Py thagoras’ T heorem
PYTHAGORAS’
THEOREM
www.mathletics.com.au
How does it work?
Solutions
Pythagoras’ Theorem
Page 3 questions
Right-angled triangles
1
E
D
b
a
y
x
z
Hypotenuse is side: y
F
Hypotenuse is side: DF
Q
c
d
k
j
l
Hypotenuse is side: k
P
R
Hypotenuse is side: PQ
2
Name the hypotenuse for each of these badly drawn triangles.
M
b
a
c
a
L
b
N
Hypotenuse is side: a
Hypotenuse is side: MN
Pythagoras’ Theorem Solutions
Mathletics Passport
© 3P Learning
I
7
SERIES
TOPIC
1
How does it work?
Solutions
Pythagoras’ Theorem
Page 5 questions
Squares and right-angled triangles
1
13 units
Area 1 = 5 units
3
5 units
5 units = 25 units2
#
Area 2 = 12 units
#
12 units = 144 units2
Area 3 = 13 units
#
13 units = 169 units2
1
Area 1 + Area 2 = 25 units2 + 144 units2
2
= 169 units2
= Area 3
12 units
2
3
10 units
Area 1 = 6 units
#
6 units = 36 units2
Area 2 = 8 units
#
8 units = 64 units2
1
6 units
2
Area 3 = 10 units
#
10 units = 100 units2
Area 1 + Area 2 = 36 units2 + 64 units2
8 units
= 100 units2
= Area 3
2
I
7
SERIES
TOPIC
Pythagoras’ Theorem Solutions
Mathletics Passport
© 3P Learning
How does it work?
Solutions
Pythagoras’ Theorem
Page 7 questions
Pythagoras’ Theorem for right-angled triangles
1
b
a
25
15
9
14
20
12
92 + 122 = 81 + 144
= 225

142 + 202 = 196 + 400
= 596
152 = 225
Right-angled
Not right-angled
Right-angled
252 = 625
Not right-angled

d
c
1.2
3.4
7.1
3.5
3.7
9.6
7.12 + 3.42 = 50.41 + 11.56
= 61.97
Right-angled

9.62 = 92.16
1.22 + 3.52 = 1.44 + 12.25
= 13.69
Not right-angled

Right-angled
3.72 = 13.69
Not right-angled
f
e
21
20
24
25
29
7
202 + 212 = 400 + 441
= 841

Right-angled
292 = 841
Not right-angled
242 + 72 = 576 + 49
= 625
252 = 625
Right-angled
Not right-angled

Pythagoras’ Theorem Solutions
Mathletics Passport
© 3P Learning
I
7
SERIES
TOPIC
3
How does it work?
Solutions
Pythagoras’ Theorem
Page 8 questions
Pythagoras’ Theorem for right-angled triangles
A
2
J
10
16

