7
PLANE GEOMETRY I
Lines and Planes
Definitions:
A Point: is an idea associated with position. It is
symbolized by a dot (.) and represents specific
location. It has neither size nor shape
E
J
Endpoint
A
B
Endpoint
A Ray: is a line segment that has only one
defined end point and one side that extends
endlessly away from the end point. A ray is
named by its end point and by the other point on
the line.
A
B
B
A
A Plane: is a flat surface which has length and
width only. A plane therefore has two
dimensions, length and width; no thickness. e.g.
A floor of a football field
M
D
G
F
Solution
Lines = ⃡
B
Q
⃡
, ⃡
Line segment =
P
K
A
Rays =
N
L
C
H
A Line: is an infinite set of points which extends
indefinitely in two directions
A Line Segment : is a set of points in a line
consisting of twodistinct ends. It represents a
collection of points inside the endpoints and it is
named by its end points
Baffour– Ba Series
⃡
,
,
,
,
Exercises 7.1
Fill in the blank spaces with the correct response
1. An idea associated with position is
called………
2. An infinite set of points extending indefinitely
in two dimensions is called …
3. A set of points in a line consisting of two
distinct end points is called ……
4. A flat surface which has length and width only
is called ………
5. A line that starts from a point and end at
infinity is called…….
A PLANE
Worked Examples
Identify three lines, rays and line segments
in the diagram below:
The Circle
A circle is a set of points in a plane which are at
the same distance from a fixed point. The fixed
point is called the centre of the circle and the set
of points forms the circumference of the circle.
Baffour – Ba series, Core Maths for Schools and Colleges
Page 190
Parts of a Circle
Segment
Chord
An angle is formed when two straight lines meet
at a point. The point where the two straight lines
meet is simply called Vertex.
Diameter
Circumference
Sector
Circumference: It is the distance around a circular
region. It is also known as the length or perimeter
of a circle.
Diameter: It is a straight line that divides a circle
into two equal parts
Semi - circle: It is half a circle
Chord:It is s a straight line that connects any two
points on a circle.
Arc: It is a portion on the circumference of a
circle
Segment: It is the area bounded by an arc and a
chord.
Radius : A line drwn from the center of a circle
to touch any part of the circumference. The plural
is radii
Sector: It is area bounded by two radii and an arc
Exercises 7.2
Complete the each with the correct answer
1. Half a circle is called …
2. A straight line drawn from the center of a
circle to touch any point on the circumference is
called …
3. The distance around a circle is called…
4. Any straight line that passes throught the
center of a circle, touching the circumference at
both ends is called …
Angles
Measurement of Angles
The instrument used to measure angles is called a
protractor. The unit of measure is the degree ( 0 ).
The scale on the protractor is divided into degrees
numbered from 00 to 1800 starting from either
end.
90
0
180
180,
0,
A
0
0
B
C
Using the Protractor
(A) Angles opening to the right.
For e.g.
a
o
b
I. Place the point “C” of the protractor on
point “o” of the angle.
II. Align the line segment “CB” of the protractor
with arm “ob” of angle a o b so that CB falls
exactly on “ob”. The arm “o a” of angle aob
points to the number
of degrees the angle measures on the protractor.
III. In order to determine the size of angles
opening to the right, the inner set of measurement
is used.
Baffour – Ba Series, Core Maths for Schools and Colleges
Page 191
a
2. Right Angle: Angle whose measure is
exactly 900.
90
a
180 ,0
0,180
b
C
o
(B) Angles opening to the left.
b
= 900
3. Obtuse Angle: Any angle whose measure is
greater than 900 but less than 1800
a
I. Place the point “C” of the protractor on
the point O of the angle.
II. Line up ̅̅̅̅ of the protractor with arm “ ” of
angle
so that AC falls exactly on arm “
of xoy. At this stage, the
of angle xoy will
point to the number of degrees the angle
measures on the protractor.
III. In order to determine the size of angles
opening to the left, use the outer values of
measure.
o
900< < 1800
b
4. Straight Angle: Any angle whose measure is
1800
a
o
= 1800
b
5. Reflex Angle: Any angle whose measure is
greater than 1800 but less than 3600.
90
a
1800< < 3600
b
180,0
0,180
C
Types of Angles
1. Acute angle: Any angle whose measure
is less than 900.
a
o
o
b
00< θ < 900
The values of the type of angles are summarized
below:
00< an acute angle < a right angle(900) < a
straight angle ( 1800 ) < a reflex angle < a
complete turn (3600 )
Pair of Angles
1. Complementary Angles: They are any two
angles that sum up to 900.
Baffour – Ba Series, Core Maths for Schools and Colleges
Page 192
a0
0
0
Worked Examples
The sizes of two angles are (100 – x)0 and (3x +
50)0. Calculate x if these two angles are
supplementary
0
a + b = 90
b0
Complementary angles
For example, in the diagram below,:
a
c
o
600
300
b
600 + 300 = 900, 600 and 300 are complementary
angles
2. Supplementary Angles: They are any two
angles that sum up to 1800.
c0
c0 + d0 = 1800
d0
Supplementary angles
For example in the diagram below;
c
0
40
a
0
0
140
o
b
0
40 and 140 are said to be supplementary angles
because 400+ 1400 = 1800
Solution
If (100 – x)0 and (3x + 50)0 are supplementary
⇒(100 – x)0 + (3x + 50)0 = 1800
1000 – x0 + 3x0 + 500 = 1800
2x0 + 1500 = 1800
2x0 = 1800 – 1500
2x0 = 300
x = 150
Exercises 7.4
A. Fill in the spaces correctly:
1. Two straight lines meet at a point to form a
figure that is called …
2. The supplementary angle of 480 is..................
3. The complement angle of 650 is ……………
4. Any angle that measures 900 is called……….
5. Two right angles equal to …………………..
6. AB and CD intersect so that pair of vertically
opposite angles are complementary. Calculate the
angles between AB and CD
Properties of Angles
1. Angles formed on a straight line sum up to 1800.
Consider the figure below;
3. Vertically opposite angle
When two lines cross each other, vertically
opposite angles are formed. Vertically opposite
angles are equal
1800
Consider the diagram below;
i. Angle e = Angle e
f
ii. Angle f = Angle f
e
e
f
Worked Examples
Find the value of in the diagrams below
1. x + 200 + 450 = 1800 (< s on a straight line)
x = 1800 – 200 – 450
Baffour – Ba Series, Core Maths for Schools and Colleges
Page 193
x = 1150
0
x 450
20
2. 4x + x + x = 1800 (< s on a straight line)
6x = 1800
Angles Connected with Parallel Lines
1. Vertically Opposite Angles
When two straight lines intersect, the opposite
angles formed are called vertically opposite
angles.
