GRE_Math_Bible_eBook
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Geometry 81
The altitude to the base of an isosceles or equilateral triangle bisects the base and bisects the vertex angle:
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Isosceles: |
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Equilateral: |
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s/2 |
s/2 |
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The angle sum of a triangle is 180°: |
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a + b + c = 180˚ |
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Example: In the figure, lines AB and D E are parallel. |
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What is the value of x ? |
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(A) |
22.5 |
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x° |
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(B) |
45 |
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(C) |
60 |
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(D) |
67.5 |
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(E) |
70 |
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Since lines AB and DE are parallel, we can equate the alternate interior angles at A and D to get A = D = x (from the figure). Also, equating vertical angles ACB and DCE yields ACB = DCE = 45 (from the figure). Now, we know that the angle sum of a triangle is 180°. Hence, DCE + CED + EDC = 180. Plugging the known values into this equation yields 45 + x + x = 180. Solving this equation for x yields x = 67.5. The answer is (D).
1
The area of a triangle is 2 bh, where b is the base and h is the height. Sometimes the base must be extended in order to draw the altitude, as in the third drawing directly below:
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In a triangle, the longer side is opposite the larger angle, and vice versa:
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50˚ |
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30˚ |
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Pythagorean Theorem (right triangles only): The square of the hypotenuse is equal to the sum of the squares of the legs.
50˚ is larger than 30˚, so side b is longer than side a.
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c2 = a2 + b2 |
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Pythagorean triples: The numbers 3, 4, and 5 can always represent the sides of a right triangle and they appear very often: 52 = 32 + 42 . Another, but less common, Pythagorean Triple is 5, 12, 13: 132 = 52 +122 .
82GRE Math Bible
Two triangles are similar (same shape and usually different sizes) if their corresponding angles are equal. If two triangles are similar, their corresponding sides are proportional:
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b
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da = be = cf
If two angles of a triangle are congruent to two angles of another triangle, the triangles are similar.
In the figure to the right, the large and small triangles are similar because both contain a right angle and they share A..
A
Two triangles are congruent (identical) if they have the same size and shape.
In a triangle, an exterior angle is equal to the sum of its remote interior angles and is therefore greater than either of them:
a
e = a + b and e > a and e > b
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In a triangle, the sum of the lengths of any two sides is greater than the length of the remaining side:
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x + y > z |
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y + z > x |
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x + z > y |
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Example: Two sides of a triangle measure 4 and 12. Which one of the following could equal the length of the third side?
(A)5
(B)7
(C)9
(D)17
(E)20
Each side of a triangle is shorter than the sum of the lengths of the other two sides, and, at the same time, longer than the difference of the two. Hence, the length of the third side of the triangle in the question is greater than the difference of the other two sides (12 – 4 = 8) and smaller than their sum (12 + 4 = 16). Since only choice (C) lies between the values, the answer is (C).
In a 30°–60°–90° triangle, the sides have the following relationships:
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30˚ |
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In general |
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60˚ |
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60˚ |
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Geometry 83
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In a 45°–45°–90° triangle, the sides have the following relationships: |
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Quadrilaterals |
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A quadrilateral is a four-sided closed figure, where each side is a straight line. |
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The angle sum of a quadrilateral is 360˚. You can |
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view a quadrilateral as being composed of two |
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180˚ |
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180˚ |
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180-degree triangles: |
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A parallelogram is a quadrilateral in which the |
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opposite sides are both parallel and congruent. Its |
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A = bh |
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area is base × height: |
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The diagonals of a parallelogram bisect each other:
A parallelogram with four right angles is a |
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rectangle. If w is the width and l is the length of a |
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A = l w |
rectangle, then its area is and its perimeter is P = |
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P = 2w + 2l |
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2w + 2l. |
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Example: In the figure, the area of quadrilateral ABCD is 75. What is the area of parallelogram EFGH ?