20
K
10.5

H
14.5
I
102 + 10.52 = 14.52
210.25 = 210.25
12
J
122 + 162 = 202
400 = 400
B
M
K
29
52
25
20
L

48
48


G
21
A
15
H
C
202 + 152 = 242
625 ! 576
N
292 + 212 = 482
1282 ! 2304
The right-angled triangles are: ΔAJK , ΔHIJ , ΔGHK
4
I
7
SERIES
TOPIC
20
Pythagoras’ Theorem Solutions
Mathletics Passport
© 3P Learning
482 + 202 = 522
2704 = 2704
How does it work?
Solutions
Pythagoras’ Theorem
Page 8 questions
Pythagoras’ Theorem for right-angled triangles
Earn an awesome passport with this one! Name all the right-angled triangles in this image and mark
where the right-angles are with the correct symbol.
3
65
R
S
15
36
P
ΔPUV
122 + 162 = 202
400 = 400
ΔQRU
152 + 362 = 392
1521 = 1521
ΔRSU
392 + 522 = 652
4225 = 4225
ΔSTU
242 + 522 = 3280
3280 = 3280
3280
T
5
The right-angled triangles are:
52
24
15
U
Q
16
12
2
V
Page 9 questions
Pythagoras’ Theorem for right-angled triangles
Assuming the scale of the page is the same as the original print, the measurements should be as follows:
NOTE: if not the same scale, the same relationship between your measurements should work.
a
a2
b
b2
c
c2
a2 + b2
1
45 mm
2025
60 mm
3600
75 mm
5625
2025 + 3600 = 5625
2
70 mm
4900
24 mm
576
74 mm
5476
4900 + 576 = 5476
3
15 mm
225
36 mm
1296
39 mm
1521
225 + 1296 = 1521
4
36 mm
1296
77 mm
5929
85 mm
7225
1296 + 5929 = 7225
5
9 mm
81
40 mm
1600
41 mm
1681
81 + 1600 = 1681
6
40 mm
1600
42 mm
1764
58 mm
3364
1600 + 1764 = 3364
Pythagoras’ Theorem Solutions
Mathletics Passport
© 3P Learning
I
7
SERIES
TOPIC
5
Where does it work?
Solutions
Pythagoras’ Theorem
Page 11 questions
Calculating the length of the hypotenuse
1
c2 = 62 + 82
a
` c2 = 36 + 64
` g2 = 64 + 225
` c2 = 100
` g2 = 289
` c =
` g =
100
c2 = 52 + 122
a
289
` g = 17
` c = 10
2
g2 = 82 + 152
b
d2 = 1.22 + 1.62
b
` c2 = 25 + 144
` d2 = 1.44 + 2.56
` c2 = 169
` d2 = 4
` c =
` d =
169
` c = 13
4
` d = 2
Page 12 questions
Calculating the length of the hypotenuse
3
h2 = 1.12 + 6.02
a
` h2 = 1.21 + 36
` n2 = 144 + 1225
` h2 = 37.21
` n2 = 1389
` h =
4
37.21 in exact square root form
c2 = (10 units) 2 + (9 units) 2
a
` n =
1389 in exact square root form
p2 = (5.9 units) 2 + (3.4 units) 2
b
` c2 = 100 units2 + 81 units2
` p2 = 34.812 units2 + 11.56 units2
` c2 = 181 units2
` p2 = 46.37 units2
` c =
6
n2 = 122 + 352
b
181 units
` p =
46.37 units
` c = 13.45362405... units
` p = 6.809552114... units
` c . 13.45 units to 2 decimal places
` p . 6.81 units to 2 decimal places
I
7
SERIES
TOPIC
Pythagoras’ Theorem Solutions
Mathletics Passport
© 3P Learning
Where does it work?
Solutions