,
Vertically opposite angles are equal to each other.
For example, in the figure below;
300
3. x – 200 + x = 1800
2x = 1800 + 200
2x = 2000
,
b
a
i. Angle a = Angle c
ii. Angle b = Angle d
x = 100
2. Angles formed in a circle add up to 3600
Consider the figure below:
b
a
a + b + c = 3600
(Vertically opp. Angles)
(Vertically opp. Angles)
Worked Examples
1. Given that (12x – 100)0 and (9x + 20)0 are
vertically opposite angles. Calculate:
i. the value of x.
ii. the value of (12x – 9x)0
iii. What is the supplementary angle of (12x – 9x )0
c
Worked Examples
Find the angles marked with letters
i. x + 900 + 1370 = 3600
x = 3600 – 900 – 1370
x = 133
d
200
0
0
c
x
0
137
ii. x – 100 + x + 400 = 3600
2x = 3600 + 100 – 400
x - 100
2x = 3300
x + 400
x = 1650
x0
Solution
i. (12x – 100)0 and (9x + 20)0 are vertically
opposite angles
(12x – 100)0 = (9x + 20)0
12x – 9x = 200 + 1000
3x = 1200
x = 40
ii. 12x – 9x
= 12(400) – 9 (400)
= 4800 – 3600
= 1200
iii. 1800 – 1200 = 600
The supplementary angle of 12x – 9x = 600
Baffour – Ba Series, Core Maths for Schools and Colleges
Page 194
2. Find the angles marked with letters:
Worked Example
Find a, b, c and d in the diagram below;
a
40
0
0
120
c
Solution
Angles a and 400 are vertically opposite angles.
Therefore
400
3. Find the values of the angles marked with
letters in the diagram below
1250
Solution
z + 1250 = 1800 (angles on straight line)
z = 1800 – 1250 = 550
z=x
(vertically opposite angles)
0
x = 55
y = 1250
b
Solution
a = 1200
c + 1200 = 1800
1800 –1200
c = 600
a
(Alternate angles)
(Straight angles)
c = b = 600
(Alternate angles)
Special Alternate Angles
Consider the diagrams below:
x1
y1
y2
Y
x2
(vertically opposite angles)
2. Alternate angles
They are angles that are formed at the corners of
a figure. They are also called Z or N or Σ angles.
Alternate angles are equal.
From the diagram above x1 and y1 are alternate
angles. Therefore, x1 = y1.
Similarly, x2 and y2 are alternate angles.
Therefore x2 = y2.
⇒y1 + y2 = x1 + x2 and y1 + y2 = y
Consider the diagram below:
e c
Worked Examples
Find the value of y in the figure below:
c e
51o
yo
i. Angle c = Angle c
(Alternate angles)
ii. Angle e = Angle e (Alternate angles)
Baffour – Ba Series, Core Maths for Schools and Colleges
35o
Page 195
2. In the figure below, find the value of y
Solution
y1
y2
y
51
35
o
60o
y0
o
25o
y1 and 51o are alternate angles so y1 = 51o
y2 and 35o are alternate angles so y2 = 35o
But y = y1 + y2
y = 51o + 35o
y = 86o
3. Corresponding angles
Consider the diagram below:
c
2. In the diagram below, find the value of angle x.
a
d
b
25o
c d
64o
x
a
Angle a corresponds with angle a
Angle b corresponds with angle b
Angle c corresponds with angle c
Angle d corresponds with angle d
But corresponding angles are equal
⇒a = a , b = b, c = c and d = d
Solution
25o
y1
y2
xo
y1 + y2 = 64o
But y1 = 25o (Alternate angles)
25o + y2 = 64o
y2 = 64o – 25o
y2 = 39o
y2 = xo
(Alternate angles)
o
o
x = 39
Exercises 7.5
1. In the figure below, if
the value of angle g
A
b
o
0
50
B
is parallel to
Worked Examples
1. Find x and y in the diagram below?
x
y
0
65
, find
Solution
x = 650
x + y = 1800
650 + y = 1800
y = 1150
( Corresponding angles)
(Angles on a straight line)
g
0
E
130
2. Find the value of m in the diagram below
D
Baffour – Ba Series, Core Maths for Schools and Colleges
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30
2. A straight line intersects three parallel lines as
shown in the diagram below. Find the value of x
0
125
0
m
Solution
m and n are corresponding angles
m = 300
Solution
Name the angles as shown below;
4. Co-interior Angles
In the diagram below:
e
x
125
0
f
y
c
d
x
d and f are co – interior angles as well as e and c.
Co – interior angles are also called supplementary
angles because they sum up to 1800.
i. d + f =1800
ii e + c = 1800
y = 1250
x + y = 1800
x + 1250 = 1800
x = 1800 – 1250
x = 550
Worked Examples
1. In the diagram, PQ is parallel to SR. Find the
value of x
3. In the figure below, find the value of the angle
named y
R
Q
(Corresponding angles)
(Angles on a straight line)
0
44
0
37
y
x + 1300
0
35
P
S
Solution
(x + 1300) and 350 are interior opposite angles. (x
+ 1300) + 350 = 1800
x + 1300 + 350 = 1800
x = 1800 – 1300 – 350
x = 150
Solution
Name the angles as shown below;
Baffour – Ba Series, Core Maths for Schools and Colleges
0
44
0
0
44
37
y
x
Page 197
x + 440 + 370 = 1800
x + 810 = 1800
x = 1800 – 810
x = 990
8. In the diagram below, PQRS is a
parallelogram. If < QPS = 2x, < RSP = 3x and <
RST = y, find the value of y
Q
x = y (corresponding angles)
But x = 990, ⇒y = 990
4. Find the value of x in the figure below;
x
20
0
Solution
Draw a parallel line througg angle x and name
S
3x y
S
2x
P
T
Solution
2x and y are corresponding angles
⇒y = 2x
3x + y = 1800 ( < s on a straight line )
3x + 2x = 1800
(But y = 2x)
0
5x = 180
x = 360
⇒ y = 2x = 2(360)
the angles as shown below:
900 + 200 + x = 3600
x = 3600 – 900 – 200
x = 2500
0
20
y = 720
9. Find the angle marked .
F
x
0
20
E
40o
5. Find the value of angle x.
A
x
C
B
E
0
130
C
D
110o
A
B
Not drawn to scale
Ѳ
D
F
Solution
x = < BCD
(Alternate angles)
0
0
< BCD + 130 = 180 (Interior opp.<s)
< BCD = 1800 – 1300
< BCD = 500
But x = < BCD = 500
Therefore x = 500
Solution
Angle ADC = 110o.
