390
11.2 Pythagorean Theorem
Introduction
The Pythagorean Theorem is a famous theorem in Mathematics. It is named after a Greek philosopher
and mathematician, Pythagoras, who discovered1 it thousands of years ago. It describes a special
relationship between the lengths of the three sides of a right-angle triangle. The theorem states that
the squares of the lengths of the two shorter sides that meet at the right-angle (called the legs of the
right triangle) equal the square of the longest side opposite the right-angle (called the hypotenuse of
the right triangle).
A
c
Hypotenuse
b
90°
legs
C
In a right-triangle, for example, ABC with the right angle
at C, the Pythagorean Theorem is written as an equation
relating the lengths of the sides of the right-triangle a, b,
and c, where a and b represent the legs, and c represents the
hypotenuse, as follows:
B
a
a2 + b2 = c2
Using this equation, if the lengths of both legs (a and b) are known, then the hypotenuse (c) can be
calculated as follows:
c = a 2 + b2
12
Similarly, if the lengths of the hypotenuse (c) and one leg
(a or b) are known, then the length of the other leg can be
calculated, as follows:
a = c -b
a=3
b=4
c=5
11
10
9
8
a = c 2 – b2 or b = c 2 – a2
c 2 = a 2 + b2
7
A set of positive integers that satisfies the Pythagorean
Theorem are known as Pythagorean triples. For example,
the integers 3, 4, and 5 are Pythagorean triples.
b2
6
5
4
3
32 + 42 = 52
a2
2
(9 + 16 = 25)
1
0
1
2
3
4
5
6
7
8
9
10 11 12 13
Exhibit 11.2-a Pythagorean Theorem
Some of the other Pythagorean triples are: (5, 12, 13),
(7, 24, 25), (8, 15, 17), (9, 40, 41), (12, 35, 37), (20, 21, 28),…
Proofs of the Pythagorean Theorem
The theorem has numerous proofs. In this section, three proofs of the theorem will be discussed.
Proof Using Similar Triangles
A
y
θ
For this proof of the Pythagorean Theorem, start with a
right-triangle, ABC, with legs, a and b, and hypotenuse, c.
Let ∠A = θ and ∠B = ϕ [where θ + ϕ = 90°]
x
b
90°
C
Draw CD = to AB
c
D
a
Let BD = x and AD = y [where x + y = c]
ϕ
B
Exhibit 11.2-b Proof of Pythagorean Theorem using Similar Triangles
1
Chapter 11 | Basic Trigonometry
A
lthough Pythagoras is credited with the discovery of the theorem, there is evidence to suggest that it was known by the ancient Babylonians,
over 1,000 years prior to Pythagoras.
391
As shown in Exhibit 11.2-b,
A
=
∆ABC and ∆CBD are similar.
c a
=
a x
Therefore,
D
θ
c
b
i.e., a2 = cx ϕ
C
=
∆ABC and ∆ACD are similar.
b y
Therefore, =
c b
i.e., b = cy
A
and
B
a
y
D
θ
c
b
ϕ
θ
C
A
C
Adding
B
a
θ
2
x
b
ϕ
ϕ
B
a
C
,
a2 + b2 = cx + cy
a2 + b2 = c(x + y) = c(c) = c2
Therefore, a2 + b2 = c2.
Proof Using a Geometric Construction
For this proof of the Pythagorean Theorem, start with a right triangle with legs, a and b, and
hypotenuse, c. Then, three additional copies of the triangle are created and lined up tip-to-tip so that
their boundary forms a square, as shown in Exhibit 11.2-c.
The outer boundary forms a square, as all angles are rightangles and all side lengths are equal to (a + b).
Also, the inner boundary forms a square, as all angles are
right-angles (by the IAT-Part 2 from Chapter 10) and all
side lengths are equal to c.
a
a ×b 
(a + b ) = c + 4 
 2 
2
2
2
a + 2ab + b = c + 2ab
2
θ + ϕ = 90°
ϕ
b
a
Area of larger square = Area of smaller square + Area of
each of the 4 triangles.
2
c
θ
b
ϕ
θ
c
ϕ
2
a+b
ϕ
a +b =c
2
a
θ
c
b
2
c
Therefore, a + b = c .