Pythagoras’ Theorem
Page 13 questions
Calculating the length of the hypotenuse
Stage 2
Stage 3
d2 = 402 + 1982
d2 = 392 + 2522
d2 = 362 + 3602
` d2 = 1600 + 39204
` d2 = 1521 + 63504
` d2 = 1296 + 129600
` d2 = 40804
` d2 = 65025
` d2 = 130896
` d =
` d =
` d =
Stage 1
5
40804
` d = 202m
65025
` d = 255m
130896
` d = 361.7955224... m
` d . 361.8m to 2 decimal places
` The total length of the 3 stage flight path . 202 + 255 + 361.8 m
. 818.8 m
Page 15 questions
Calculating the length of a short side
1
a
a2 = 262 - 242
` a2 = 676 - 576
` b2 = 72.25 - 1.69
` a2 = 100
` b2 = 70.56
` a =
` b =
100
` a = 10
2
a
b2 = 8.52 - 1.32
b
70.56
` b = 8.4
j2 = 702 - 562
b2 = (18.1 units) 2 - (18 units) 2
b
` j2 = 4900 - 3136
` b2 = 327.61 units2 - 324 units2
` j2 = 1764
` b2 = 3.61 units
` j =
` b =
1764
3.61 units
` b = 1.9 units
` j = 42
Pythagoras’ Theorem Solutions
Mathletics Passport
© 3P Learning
I
7
SERIES
TOPIC
7
Where does it work?
Solutions
Pythagoras’ Theorem
Page 16 questions
Calculating the length of a short side
3
b2 = 172 - 112
a
` b2 = 289 - 121
` w2 = 203.0625 + 138.0625
` b2 = 168
` w2 = 65
` b =
4
8
w2 = 14.252 + 11.752
b
168 in exact square root form
y2 = 13.82 - 8.32
a
` w =
65 in exact square root form
x2 = 41.082 - 23.422
b
` y2 = 190.44 - 68.89
` x2 = 1687.5664 - 548.4968
` y2 = 121.55
` x2 = 1139.07
` y =
` x =
121.55
1139.07
` y = 11.02497166 ...
` x = 33.7501
` y . 11.0 to 1 decimal point
` x . 33.8 to 1 decimal point
I
7
SERIES
TOPIC
Pythagoras’ Theorem Solutions
Mathletics Passport
© 3P Learning
Where does it work?
Solutions
Pythagoras’ Theorem
Page 17 questions
Combination of hypotenuse and short side calculations
The special name given a right-angled triangle which is exactly one half of an equilateral triangle:
H
=
E
M
I
E
Q
triangle
123456
c
15
14.12
20
545
2
a
2.9
544
4
42
M
I
73.4
42.1
b
d
E
5
41.9
25
Q
53.2
e
32
3
67.7
E
68
6
1
h
20
33
H
14.16
30
g
60
67
Pythagoras’ Theorem Solutions
Mathletics Passport
© 3P Learning
I
7
SERIES
TOPIC
9
Where does it work?
Solutions
Pythagoras’ Theorem
Page 19 questions
Applications of Pythagoras’ Theorem
1
x2 = (13 m) 2 - (12 m) 2
` x2 = 169 m2 - 144 m2
13 m
` x2 = 25 m2
12 m
` x =
x
25 m
` x = 5m
42 cm
2
cut2 = (42 cm) 2 + (34 cm) 2
34 cm
` cut2 = 1764 cm2 + 1156 cm2
cut
` cut2 = 2920 cm2
` cut =
2920 cm
` cut = 54.03702434 cm
` cut . 54 to nearest whole cm
1.7 km
3
Start
(i)
d2 = (1.7 km) 2 + (3.9 km) 2
` d2 = 2.89 km2 + 15.21 km2
` d2 = 18.1 km2
3.9 km
` d =
18.1 km
` d = 4.254409477 ... km
d
` d . 4.25 km to 2 decimal points
Finish
(ii) To avoid the swamp, Mila walked 3.9 km + 1.7 km = 5.6 km