(Opposite interior angle of a parallelogram)
Angle EDF = 110o (opposite angle)
θ + 40o + 110o = 180o
θ + 150o = 180o
θ = 1800 – 1500
θ = 300
Baffour – Ba Series, Core Maths for Schools and Colleges
Page 198
Exercises 7.6
A. Find the value of the variables: by variables
in the diagrams below:
0
1.
52
Mark in the sizes of the angles x , y and z
R
S
o
0
y
z
0
70
y
0
x
30
70
P
0
52
T
Q
0
60
2.
3. In the figure below, DE// BC. Mark in the sizes
of angles x and y
350
A
e
82
0
40
0
x
D
3.
C
D
x
A
B
B
1180
4.
a
0
20
b
c
40
5.
10x
x
T
A
V
0
100
Q
R
B
S
2. In the figure below, PQRS is a parallelogram.
base × height
Types of Triangles
I. Right – angled Triangle
It is a triangle in which one angle is a right angle
or 900
0
80
F
4. ABCD is a parallelogram in which angle A =
720 and AB is equal in length to diagonal BD.
Calculate the sizes of all the angle in the figure
A =
B. 1. In the figure below,< PRS = 1000, < TPV =
800 and PV is parallel to QS. Explain why ∆PQR
is isosceles
P
C
Triangles
A triangle is a plane figure bounded by three
straight lines. A triangle has three interior angles
that sum up to 1800. The area of a triangle,
0
x + 100
1000
y
y
E
Note:
The symbol
Baffour – Ba Series, Core Maths for Schools and Colleges
C
represents a right angle or 900
Page 199
II. Obtuse – angled Triangle
It is a triangle in which one angle is an obtuse
angle (greater than 900, but less than 1800)
B
A
A
C
B
III. Acute - angled Triangle
It is a triangle in which each interior anlge is less
than 900
A
400
800
B
600
Congruent Triangles
They are two or more triangles that have the same
shape or the same size and angles. In other words,
triangles are congruent if:
I. 3 sides = 3 sides
II. 2 sides, included angle = 2 sides, included
angle
III. 2 angles, one side = 2 angles, corresponding
side
Example of a congruent triangle is shown below
C
IV. Equilateral Triangle
It is a triangle in which all the three sides are
equal
C
B
G
E
C
A
F
∆ABC is congruent to ∆ EFG because they have
equal corresponding sides
Properties of a Triangle
1. The sum of angles in a triangle is equal to 1800
In an equilateral triangle, each angle is 600
V. Isosceles Triangle
It is a triangle in which two base angles and
corresponding sides are equal
a
b
b
a
2. For all Isosceles triangles, the base angles are
equal. The base angles are the angles that face the
two equal sides
c
VI. Scalene Triangle
It is a triangle in which all the three sides and
angles are unequal
a + b + c = 1800
c
c
a
b
a = b (base angles of an isosceles ∆)
Baffour – Ba Series, Core Maths for Schools and Colleges
Page 200
The Exterior Angle Theorem
The exterior angle theorem of triangles states that
the exterior angle of a triangle is equal to the sum
of the two interior angles opposite to it (exterior
angle).
d
a
C
y
z
Find the values of angles x, y
and z in the figure below;
e
b
External angle
External angle
1350
From the figure above, the theorem
summarized as: d = c + b and e = a + c
1. t + 700 + 500 = 1800
500
t = 1800 – 700 – 500
t = 600
700
t
2. m = 700 ( base angles of an isosceles ∆)
n + 700 + 700 = 1800
n
n = 1800 – 700 – 700
n = 400
700
3. a + 700 + 600 = 1800
a = 1800 – 700 – 600
a = 500
700
0
m
0
a
b
b = 70 + 60 ( interior opposite angles)
b = 1300
5. 2x + 4x + 3x = 1800
9x = 180
x = 200
is
∆ CDA is an isosceles triangle
Therefore n = z (Base <s of a triangle)
n + n + 1350 = 1800
2n + 1350 = 1800
2n = 1800 – 1350
C
2n = 450
n=
A
y
n = 22.50
⇒z = 22.50
z
m 1350
B
60
x
A
B
Solution
Name the vertices and the angles as shown
below;
Worked Examples
Find the angles marked with letters:
0
6.
Interior opposite
angles
c
a = 2x + 4x (Interior opposite angles)
a = 2(200) + 4(200) (Put x = 200)
a = 400 + 800
a = 1200
D
n
x
A
n + x = 1800 ( < s on a straight line)
But n = 22.50
22.50 + x = 1800
x = 1800 – 22.50
x = 157.50
m + 1350 = 1800
m = 1800 – 1350 = 450
∆ CBD is an isosceles triangle
Baffour – Ba Series, Core Maths for Schools and Colleges
Page 201
⇒m = y ( Base < s of a triangle)
y = 450
7. In the figure below, AC is protruded to D. Find
the value of x and < ACB
3y = 180o
A
m
a
y=
y = 60o
E
b
e
B
C
D
y + e = 180o
(Angles on a straight line)
o
o
o
60 + e = 180
eo = 180o – 60o = 120o
9x0
C
A
y
y
2x+ 70
x + 170
x
a
y
B
Solution
x + 170 + 2x + 70 = 9x0 (Exterior < theorem )
170 + 70 = 9x0 – x0 – 2x0
240 = 6x0
x0 =
a + a + eo = 180o
2a + 120 = 180o
2a = 180o – 120o
2a = 60o
a = 30o
(< in an isosceles ∆)
(But e =120o )
x = 40
< ACB + 9x0 = 1800 ( < s on a straight line)
< ACB + 9(40) = 1800
(But x = 40)
< ACB + 360 = 1800
< ACB = 1800 – 360
< ACB =1440
y + a + x = 180o (Angles on a straight line)
But y = 60o and a = 30o
60o + 30o + x = 180
x = 180o – 60o – 30o = 90o
2. In the diagram below, find the value of angles
a, b and c.