2
2
b
2
a
c
θ
ϕ
θ
a
b
a+b
Exhibit 11.2-c P roof of Pythagorean Theorem using
a Geometric Construction
11.2 Pythagorean Theorem
392
Proof Using Congruent Triangles and Areas
For this proof of the Pythagorean Theorem, first draw ∆ABC, so that point C is on the right-angle
and label the opposite sides using the same, small-case letters, so that c is the hypotenuse and a and
b represent the legs. Then, draw squares on each side of the triangle. Following this, draw a line from
point C to the far side of the square on side c, perpendicular to AB, which intersects AB at ‘K’. Label the
diagram, as in Exhibit 11.2-d.
F
H
Area 1
G
C
b
A
Area 3
c
Area 2
K
C
a
b
a
J
b
A
B
Area 1 = Area 2
Area 3 = Area 4
Area 4
a
K
c
B
c
E
D
L
Exhibit 11.2-d Proof of Pythagorean Theorem using Congruent Triangles and Areas
Join BG and CE.
h2 = AK
∆ABG and ∆ACE are congruent, (SAS Property).
Area of ∆ABG = Area of ∆ACE 1
1
AG . h1 = AE . h2
2
2
1
1
AG . AC = AE . AK
2
2
b
1
using Area of ∆ = Base × Height,
2
A
G
c
b
A
1
1
Area 1 = Area 2
2
2
B
c
E
h1 = AC
Area 1 = Area 2. Similarly, it can be proven that Area 3 = Area 4. Adding
+
, we get:
Area 1 + Area 3 = Area 2 + Area 4
i.e., b2 + a2 = c2
Therefore, a2 + b2 = c2.
Determining the Unknown Length of One Side of a Right Triangle
If the lengths of any two sides of a right triangle are given, the Pythagorean Theorem can be used to
determine the length of the third side.
If the length of the hypotenuse is unknown, use the re-arrangement of the formula, c2 = a2 + b2 , and
perform the square root on both sides to determine c.
Similarly, if the length of the hypotenuse is known, the formula can be rearranged from a2 + b2 = c2 to
a2 = c2 − b2 (or b2 = c2 − a2) and the square root can be performed on both sides to determine a or b.
Example 11.2-a
Calculating the Length of the Hypotenuse of a Right Triangle
Using the Pythagorean Theorem, calculate the length (rounded to the nearest hundredth, as needed)
of the hypotenuse, c, of the following right triangles, given the lengths of the two legs, a and b.
(i) a = 3 m and b = 4 m
Chapter 11 | Basic Trigonometry
(ii) a = 10 cm and b = 12 cm
C
393
Solution
(i) Using a2 + b2 = c2,
c2 = a2 + b2 = 32 + 42 = 9 + 16 = 25
A
c
b=3
c == 25 ==55m
m
C
(ii) Using a2 + b2 = c2,
c2 = a2 + b2 = 102 + 122 = 100 + 144 = 244
B
A
c = 244 = 15.620499... ≈ 15.62 cm
c
b = 12
C
Example 11.2-b
a=4
a = 10
B
Calculating the Length of One of the Legs of a Right-Triangle
Using the Pythagorean Theorem, calculate the length (rounded to the nearest hundredth, as needed)
of the missing leg, of the following right triangles, given the lengths of the hypotenuse, c, and one leg:
(i) a = 5 cm and c = 13 cm
Solution
(ii) b = 8 m and c = 16 m
(i) Using a2 + b2 = c2
b2 = c2 − a2 = 132 − 52 = 169 – 25 = 144
A
b = 144 = 12 cm
c = 13
b
C a=5 B
(ii) Using a2 + b2 = c2
a2 = c2 − b2 = 162 – 82 = 256 – 64 = 192
A
a = 192 = 13.856406... ≈ 13.86 m
c = 16
b = 13
a
C
Example 11.2-c
B
Calculate the Unknown Lengths (x and y) in the Following Diagrams
In the following figures, calculate the lengths of x and y (rounded to the nearest tenth, as needed):
(i)
y
D
6m
B
Solution
(i)
(ii)
A
D
6m
B
y
17 m
B x
10 m
x
4m
D
40 m
4m
C
41 m
A
5m
D
y B
x
10 m
C
C
10 m
x
(iii)
A
C
I n the right triangle BCD, using the Pythagorean Theorem,
x2 + 62 = 102
x2 = 102 − 62
= 100 − 36 = 64
x = 64 = 8 m
11.2 Pythagorean Theorem
394
In the right triangle ABC, using the Pythagorean Theorem,
(y + 6)2 + x2 = 172
A
Solution
continued
(y + 6)2 = 172 − 82
17 m
y+6m
= 289 − 64 = 225
(y + 6) = 225 = 15
B
x=8m
C
y = 15 − 6 = 9 m
Therefore, x = 8 m and y = 9 m.