` Mila walked a further 5.6 km - 4.25 km . 1.35 km
10
I
7
SERIES
TOPIC
Pythagoras’ Theorem Solutions
Mathletics Passport
© 3P Learning
Where does it work?
Solutions
Pythagoras’ Theorem
Page 20 questions
Applications of Pythagoras’ Theorem
4
(i)
17 m
2.6 m
base2 = (17 m) 2 - (2.6 m) 2
` base2 = 289 m2 - 6.76 m2
` base2 = 282.24 m2
` base =
base
282.24 m
` base = 16.8 m
(ii)
Area = (base
#
height) ' 2
` Area = (16.8 m # 2.6 m) ' 2
` Area = 43.68 m2 ' 2
` Area = 21.84 m2
C
5
AB2 = (18 m) 2 + (3.3 m) 2
` AB2 = 324 m2 + 10.89 m2
` AB2 = 334.89 m2
B
54.4 m
` AB =
3.3 m
A
18 m
334.89 m
` AB = 18.3 m
BC2 = (54.4 m) 2 + (3.3 m) 2
Diagram not drawn to scale.
` BC2 = 2959.36 m2 + 10.89 m2
` BC2 = 2970.25 m2
` BC =
2970.25 m
` BC = 54.5 m
Distance around wall = 79m
Distance AC = AB + BC
= 18.3 m + 54.5 m
= 72.8 m
Shortest path
Pythagoras’ Theorem Solutions
Mathletics Passport
© 3P Learning
I
7
SERIES
TOPIC
11
Where does it work?
Solutions
Pythagoras’ Theorem
Page 20 questions
Applications of Pythagoras’ Theorem
6
(i)
172 cm
y2 = (172 cm - 137 cm) 2 + (120 cm) 2
y2 = (35 cm) 2 + (120 cm) 2
120 cm
y
` y2 = 1225 cm2 + 14400 cm2
` y2 = 15625 cm2
` y =
137 cm
15625 cm
` y = 125 cm
(ii) Perimeter of the trapezium = 172 cm + 120 cm + 137 cm + 125 cm
= 554 cm
Page 21 questions
Applications of Pythagoras’ Theorem
7
WY2 = YZ2 - WZ2
Y
65
= 342 - 162
= 63 - 30
= 4225 - 256
= 1156 - 256
= 33
= 3969
= 900
3969
WY = 63
` WX =
12
16
I
7
SERIES
TOPIC
Pythagoras’ Theorem Solutions
Mathletics Passport
900
WX = 30
34
W
XY = WY - WX
= 652 - 162
` WY =
X
WX2 = XZ2 - WZ2
© 3P Learning
Where does it work?
Solutions
Pythagoras’ Theorem
Page 21 questions
Applications of Pythagoras’ Theorem
8
Calculate
the length of the cable support BD on the crane picture below if CD = 9.5 m, AB = 6 m
and BC = 18.5 m
C
18.5 m
B
9.5 m
6 m
D
A
AC2 = BC2 - AB2
AD = AC - DC
BD2 = AD2 + AB2
= 18.52 - 62
= 17.5 - 9.5
= 82 + 62
= 342.25 - 36
= 8m
= 64 + 36
= 306.25
` AC =
= 100
` BD =
306.25
AC = 17.5 m
BD = 10 m
Pythagoras’ Theorem Solutions
Mathletics Passport
100
© 3P Learning
I
7
SERIES
TOPIC
13
What else can you do?
Solutions
Pythagoras’ Theorem
Page 23 questions
Pythagorean triads
1
a
26
b
20
12
10
24
16
" 12 , 16 , 20 ,
" 10 , 24 , 26 ,
psst! Note that they are written in order of size.
35
d
c
9
41
12
37
40
" 12 , 35 , 37 ,
2
" 9 , 40 , 41 ,
Show whether these sets of positive integers form a Pythagorean triad or not.
a
" 7 , 24 , 25 ,
b
" 14 , 48 , 50 ,
142 + 482 = 502 ?
122 + 342 = 362 ?
49 + 576 = 625?
196 + 2304 = 2500?
144 + 1156 = 1296?