Some Solved Past Questions
1. In the diagram below, |AE| = |ED| = |DC| = |CE|
O
c
B
A
75O
aO
56O
bc
19O
x
E
C
D
Calculate the size of the angle marked x.
Solution
Name the angles as shown below
y + y + y = 1800
(Equilateral triangle)
Solution
bo + 560 + 560 = 180o
(Base < s of an isosceles triangle )
b = 180o – 56o – 56o = 68o
ao + 56o = 180o(< on a straight line)
a = 180o – 56o = 124o
c + 56o + 19o = 180o(<s on a straight line)
c = 180o – 56o – 19o = 1050
Baffour – Ba Series, Core Maths for Schools and Colleges
Page 202
3. Find the value of the angle marked x in the
diagram below:
1.
2.
540
2x
A
55o
4x
5y
10y
720
p
C
D
3.
xo
3x + 600
E
B
Solution
Name the angles as shown below;
a + a + 55o = 1800
A
o
0
2a + 55 = 180
o
55
0
0
2a = 180 – 55
a C
D a
b
2a = 125o
2a = 125o
x
0
o
E
B
a = 62.5
a + b = 180o (angles on a straight line)
62.50 + b = 180o
b = 180o – 62.50 = 117.50
2x
(x + 17)0
(x + 21)0
4.
(3x – 18)0
(x + 7)0
(x – 9)0 y
0
C
5.
3y0
D
7y0 5y0
B
6.
2x
b and x are alternate angles
⇒b = x. But b = 117.50.
Therefore x = 117.50
x0
A
x + 750
E
1500
4. Find the value of p
in the diagram below;
30o
c e
Po
D
b
7.
150
70o
Solution
p + 30o = 70o(external angle theorem)
p = 70o – 30o = 40o
d
a
A
B
C
B. 1. Find the value of the reflex angle marked y
in the diagram below:
Exercises 7.7
A. Find the values of the angles marked with
variables in the figures below:
Baffour – Ba Series, Core Maths for Schools and Colleges
A
B
C
40o
T
D
40o
E
y
F
Page 203
2. The diagram below is right angled triangle.
3. In the figure below, AB is parallel to DC.
A
x
(5x + 18)
7x
0
B
Find the size of the smallest angle in the triangle
C
55
3. In the figure below, find the values of angles x
and y
y
g
400
600
D
x
450
0
F
y
x
4. In the figure below ABC is an isosceles
triangle. Triangle ABD has a right angle at B. <
ADB = 400 , < CBE = 500. Work out for the sizes
of angles x and y
D
0
40
C
The Right – Angled Triangle
Any triangle which has one right – angle is called
a right-angled triangle.This means that in a rightangled triangle, an angle is 900
The Pythagoras Theorem
(Hypotenuse Rule)
It states that the square on the hypotenuse of a
right- angled triangle is equal to the sum of the
lengths of the squares on the other two sides
E
This is illustrated in the diagram below:
0
50
x
y
A
B
650
5. Determine
the values of
angles d, e, f
f
a
b
a2 = b2 + c2
e
c
1320
This is illustrated in the diagram below;
d
Challenge Problems
g
e
1. In the diagram below, determine the value of x
C
1.
D
100
700
A
x
2.
x
E
3
5
4
100
200
600
B
Baffour – Ba Series, Core Maths for Schools and Colleges
Page 204
By Pythagoras theorem, 52 = 32 + 42
Worked Examples
Find the length of the side marked with letters in
the triangle below:
1.
15cm
Solution
2
2
2
a 15 20
a2 225 400
a2 = 625,
a=√
a = 25cm.
/XZ/2 =132 − 52
/XZ/2 = 169 – 25
/XZ/2 = 144
/XZ/ = √
/XZ/ = 12cm
X
5. XYZ is right angled
triangle, with length
z
of sides shown below.
20cm
y
Y
Z
x
Express z in terms of x and y
2. 522 = 202 2
k2 = 522 – 202 20cm
k2 = 2704 – 400
k2 = 2304
52cm
Solution
From Pythagoras
theorem,
y2 = x2 + z2
z2 = y2 – x2
z=√
√
3.
37cm
p
Solution
372 = p2 + 212
p2 = 372 – 212
P2 = 1369 – 441
P2 = 928 cm
P= √
cm
P = 30.5cm (3 s. f)
6. In the quadrilateral ABCD below, /AB/ = 3cm.
/BC/ = 4cm, /CD/ = 12cm, angle ABC = 900 and
angle ACD = 900. Calculate:
i. the perimeter of ABCD
D
ii. the area of ABCD
12cm
21cm
Solution
i. By Pythagoras theorem,
/AC/2 = /AB/2 + /BC/2
/AC/ = √
32 + 42
Y
4.In the diagram below,
what is the length
13cm
of XZ?
Solution
X
From Pythagoras theorem,
132 = /XZ/2 + 52
5cm
Z
/AC/ = √
/AC/ = 5cm
C
4cm
A
3cm
B
Again by Pythagoras theorem,
/AD/ 2 = /AC/ 2 + /CD/ 2
But /AC/ = 5cm and /CD/ = 12cm
/AD/ 2= 52 + 122
/AD/ 2 = 25 + 144
/AD/ 2 = 169
Baffour – Ba Series, Core Maths for Schools and Colleges
Page 205
A
5.