(A)96
(B)153
(C)157
(D)165
(E)171
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x + 3 |
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x + 7 |
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In the figure, ∆ACD is right angled. Hence, by The Pythagorean Theorem, AC2 = AD2 + DC2 = x2 + x2 = 2x2. By square rooting the sides, we have AC = x
2 .
The formula for the area of a triangle is 1/2 × base × height. Hence, the area of the right-triangle ACD is
1/2 · x · x, and the area of triangle ABC is 1 AC (altitude from B on AC) = 1 x
2 5
2 = 5x . Now, the
2 2
area of quadrilateral ABCD, which is given to be 75, is the sum of areas of the two triangles: x2/2 + 5x. Hence, x2/2 + 5x = 75. Multiplying both sides by 2 yields x2 + 10x = 150.
Now, the formula for the area of a parallelogram is base × height. Hence, the area of EFGH is (x + 3)(x + 7) = x2 + 10x + 21 = (x2 + 10x) + 21 = 150 + 21 (since x2 + 10x = 150) = 171. The answer is (E).
84 |
GRE Math Bible |
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If the opposite sides of a rectangle are equal, it is a |
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square and its area is A = s2 and its perimeter is |
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P = 4s, where s is the length of a side: |
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A = s2 |
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P = 4s |
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The diagonals of a square bisect each other and are |
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perpendicular to each other: |
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A quadrilateral with only one pair of parallel sides |
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is a trapezoid. The parallel sides are called bases, |
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and the non-parallel sides are called legs: |
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The area of a trapezoid is the average of the two bases times the height:
base |
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Volume
The volume of a rectangular solid (a box) is the product of the length, width, and height. The surface area is the sum of the area of the six faces:
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S = 2wl + 2hl + 2wh
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If the length, width, and height of a rectangular solid (a box) are the same, it is a cube. Its volume is the cube of one of its sides, and its surface area is the sum of the areas of the six faces:
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S = 6x2 |
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Geometry 85
Example: The length, width, and depth of a rectangular box are 6 feet, 5 feet, and 7 feet, respectively. A hose supplies water at a rate of 6 cubic feet per minute. How much time in minutes would it take to fill a conical box whose volume is three times the volume of the rectangle box?
(A)105
(B)125
(C)205
(D)235
(E)322
The volume of a rectangular tank is length × width × depth = 6 feet × 5 feet × 7 feet. Hence, the volume of the conical box, which is 3 times the volume of rectangular box, is 3(6 × 5 × 7). Now, the time taken to fill a tank equals the (volume of the tank) ÷ (the rate of filling) = 3(6 × 5 × 7) feet/6 cubic feet per minute = 105 minutes. The answer is (A).
The volume of a cylinder is V = πr2h, and the lateral surface (excluding the top and bottom) is S = 2πrh, where r is the radius and h is the height:
hV = πr2h
S = 2πrh + 2πr2
r
Circles
A circle is a set of points in a plane equidistant from a fixed point (the center of the circle). The perimeter of a circle is called the circumference.
A line segment from a circle to its center is a radius.
A line segment with both end points on a circle is a chord.
A chord passing though the center of a circle is a diameter.
A diameter can be viewed as two radii, and hence a diameter’s length is twice that of a radius.
A line passing through two points on a circle is a secant.
A piece of the circumference is an arc.
The area bounded by the circumference and an angle with vertex at the center of the circle is a sector.
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secant |
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A tangent line to a circle intersects the circle at only one point. The radius of the circle is perpendicular to the tangent line at the point of tangency:
Two tangents to a circle from a common exterior |
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point of the circle are congruent: |
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OAB AC
C
86GRE Math Bible
An angle inscribed in a semicircle is a right angle:
A central angle has by definition the same measure as its intercepted arc:
60˚
60˚
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intercepted arc: |
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The area of a circle is πr2 , and its circumference |
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On the GRE, π ≈ 3 is a sufficient approximation for π. You don’t need π ≈ 3.14.