2500 = 2500
Yes

No
" 15 , 36 , 39 ,
e
Yes
1300 ! 1296
No
" 16 , 60 , 63 ,
Yes
f

2
162 + 602 = 632 ?
122 + 302 = 31 ?
225 + 1296 = 1521?
256 + 3600 = 3969?
144 + 900 = 961?

Yes
3856 ! 3969
I
7
SERIES
TOPIC
No
Yes

1044 ! 961
No
Pythagoras’ Theorem Solutions
Mathletics Passport
© 3P Learning
No
" 12 , 30 , 31 ,
152 + 362 = 392 ?
1521 = 1521
14
" 12 , 34 , 36 ,
72 + 242 = 252 ?
625 = 625
d
c
Yes

No
What else can you do?
Solutions
Pythagoras’ Theorem
Page 25 questions
Euclid’s formula for Pythagorean triads
1
q
p2 - q2
2pq
p2 + q2
Triad
2
1
22 - 12 = 3
2 # 2 #1 = 4
22 + 12 = 5
{3,4,5}
3
1
32 - 12 = 8
2 # 3 #1 = 6
32 - 12 = 10
{ 6 , 8 , 10 }
5
2
52 - 22 = 21
2 # 5 # 2 = 20
52 + 22 = 29
{ 20 , 21 , 29 }
7
6
72 - 62 = 13
2 # 7 # 6 = 84
72 + 62 = 85
{ 13 , 84 , 85 }
11
3
112 - 32 = 112
2 # 11 # 3 = 66
112 + 32 = 130
{ 66 , 112 , 130 }
21
18
212 - 182 = 117
2 # 21 # 18 = 756
212 + 182 = 756
{ 117 , 756 , 765 }
p
2
(i) Find a Pythagorean triad in which p = 7 and p2 - q2 is equal to 33
p2 - q2
72 - q2
49 - q2
49 - 33
16
` q
=
=
=
=
=
=
33
33
33
q2
q2
4
2pq = 2 # 7 # 4
= 56
p2 + q2 = 72 + 42
= 65
Pythagorean triad is { 33 , 56 , 65 }
(ii) Find a Pythagorean triad in which q = 5 and p2 + q2 is equal to 61
p2 - q2
p2 - 52
p2 - 25
p2
p2
` p
=
=
=
=
=
=
61
61
61
61 - 25
36
6
2pq = 2 # 6 # 5
= 60
p2 + q2 = 62 - 52
= 11
Pythagorean triad is { 11 , 60 , 61 }
Pythagoras’ Theorem Solutions
Mathletics Passport
© 3P Learning
I
7
SERIES
TOPIC
15
What else can you do?
Solutions
Pythagoras’ Theorem
Page 26 questions
Pythagorean triads
3 Find a group of three integers that includes the number 14 and forms a Pythagorean triad.
{ p2 - q2 , 2pq , p2 + q2 }
2#
2pq = 14
p # q = 14
p # q= 7
p = 7 and q = 1
` p2 + q2 = 72 + 12
= 50
(p 2 q)
` p2 - q2 = 72 - 12
= 48
Pythagorean triad is: { 14 , 48 , 50 }
{ p2 - q2 , 2pq , p2 + q2 }
4
small integer
other small integer
largest integer
hint: Pythagorean triads can be made using positive integers only.
Formal explanation:
From hint, Pythagorean triads are made using positive integers. ie. positive whole numbers only.
One of the smaller integers is found using p 2 - q 2
If the value of p was smaller than the value of q, then the answer would be negative. So this
could not be used because only positive whole numbers are allowed.
Showing using chosen values p = 2 and q = 1 :
When p = 2 and q = 1 , p 2 - q 2 = 2 2 - 1 2 = 3 (this is a positive integer and is allowed)
If we swap these around, so p = 2 and q = 2
p 2 - q 2 = 1 2 - 2 2 = - 3 (this is a negative integer and is not allowed)
This will always happen if the value of p is smaller than the value of q when using Euclid’s formula.
Negative numbers are not allowed because each integer represents the length of the side of a
right-angled triangle. So a side length of -3 does not make sense.
16
I
7
SERIES
TOPIC
Pythagoras’ Theorem Solutions
Mathletics Passport
© 3P Learning
What else can you do?
Solutions
Pythagoras’ Theorem
Page 28 questions
Wheel of Theodorus
2
2
2
12
8
2
16
20
2
2
2
24
and so on
Pythagoras’ Theorem Solutions
Mathletics Passport
© 3P Learning
I
7
SERIES
TOPIC
17
Jigsaw
What else
Puzzle
can you do?
Solutions
Pythagoras’ Theorem
Page 31 questions
Squares and right-angled triangles: Jigsaw Puzzle
2
1
55
4
3
5
3
4
1
2
18
I
7
SERIES
TOPIC
Pythagoras’ Theorem Solutions
Mathletics Passport
© 3P Learning
Pythagoras’ Theorem
Notes
Pythagoras’ Theorem Solutions
Mathletics Passport
© 3P Learning
I
7
SERIES
TOPIC
19
Pythagoras’ Theorem
20
I
7
SERIES
TOPIC
Notes
Pythagoras’ Theorem Solutions
Mathletics Passport
© 3P Learning
ANG
LED
./20...
EM
OR
HE
’T
AS
OR
AG
TH
PY
H
./20...
.
..../...
RIG
LED
NG
T-A
ANG
..../....
ES
HT-
OF
OF
NS
O
I
AT
IC
L
P
AP
NGL
TRI
A
TRI
RIG
NS
IO
AT
IC
PL
AP
EM
OR
HE
T
’
AS
OR
G
A
TH
PY
LES
2
IDS
RA
2
2
..../...../20...
q 2,
ND RIGHT
SQUARES A
I A N G L ES
AN G L E D T R
EU
* " p - q , 2pq , p +
..../...../20...
IANGLES
A NGLED TR
A FOR PY
UL
ID’S FORM
CL
RE
AGO AN T
TH
.
/20..
.....
*A
..../
WE
E
SOM *
*
D RIGHTSQUARES AN
A
E
SOM *
E
W