/AD/ = √
/AD/ = 13cm
17
Perimeter of ABCD
= /AB/ + /BC/ + /CD/ + /DA/
= 3cm + 4cm + 12cm + 13cm
= 32cm
25
8
D
x
Pythagorean Triples
The Pythagorean triples consist of a set of 3
positive integers that obey the Pythagoras
theorem. This means that the sum of the squares
of two of them equals the square of the other
number. For e.g. {5, 12, 13} are Pythagorean
triples because 52 + 122 = 132. Similarly, {3, 4,
5} are Pythagorean triples because 32 + 42 = .
ii. Area of ABCD;
= area of ∆ ABC + area of ∆ ACD
Area of ∆ ABC = bh
= × /AB/ × /BC/
= ×3×4
To investigate whether a given set of numbers are
Pythagorean triples, equate the sum of the squares
of the two smaller integers to the square of the
bigger integer. In other words, equate the sum of
the squares of the first two integers to the square
of the third integer. For example, {3, 4, 5} is
investigated as; 32 + 42 = 52 and not 32 = 42 + 52
= 6cm2
Area of triangle ACD;
= × /CD/ × /AC/
= × 12 × 5
= 30cm2
The following sets of integers are Pythagorean
triples.
1. {3, 4, 5}, since 52 = 32 + 42
2. {5, 12, 13}, since 132 = 122 + 52
3. {8, 15, 17}, since 172 = 152 + 82
4. {7, 24, 25}, since 252 = 72 + 242
5. {6, 8, 10}, since 102 = 82 + 102
Area of ABCD;
= 6cm2 + 30cm2
= 36cm2
Exercises 7.8
A. Find the unknown lengths:
1.
2.
a
20cm
26cm
10cm
m
b
16cm
3.
4.
15cm
For any two positive integers, a and b, where
a > b, the three sides of the right angled – triangle
can be expressed in terms of a and b to generate
Pythagorean triplets.
d
a2 + b2
2ab
45cm
39cm
27cm
Baffour – Ba Series, Core Maths for Schools and Colleges
a2 - b2
Page 206
Worked Examples
1. Show that the numbers 9, 12 and 15 represents
the sides of a right-angled triangle.
Solution
Let a = 15cm, b = 9cm and c =12 cm
By Pythagoras theorem,
2
= b2 + c2
152= 92+ 122
225 = 81 + 144
225 = 225
The triangle is a right-angled triangle.
2. A triangle has sides 3, 4 and 5 units. Show that
it is a right-angled triangle.
Solution
Let a = 5, b = 4 and c = 3,
By Pythagoras theorem,
a2 = b2 + c2
52 = 42 + 32
25 = 16 + 9
25 = 25
⇒The triangle is a right-angled triangle
3.Find the Pythagorean triplets, if x = 3 and y = 2.
Solution
2xy
x2 + y2
x2 = 32 = 9
y2 = 22 = 4
x2 - y2 = 9 – 4 = 5
x2 - y2
2xy = 2(3) (2) = 12
x2 + y2 = 9 + 4 = 13
Therefore the triplets are 5, 12 and 13
Exercises 7.9
A. Identify the set of numbers that form the
sides of a right angled triangle.
1. {9, 12, 15} 2. {8, 15, 17} 3. {7, 24, 25}
4. {12, 15, 19} 5. {5, 8, 17} 6. {7, 8, 15}
B. 1. A triangle has sides AB, BC and CA
measuring 14, 48 and 50 units.
i. Prove that the triangle is right – angled, and
calculate its area.
Ans: A = 336 sq. units
ii. Calculate the length of the altitude from B to
CA
2. The points A, B and C have coordinates (-3, 1),
(2, -1) and (4, 2) respectively. Show that the
angle ABC is a right angle. Find the coordinates
of D if :
i. ABCD is a rectangle,
ii. ABCD is a parallelogram,
iii. D is the centre of triangle ABC.
3. Prove that the triangle with sides PQ = 6cm,
PR = 2.5cm and QR = 6.5cn is a right - angled
4. Mr. Green tells you that a right angled triangle
has a hypotenuse of 13 and a leg of 5. If he asks
you to find the other leg of the triangle without
using a paper and pencil, what will be your
answer?
Generating the Pythagoras Triples
If only one side of a right – angled triangle is
known, then Pythagorean triples can be
generated using the sides:
{ a, [ (a2 – 1) ], [ (a2 + 1) ]}
Pythagorean triples are generated by the formula:
[ (a2 + 1)]2 = [ (a2 – 1)]2 + a,
where a is an odd integer greater than one.
Prove
(x + 1)2 – (x – 1)2 = x2 + 2x + 1 – (x2 – 2x + 1)
(x + 1)2 – (x – 1)2 = x2 + 2x + 1 – x2 + 2x - 1)
(x + 1)2 – (x – 1)2 = 4x
Baffour – Ba Series, Core Maths for Schools and Colleges
Page 207
(x + 1)2 = (x – 1)2 + 4x
S
(x +1)2 = (x – 1)2 +
R
(Divide through by 4)
15cm
[ (x +1)]2 = [ ( x - 1)]2 + x
Substitute x = a2 to obtain:
P
[ (a2 +1)]2 = [ (a2 – 1)]2 + a2 ………(1)
P
Applying Pythagorean triples for odd numbers:
This identity can be used to generate Pythagorean
triples a, an odd integer greater than 1. For
example, when a = 3,
[ (32 +1)]2 = [ (32 – 1)]2 + a2
[ (10)]
2
2
2
2
2
= [ (8)] + a
Worked Examples
1. The width of a rectangle is 7cm. What are the
sizes of the length and diagonals if they are whole
numbers in centimeters?
Solution
Applying Pythagorean triples for odd numbers:
[ (a +1)]
2
2
When a = 15,
[ (152 +1)]2 = [ (152 – 1)]2 + 152
[ (226)]2 = [ (224)]2 + 152
1132 = 1122 + 152
⇒length = 112cm, breadth = 15cm and diagonal
= 113cm
2
5 =4 +3
2
[ (a2 +1)]2 = [ (a2 – 1)]2 + a2
2
2
= [ (a – 1)] + a
Area = Length × Breadth
= 112 cm × 15 cm
= 1680 cm2
Exercises 7.10
Generate Pythagoras triples using
following pair of positive integers:
1. 3 and 2
2. 5 and 2 3. 4 and 1
the
When a = 7,
[ (72 +1)]2 = [ (72 – 1)]2 + 152
[ (50)]2= [ (48)]2 + 152
252 = 242 + 72
Therefore, the sizes of the length and diagonal are
24 and 25 respectively
2. A Student was asked to prepare a rectangular
model in the form of two joined right angled
triangles with its shortest length being 15cm. Find
the length of the other two sides and the area of
the rectangle if the sides are all integers.