(ii)
In the right triangle ABC, using the Pythagorean Theorem,
y2 + 402 = 412
A
40 m
y
B
C
41 m
4m
B x
y = 81 = 9 m
In the right triangle ABD, using the Pythagorean Theorem,
x2 + 42 = 92
A
y=9m
y2 = 412 − 402
= 1,681 − 1,600 = 81
x2 = 92 − 42
= 81 − 16 = 65
D
x = 65 = 8.062257... ≈ 8.1 m
Therefore, x = 8.1 m and y = 9 m.
(iii)
In the right triangle ADB, using the Pythagorean Theorem,
y2 + 42 = 52
A
4m
5m
D
y B
y2 = 52 − 42
= 25 − 16 = 9
y=
In the right triangle ADC, using the Pythagorean Theorem,
42 + 132 = x2
A
x
4m
D
3 + 10 = 13 m
9 =3m
x2 = 16 + 169 = 185
C
x = 185 = 13.601470... ≈ 13.6 m
Therefore, x = 13.6 and y = 3 m.
Calculating the Distance Between Two Points
In Chapter 9, the concept of the distance between two points was introduced; however, it was limited
to points that are on the same vertical line (sharing the same x-coordinate) or horizontal line (sharing
the same y-coordinate). If the two points share neither the same x-coordinate nor y-coordinate, the
calculation becomes more difficult. Certainly, the distance is at most, the sum of the horizontal and
vertical distances between the two points, but there is a shorter distance – the line segment joining
the two points.
In this section, the method to calculate the shortest distance between two points having coordinates
P(x1 , y1) and Q(x2 , y2) will be demonstrated.
Chapter 11 | Basic Trigonometry
395
Y
Q = (x2, y2)
y2
d
Every horizontal line and every vertical line meet at a
right-angle. Therefore, the shortest distance between
two points is related to the horizontal and vertical
distances between the points by the Pythagorean
Theorem. This forms the equation for the shortest
distance:
d2 = (∆x)2 + (∆y)2,
∆y = (y2 – y1)
y1
P = (x1, y1)
∆x = (x2 – x1)
X
x2
x1
where ∆x is the horizontal distance between the two
points and ∆y is the vertical distance between the
two points.
Exhibit 11.2-e Euclid’s Proof of the Pythagorean Theorem using Congruent Triangles
i.e., the equation for the shortest distance, d, between the two points P(x1, y1) and Q(x2, y2) is:
d2 = (x2 − x1)2 + (y2 − y1)2
Performing the square root on both sides, the shortest distance, d, between the two points P(x1, y1) and Q(x2, y2) is:
d = ( x 2 – x1) 2 + ( y2 – y1) 2
Example 11.2-d
Calculating the Distance Between Two Points in the Cartesian Plane
Calculate the distance (rounded to the nearest tenth of a unit, as needed) between the following points:
(i) A(2, 1) and B(7, 8)
(ii) P(–3, 7) and Q(3, –1)
(iii)X(4.5, –1.2) and Y(–7.3, 2.8)
Solution
Using d2 = (x2 – x1)2 + (y2 – y1)2
(i) d2 = (7 – 2)2 + (8 – 1)2
= 52 + 72 = 25 + 49 = 74
d = 74 = 8.602325... ≈ 8.6 units
(ii) d2 = [3 –(–3)]2 + [(–1) – 7]2
= 62 + (–8)2 = 36 + 64 = 100
d = 100 = 10 units
(iii) d2 = [(–7.3) – 4.5]2 + [2.8 –(–1.2)]2 = (–11.8)2 + (4.0)2 = 139.24 + 16 = 155.24
d = 155.24 = 12.459534... ≈ 12.5 units
Applications of the Pythagorean Theorem
Example 11.2-e
Calculating the Distance Between Two Cities
Toronto is 45 km north and 26 km east of Hamilton. Find the shortest flying distance (rounded to the
nearest kilometre) between the two cities.