A
Example: In the circle shown in the figure, the length of the arc ACB is 3 times the length of the arc AB. What is the length of the line segment AB ?
(A)3
(B)4
(C)5
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Since the length of the arc ACB is 3 times the length of the arc AB, the angle made by the arc ACB must be three times the angle made by the arc AB. Now, the two arcs together make 360° at the center of the circle. Hence, the smaller angle, the angle made by the arc AB, must equal one-quarter of the full angle, which is 360°. One-quarter of 360° is 90°. Hence, AOB = 90°. Hence, triangle AOB is a right triangle, and applying The Pythagorean Theorem to the triangle yields
AB2 = OA2 + OB2
= 32 + 32 = 9 + 9 = 18 OA = OB = radius of circle = OC = 3 (from the figure)
AB = 
18 = 3
2
The answer is (E).
Geometry 87
Shaded Regions
To find the area of the shaded region of a figure, subtract the area of the unshaded region from the area of the entire figure.
Example: In the figure, ABCD is a rectangle. What is the area of the shaded region in the figure?
(A)18
(B)20
(C)24
(D)28
(E)32
From the figure, the area of the shaded region equals
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The figure is not drawn to scale.
(Area of ∆ABD) – (Area of ∆AFE)
Now, the area of ∆ABD, by the formula 1/2 × base × height, equals
1/2 × AB × AD = (1/2)(AF + FB)(AE + ED) = (1/2)(8 + 4)(6 + 1) = (1/2)(12)(7) = 6 × 7 = 42
and the area of ∆AFE equals |
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1/2 × AF × AE = 1/2 × 8 × 6 = 4 × 6 = 24 |
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Hence, the area of the shaded region equals |
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(Area of ∆ABD) – (Area of ∆AFE) = |
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42 – 24 = |
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The answer is (A). |
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Example: What is the area of shaded region in the figure? |
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(A)10π + 27
3
(B)10π + 274 
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30π + 27 |
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30π + 9 |
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(E)36π + 27
3
From the figure, we have
The area of the Shaded region = (Area of Circle) – (Area of Sector AOB) + (Area of ∆AOB).
By the formula of the area of a circle, the area of the circle in the figure is π × radius2 = π(6)2 = 36π.
The formula for the area of a sector is
(Angle made by sector/360°) × (Area of the circle) =
60/360 × 36π =
1/6 × 36π =
6π
88 GRE Math Bible
Also, since OA and OB are radii, angles opposite them are equal. Hence, AOB is an isosceles triangle with one angle (OAB =) 60°. An isosceles triangle with one angle measuring 60° is always an equilateral triangle.
Now, the formula for the area of an equilateral triangle is |
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• side2 . Hence, the area of ∆AOB is |
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43 • 62 = 
43 • 36 = 9
3
Hence, the area of the shaded region is 36π – 6π + 9
3 = 30π + 9
3 . The answer is (D).
“Birds-Eye” View
Most geometry problems on the GRE require straightforward calculations. However, some problems measure your insight into the basic rules of geometry. For this type of problem, you should step back and take a “birds-eye” view of the problem. The following example will illustrate.
Example: In the figure to the right, O is both the center of the circle with radius 2 and a vertex of the square OPRS. What is the length of diagonal PS ?
(A)1/2
(B)
2
2
(C)4
(D)2
(E)2
5
S R
O P
The diagonals of a square are equal. Hence, line segment OR (not shown) is equal to SP. Now, OR is a radius of the circle and therefore OR = 2. Hence, SP = 2 as well, and the answer is (D).
Geometry 89
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
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
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)12
2
(E)18
A
D
3
O
3
C B
90GRE 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
y° t°
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z° s°
m
x°
5.What is the value of y in the figure?
(A)20
(B)30
(C)35
(D)45
(E)50
3a°
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6.In the triangle, what is the value of x ?
(A)25
(B)55
(C)60
(D)77
(E)85
y° + x°
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60° + x° |