Solution
Challenge Problems
If three sides of a triangle are x2 – y2, 2xy and
x2 + y2, show that the triangle is a right – angled
triangle
Application of Pythagoras Theorem
Some practical problems as well as real life
situations are solved by application of Pythagoras
theorem once a right – angled triangle is formed
or produced. For e.g. when a ladder on a vertical
wall. The lengths of the involving sides are
calculated by applying the theory of Pythagoras.
This is just a tip of the iceberg. Some other
applications are discussed below:
Baffour – Ba Series, Core Maths for Schools and Colleges
Page 208
A. Placing a Ladder against a Wall
When a ladder is placed against a vertical wall,
the foot of the ladder and the ground floor meet at
a right angle. Knowing either how high the ladder
is or the height of the wall or the distance
between the legs, an unknown side can be
calculated using the Pythagoras theorem.
By Pythagoras theorem,
172 82 x2
2
289
2
289
2
x
225
,
Worked Examples
1. A ladder leans against a vertical wall of length
5m. The distance between the foot of the ladder
and the wall is 7m. Find the length of the ladder.
Exercises 7.11
1. A ladder leans against a vertical wall of height
12m. If the foot of the ladder is 5m away from the
wall, calculate the length of the ladder.
Solution
By Pythagoras theorem,
2
52 + 72
2. A ladder leans against a vertical wall of height
16m. If the foot of the ladder is 8m away from the
wall, calculate the length of the ladder.
2
5
m
2
74,
l = √ m = 8.60m ( 3 s. f.)
7
m
2. A ladder leans against a vertical wall of length
9cm. If the length of the ladder is 12cm, find the
distance between the foot of the ladder and that of
the wall.
Solution
By Pythagoras theorem,
122 92 f 2
2
144
2
144
9cm
2
63
,
f √ cm = 7. 94 (3 s. f)
x √
m
x = 15m
3. A ladder is 8m long. The foot of the ladder is
3m away from the base of the wall. How far up
the wall is the top of the ladder?
4. If a ladder to a slide is 8 feet and the ground
from the ladder to the slide is 4 feet, then how far
down then how far down will the child slide?
5. A 16 ft ladder leans against a wall with its base
4 ft from the wall. How far off the floor is the top
of the ladder?
12cm
3. A ladder which is 17m high is placed against a
vertical wall. If the distance separating the foot of
the ladder and the wall is 8m, what is the height
of the wall?
Solution
Let the height of the wall be x
17
Challenge Problems
1. A ladder 16m long is placed so that its foot is
3m from a building. How much further must the
foot of the ladder be moved from the building in
order to lower the top of the ladder by 2m?
2. i. The greatest length of an extending ladder is
10m. Calculate the greatest distance up a vertical
wall the ladder can reach when the foot of the
ladder is 6m from the foot of the wall.
Baffour – Ba Series, Core Maths for Schools and Colleges
Page 209
ii. When the ladder is adjusted to 8.5m, it reaches
a point 7.5 m above the ground. Calculate the
distance of the foot of the ladder
from the foot of the wall.
Length of the Diagonals of a Plane figure
I. Make a sketch of the figure
II. Represent the unknown side (diagonal) by any
preferred variable
III. Apply Pythagoras theorem to find the length
of the diagonal
B. The Diagonals of a Rectangle and a Square
The sides of a rectangle meet at a right angle. A
diagonal drawn from one corner to another
divides the rectangle into two equal triangles.
Given the dimensions of the rectangle, the length
of the diagonal can be calculated by application
of Pythagoras theorem. Similarly, given the
length of the diagonal and how long or how wide
the rectangle is, the other or unknown side can be
calculated
From Pythagoras
theorem,
d2 = l2 + a2
l2 = d2 – a2
a2 = d2 – l2
d
a
l
For a square of side a, as shown below;
Using Pythagoras theorem:
I. The length of the
d
diagonal, d is
a
calculated as:
a
d2 = a2 + a2
d2 = 2a2
d=√ (
II. The length of a side, a, is calculated as:
(
a=√
)
Worked Examples
1. A rectangle has length of 8cm and a breadth of
6cm. How long is its diagonal?
Solution
From Pythagoras theorem,
d2 = 82 + 62
d2 = 64 + 36
d
d2 = 100
d=√
8cm
d = 10cm
6cm
2. Find the length of the diagonal of a rectangle
which is 8cm long and 5cm wide
Solution
Let x be the length of the unknown side
From Pythagoras theorem,
x2 = 52 + 82
d2 = 25 + 64
x cm
d2 = 89
5cm
d=√
8cm
d = 9.4cm
3. The diagonal of a rectangle is 20cm long. If the
length of the rectangle is 17cm, how long is the
breadth?
Solution
Let x represent the breadth of the rectangle
By Pythagoras theorem,
202 = 172 + x2
x2 = 202 – 172
20cm
x2 = 111
xcm
x=√
17cm
x = 10.5cm
Baffour – Ba Series, Core Maths for Schools and Colleges
Page 210
4. A square has a side 12cm. What is the length
of the diagonal of the square?
2. An equilateral triangle has each side 2a meters
long. Find the length of an altitude of the triangle,
and hence find the area of the triangle in terms of a
Solution
Let the diagonal of the square be x
12
x2 = 122 + 122
2
x = 288
x
12
12
x=√
x = 16.97 cm
3. The sides of a rectangular floor are x m and
(x + 7) m. The diagonal is (x + 8) m. Calculate:
i. the value of x
Ans = 5
ii. the area of the floor.
Ans : 60cm2
Exercises 7.12
1. The length and breadth of a rectangle is 20 cm
and 11cm respectively. How long is the diagonal
of the rectangle?
C. The Diagonals of a Rhombus
The diagonals of a rhombus bisect each other at
900. Given the length of the diagonals, the side of
the rhombus can be calculated by applying the
Pythagoras theorem as shown below;
By Pythagoras theorem,
12
2. The diagonal of a rectangle is 61cm long. If the
breadth is 11cm, find its length.
3. The diagonal of a rectangle is 12 inches and its
width is 6 inches. Find its length.
4. A rectangle which is 8cm by 6cm is divided
into two right – angled triangles. What are the
lengths of the bases and the altitudes of the two
different isosceles triangles that can be formed
from these triangles?