Solution
Toronto
45 km
d
26 km
Hamilton
Using the Pythagorean theorem,
d2 = 452 + 262
= 2,025 + 676
= 2,701
d = 2,701
= 51.971145...
≈ 52 km
Therefore, the shortest flying distance between the two cities is 52 km.
11.2 Pythagorean Theorem
396
Example 11.2-f
Calculating the Length of a Guy Wire
A guy wire is tied to an antenna tower 12 m above the ground and the other end of the guy wire is
tied to the ground 15 m away. Determine the length of the guy wire, rounded to the nearest tenth of
a metre.
Solution
Let l be the length of the guy wire. Using the Pythagorean
Theorem, l2 = 152 + 122
= 225 + 144 = 369
Antenna
Tower
l = 369
= 19.209372...
≈ 19.2 m
Therefore, the length of the guy wire is 19.2 m.
Example 11.2-g
12 m

Guy wire
15 m
Ground
Determining the Dimensions of a Television
A 42″ television, with a length to height ratio 16 : 9, measures 42 inches across the diagonal. Find the
length and the height of the TV, to the nearest tenth of an inch.
Solution
Since the ratio of the length of the TV to the height of the TV is 16 : 9, let 16x represent the length
of the TV and 9x represent the height of the TV.
Using the Pythagorean Theorem,
422 = (16x)2 + (9x)2
1,764 = 256x2 + 81x2
1,764 = 337x2
x2 =
s
che
in
42
9x
16x
1,764
= 5.234421...
337
x = 5.234421... = 2.287885...
Length of the TV = 16x = 16(2.287885...) = 36.606172... ≈ 36.6 inches
Height of the TV = 9x = 9(2.287885...) = 20.590972... ≈ 20.6 inches
Therefore, the length of the TV is 36.6 inches and the height is 20.6 inches.
Example 11.2-h
Calculating the Height, Surface Area, and Volume of a Pyramid
The Great Pyramids of Giza in Egypt have certain special properties: the ratio of the slant height
(S) of the pyramid to the semi-base (b1) of the pyramid adheres to the “Golden Ratio,” which is
approximately 1.618 : 1.
Jorge wishes to know the height (h) and volume (V) of the
largest pyramid. If he measures the length of one side of the
base to be 230 m, find the height of the pyramid, rounded to
the nearest metre. Then, find the surface area and volume of
the pyramid, rounded to the nearest square metre and cubic
metre, respectively.
h
S
b1
230
Chapter 11 | Basic Trigonometry
230
397
Solution
Since the base is 230 m, bb1 =
230
= 115 m.
2
Step 1: Calculate the slant height of the pyramid, s, using the Golden Ratio
S = 1.618(115) = 186.07 ≈ 186 m,
Therefore, the slant height of the pyramid is 186 m.
S
= 1.618,
b1
Step 2: Calculate the height of the pyramid using the Pythagorean Theorem.
S2 = b12 + h2
(186)2 = (115)2 + (h)2
34,596 = 13,225 + h2
h2 = 21,371
h = 21,371 = 146.188234... ≈ 146 m
Therefore, the height of the pyramid is 146 m.
Step 3: C
alculate the surface area of the 4 equal triangular sides using slant height s = 186 m and
b = 230 m.
 (230)(186) 
SA = 4 × 
= 85,560 m2

2


Step 4: C
alculate the volume of the pyramid using the formula V =
pyramid.
V=
b2 × h
for a square-based
3
(230) 2 (146)
≈ 2,574,
467 m3 ≈ 2,574,467 m3
=
2,574,466.667
3
Therefore, the surface area of the pyramid is 85,560 m2 and the volume of the pyramid is
2,574,467 m3.
11.2 Exercises
Answers to odd-numbered problems are available at the end of the textbook.
In Problems 1 to 4, use the Pythagorean Theorem to determine the length of the missing side in the given right-angled triangles,
where a and b represent the legs of the triangle and c represents the hypotenuse of the triangle. Express the answers rounded to the
nearest hundredth, wherever applicable.
