accounts for
140
360
,or
7
18
of the circle’s circumference,
we have the length of minor arc BC 5
S
7
18
D
~9!5
7
2
.
Circles and Inscribed Polygons
A polygon is inscribed in a circle if each vertex of the polygon lies on the circle’s
circumference.
A test question might require you to visualize the possible shapes or proportions of a
type of triangle or quadrilateral inscribed in a circle.
17. An equilateral triangle is inscribed in a circle such that
each vertex of the triangle lies along the circle’s circumference.
Column A
Column B
The length of any side of
the triangle
The circle’s diameter
(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). If any of the triangle’s sides were the length of the
circle’s diameter, that side would pass through the circle’s center. In this case,
however, it would be impossible to construct an inscribed equilateral triangle:
Each of the other two sides would need to extend beyond the circle’s circum-
ference.
The next figure shows an inscribed square. The square is partitioned into four
congruent triangles, each with one vertex at the circle’s center (O).
A
B
O
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Look at any one of the four congruent triangles—for example, DABO. Notice that
DABO is a right triangle with the 90° angle at the circle’s center. The length of each of
the triangle’s two legs (
AO and BO) equals the circle’s radius (r). Accordingly, DABO is
a right isosceles triangle, m∠OAB 5 m∠OBA 5 45°, andAB 5 r
=
2. (The ratio of the
triangle’s sides is 1:1:
=
2.) Since AB is also the side of the square, the area of a
square inscribed in a circle is ~r
=
2!
2
,or2r
2
. (The area of DABO is
r
2
2
, or one fourth
the area of the square.)
You can also determine relationships between the inscribed square and the circle:
• The ratio of the inscribed square’s area to the circle’s area is 2:p.
• The difference between the two areas—the total shaded area—is pr
2
2 2r
2
.
• The area of each crescent-shaped shaded area is
1
4
~pr
2
2 2r
2
!.
The next figure shows a circle with an inscribed regular hexagon. (In a regular
polygon, all sides are congruent.) The hexagon is partitioned into six congruent
triangles, each with one vertex at the circle’s center (O).
B
A
O
60º
Look at any one of the six congruent triangles—for example, DABO. Since all six
triangles are congruent, m∠AOB 5 60° (one sixth of 360°).You can see that the length
of
AO and BO each equals the circle’s radius (r). Accordingly, m∠OAB 5 m∠OBA 5
60°, DABO is an equilateral triangle, and length of
AB 5 r.
Applying the area formula for equilateral triangles: Area of DABO 5
r
2
=
3
4
. The area
of the hexagon is 6 times the area of DABO, or
3r
2
=
3
2
. You can also determine
relationships between the inscribed hexagon and the circle. For example, the dif-
ference between the two areas—the total shaded area—is pr
2
2
3r
2
=
3
2
.
PART IV: Quantitative Reasoning282
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. Thus, the
area of the cut-out is 11 3 10 5 110. Accordingly, the area of the shaded region is
238 2 110 5 128.
Another way to solve the problem is to partition. triangle conforms to the 30 -6 0 -9 0° Pythagorean angle triplet.
Thus, the ratio of the three sides is 1:
=
3:2. The hypotenuse is given as 3,
so the trapezoid’s