5. Find the area and perimeter of a square whose
diagonal is 12cm long.
c2 = a2 + b2
a2 = c2 – b2
2
2
a
b
b
2
b =c –a
Perimeter of the rhombus,
P = c + c + c + c = 4c
C
a
C
Worked Examples
1. The length of the diagonals of a rhombus are
10cm and 24cm. Find:
i. the side of the rhombus
i. the perimeter of the rhombus
Solution
Let the shorter diagonal be
longer diagonal be CD = 24cm
6. The length of a rectangle is 1 cm longer than
its width. If the diagonal of the rectangle is 4cm,
what are the dimensions of the rectangle?
= × 24cm = 12cm
By Pythagoras theorem,
=
+
Baffour – Ba Series, Core Maths for Schools and Colleges
= 10cm and the
C
12cm
= × 10cm = 5cm
A
Challenge Problems
1. In ∆ABC, AB = AC = 12cm and BC = 8cm.
Express the length of the altitude from A to BC as
a surd in its simplest form.
C
C
5cm
O 5cm
B
12cm
D
Page 211
Similarly, in an isosceles triangle, the diagonal
touches the base and divides it into at a right
angle. Given the height and the two equal sides,
the third side can be found by applying
Pythagoras theorem. This is shown in the diagram
below;
=
+
=√
= 13cm
ii. Perimeter of rhombus
P=
+
+
+
=
=
=
= 13cm
P = 13 + 13 + 13 + 13 = 52cm
By Pythagoras theorem,
a
h2 = a2 + b2
b2 = h2 – a2
Alternatively,
P = 4 × 13cm
P = 52cm
a
h
b
b
2b
Exercises 7.13
1. The length of the diagonals of a rhombus is
48cm and 14cm. Find the perimeter of the
rhombus.
2. The sides of a rhombus are 8cm long. One of
its diagonal is 12cm. How long is the other
diagonal?
3. Find the length of a side of a rhombus whose
diagonals are 15cm and 22cm.
D. Triangles
In an equilateral triangle, the lengths of the three
sides are equal. The diagonal bisects the base at a
right angle. Given the length of a side, the
altitude (height) of an equilateral triangle can be
calculated by the use of Pythagoras theorem.
By Pythagoras theorem,
a2 = h2 + b2
In general, if the triangle is isosceles, then drop a
perpendicular before you apply the Pythagoras
theorem.
Worked Examples
1. PQR is an equilateral triangle of side 14cm.
Find the length of the perpendicular from P to QR
and hence, find the area of the triangle.
Solution
Let the perpendicular from P to QR be h
By Pythagoras theorem,
142 = 72 + h2
P
h2 = 142 – 72
h2 = 147
14cm
h=√
14cm
h
h = 12.12cm
7
Q
7
14cm
R
A = bh, but b = 14cm and h = 12cm
a
a
h
A=
× 14 × 12
A = 84cm2
b
b
a
Baffour – Ba Series, Core Maths for Schools and Colleges
Page 212
2. Calculate the length QR in the triangle PQR
P
13cm
Q
12cm
15cm
/AD/ = √
/AD/ = √
/AD/ = 5.7446
R
N
Solution
By Pythagoras theorem,
72 = 42 + /AD/2
/AD/2 = 72 – 42
Solution
/QN/2 = 132 – 122
169 – 144 = 25
/QN/ = √ = 5
/NR/2 = 152 – 122
/NR/ = √
/NR/ = 9cm
A
7
7
B
4
D
C
4
Area of a triangle = × BC × AD
= × 8 × 5.7
= 23.0cm3
Exercises 7.14
1. The length of the side of an equilateral triangle
is 30cm. Find the height of the triangle.
But /QR/ = /QN/ + /NR/
=5+9
= 14cm
2. ∆PQR is a triangle with PQ = QR and PR =
6m. If the height of the triangle is 7m, find PQ
3. A triangle has sides 17cm, 17cm, 16cm.
Calculate the area of the triangle
Solution
By Pythagoras theorem,
172 = 82 + h2
h2 = 172 – 82
h2 = 289 – 64
17cm
h=√
h = 15cm
3. ∆ LMN has LM = LN, MN = 12cm and area is
300cm2. Find the height of the triangle and hence
find LM.
4. Find the perimeter of the triangle below
correct to three significant figures?
h
S
17cm
14cm
8cm
A = bh,
16cm
R
9cm
T
Substitute b = 16cm and h = 25cm
A = × 16 × 15
5. Find the altitude of each triangle;
A = 120cm2
a.
b.
5
5
4. The triangle ABC has AB = AC = 7 cm, and BC
= 8cm. Find the area of the triangle, giving your
answer to 3 significant figures.
Baffour – Ba Series, Core Maths for Schools and Colleges
6
9
9
10
Page 213
D. Involving Quadratic Equations
Given the value of a side of a right – angled
triangle, the next side as a variable and the third
side as an increment or decrement in the variable,
the actual dimensions (values of the variable) can
be calculated.
I. Apply Pythagoras theorem,
II. Expand the involving brackets,
III. Equate the equation to zero,
IV. Solve the quadratic equation by using the
quadratic method, ignoring negative answers,
V. Substitute the value of the variable to obtained
the dimensions of the right – angled triangle.
Worked Examples
The length of one leg of a right – angled triangle
is 2cm more than the other. If the length of the
hypotenuse is 6cm, what are the lengths of the
two legs?
Solution
Draw a sketch of the problem, labeling the
known and unknown lengths. If one leg is
represented by x, the other is represented by
x+2
Use Pythagoras theorem
to form the equation,
x2 + (x + 2 )2 = 62
x2 + x2 + 4x + 4 – 36 = 0
2x2 + 4x – 32 = 0
x=
x=
√
( (
√
(
⇒x = –1 + √
x = 3.123
= –1 + √
2. The length of the three sides of a right –
angled triangle form a set of consecutive even
integers when arranged from least to greatest. If
the second largest side has a length of x, form an
equation and hence, solve for the three sides.