1.
3.
a.
b.
c.
a
15 cm
2.5 cm
?
b
20 cm
?
6 cm
c
?
6.5 cm
6.25 cm
2.
a.
b.
c.
a
12 cm
8 cm
?
b
15 cm
?
15 cm
c
?
17 cm
16 cm
4.
a.
b.
c.
a
7 cm
7.5 cm
?
b
24 cm
?
20 cm
c
?
12.5 cm
20.5 cm
a.
b.
c.
a
16 cm
20 cm
?
b
18 cm
?
17 cm
c
?
29 cm
23 cm
11.2 Pythagorean Theorem
398
In Problems 5 to 10, calculate the length of the missing side for each of the diagrams:
5.
6.
1.5m
2.2m
1.75m
x
7.
x
14cm
16cm
8.
11m
5m
8m
x
x
1 cm
9.
10.
x
7 cm
4m
x
10 cm
4m
3m
In Problems 11 to 14, calculate the perimeter and area of each of the given diagrams:
11.
12.
37 cm
71°
35 cm
60°
71°
60°
15 cm
13.
14.
1.8 m
6.4 m
2x
3.6 m
x
In Problems 15 to 22, calculate the length of the line segments joining the pairs of points.
15. A(-2,5) and B(4,7)
16. C(-6,1) and D(2,5)
17. E(4,-3) and F(-1,5)
18. G(1,-6) and H(-4,3)
19. M(-3,-3) and N(-7,2)
20. P(-4,-1) and Q(3,-5)
21. S(0,4) and T(-3,0)
22. U(2,0) and V(0,-6)
Chapter 11 | Basic Trigonometry
399
23. A laptop screen measures 31 cm long by 17.5 cm high. Determine the diagonal length of the computer screen, rounded
to the nearest tenth of a cm.
24. From a point ‘X’, a person walked 850 m due west and then turned and walked for another 1.7 km due south to reach
point ‘Y’. Calculate the shortest distance between X and Y, rounded to the nearest hundredth of a km.
25. A 2.5 m tent pole is secured using a 4.3 m long guy rope from the top of the pole. How far away from the base of the
pole will the rope need to be secured to the ground, assuming it is pulled taut?
26. A 5 m ladder is leaned up against a wall. If the base of the ladder is placed on the ground 1.7 m away from the wall, how
high up against the wall will the ladder reach, rounded to the nearest tenth of a metre?
27. A skateboard ramp that is 3.5 m long is built with a slope of 3/5. Determine the maximum height of the ramp, rounded
to the nearest cm.
28. A wheelchair ramp is to be constructed to the top of a set of stairs that is 1.75 m tall, with a maximum slope of 1/12.
Determine the minimum ramp length required, in order for the ramp to be built according to specifications. Can you
suggest a way to build the ramp that would save space?
29. A towel rack that is 1 m long is to be placed in a box measuring 75 cm × 60 cm × 45 cm. Will the towel rack fit along the
diagonal at the bottom of the box? Will it fit in the box if placed on the 3-dimensional diagonal?
30. Will a 16-foot-long piece of lumber fit in a truck with interior cargo dimensions of 12.5 feet by 8 feet by 7.5 feet?
In Problems 31 to 36, calculate the perpendicular height (to the nearest tenth), surface area (to the nearest whole number), and
volume (to the nearest whole number) of the solids.
31. A cone with a base diameter of 24 cm and a slant height of 30 cm.
30 cm
24 cm
32. A cone with a base diameter of 64 mm and a slant height of 105 mm.
64 mm
105 mm
33. A square pyramid with a base length and corner edge length all equal to 98 m.
98 m
98 m
98 m
34. A
pyramid with a rectangular base that has a length of 50 cm and a width of
36 cm, and a corner edge length of 45 cm.
45 cm
50 cm
35. A
truncated cone with a top diameter of 24 cm, a base diameter of 40 cm,
and a slant height of 18 cm.
36. A
truncated square pyramid with a top side length of 75 m, a base side length
of 225 m, and a slant height of 120 m.
36 cm
24 cm
18 cm
40 cm
75 m
120 m
225 m
11.2 Pythagorean Theorem