Solution
The 3 sides of the right – angled triangle when
arranged from the least to the greatest are:
(x – 2), x and (x + 2)
(x – 2)2 + x2 = (x + 2)2
(x – 2) (x – 2) + x2 = (x + 2) (x + 2)
x(x – 2) – 2(x – 2) + x2 = x (x + 2) + 2(x + 2)
2
x – 2x – 2x + 4 + x2 = x2 + 2x + 2x + 4
2
x – 4x + 4 + x2 = x2 + 4x + 4
2x2 – x2 – 4x – 4x + 4 – 4 = 0
x2 – 8x = 0
x2
x(x – 8) = 0
⇒x = 0 or x = 8
(x + 2)
(x – 2)
When x = 0, the first side is negative. That is:
0 – 2 = – 2 so ignore the answer x = 0
6
a = 2, b = 4 and c = -32 (
If x = 3.123, then x + 2 = 3.123 + 2 = 5.123
The lengths of the legs are approximately
3.123cm and 5.123cm.
x
When x = 8,
The 1st side is 8 – 2 = 6units,
The second side is 8units,
The third side is : 8 + 2 = 10units
x+ 2
*
Exercises 7.15
1. The hypotenuse of a right – angled triangle is
(2x + 3) cm long, and the other two sides have
lengths x cm and (x + 7) cm. Find x, and calculate
the area of the triangle.
Baffour – Ba Series, Core Maths for Schools and Colleges
Page 214
2. The sides of a right – angled triangle in
ascending order of magnitude are 8cm, (x – 2) cm
and x cm. Find x.
2. In a right – angled triangle, the hypotenuse is
39cm and the ratio of the other two sides is
12: 5. Find the sides.
3. The lengths in cm of the sides of a right –
angled triangle are x, (x + 2) and (x + 1), where x
> 0, find x.
Solution
Let the sides be (12x) cm, (5x)cm and 39 cm
By Pythagoras theorem,
392 = (12x)2 + (5x2)
39cm
12x
1521 = 144 x2 + 25x2
1521 = 169x2
5x
x2 = 9
x=√
x = 3cm
⇒12x = 12(3) = 36cm
5x = 5(3) = 15cm
The sides of the triangle are 15cm, 36cm, 39cm
4. The length of one leg of a right triangle is 1cm
more than the other. If the length of the
hypotenuse is 3cm, what are the lengths of the
legs?
5. The length of one leg of a right triangle is 3cm
more than the other. If the length of the
hypotenuse is 8m, what are the lengths of the
legs?
6. Find the lengths of the sides of the right –
angled triangle
(x + 2)cm
(x + 4)cm
x cm
Solving Other Applications
Worked Examples
1. How long must a guywire be to reach from the
top of a 30ft pole to a point on the ground 20ft
from the base of the pole?
Solution
Let the length of the guywire be x,
x2 = 302 + 202
x2 = 900 + 400
x
30
x2 = 1,300
x=√
20
x = 36 ft
4. A man standing 40m away from a tower
notices that the distances from the top and bottom
of a flagstaff on top of the tower are 50m and
45m respectively. Find the height of the flagstaff.
Solution
Let h and x be the heights of the flagstaff and
that of the tower respectively as shown below;
From ∆ ABC,
D
402 + x2 = 452
h
x2 = 452 – 402
2
x = 425
C
45m
50m
x=√
x
x = 20.6m
A
40m
B
From ∆ ABD
(
+ 402 = 502
(
= 502 – 402
(
900
h+x=√
Baffour – Ba Series, Core Maths for Schools and Colleges
Page 215
h + x = 30,
h = 37.7m
The height, BD of the pole is 37.7m
But x = 20.6
h + 20.6 = 30
h = 30 – 20.6
h = 9.4m
The height of the flagstaff is 9.4m
5. Two points A and C are on the same level
ground as the foot of a pole B. The distance
between A and C is 70m and A and C are on the
opposite sides of the vertical pole. The distances
from the top of the pole, D, to A and C are 45m
and 59m respectively. Find :
i. the distance between the foot of the pole, B, and
the point A,
ii. the height, BD of the pole.
Solution
i. From ∆ ABD,
x2 + h2 = 45…….(1)
85m
A
Solution
Let /BD/ = h and /AB/ = x
From ∆ ABD,
x2 + h2 = 532……….(1)
85m
D
h
C
59m
B
From ∆ BCD,
x
70m - x
2
2
(
+ h = 59 ……….(2)
eqn (2) – eqn (1);
(
– x2 = 592 – 452
702 – 140x + x2 – x2 = 1456
-140x = 1456 – 702
-140x = -3444
x = 24.6 m
The distance between B and A = 24.6m
ii. x2 + h2 = 452
⇒(
+ h2 = 452
h2 = 452 – (
h=√
6. Two points A and C, are on the same level
ground as the foot of the pole, B. The distance
between A and C is 40m and A and C are on the
same sides of the vertical pole. The distances
from the top of the pole, D to A are 53m and 85m
respectively. Find correct to one decimal place;
i. the distance between the foot of the pole, B and
the point A,
ii. the height, BD, of the pole.
(But x = 24.6)
C
53m
A
h
x
B
From ∆ BCD,
(
+ h2 = 852 ……….(2)
eqn (2) – eqn (1)
(
– x2 = 852 – 532
402 + 80x + x2 – x2 = 4,416
80x = 4,416 – 402
80x = 2,816
x = 35.2 m
The distance between, B and A = 35.2m
ii. From ∆ ABD
x2 + h2 = 532
⇒(
+ h2 = 532
h2 = 532 – (
(But x = 35.2)
h=√
h = 39.6m
Baffour – Ba Series, Core Maths for Schools and Colleges
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The height, BD of the pole is 37.7m
7. Tom and Jerry meet at a corner. Tom turns 900
left and walks 9 paces; Jerry continues straight
and walks 12 paces. Find the distance between
the two of them.
Solution
Let the distance between Tom and Jerry be x
By Pythagoras theorem,
x2 = 92 + 122
T
x2 = 81 + 144
x
x2 = 225
9
x= √
p
a
x=√
12
J
c
x = 15 paces
pace
e
The distance between
Tom
and Jerry is 15 paces
s
s
Exercises 7.16
1. How long must a guywire be to run from the
top of a 20ft to point on the ground 8ft from the
base of the pole?
2. If the hypotenus of a right – angled isosceles
triangle is 6m, what is the length of each of the
other side‟s?
3. Two foggers run 8 miles north and then 5 miles
west. What is the shortest distance, to the nearest
tenth of a mile they must travel to return to their
starting point?
measurements side A = 3cm, side B = 4cm. What
is the length of side C?
6. A man starting from point A, walks 5km due
east and then 4km due north to point B. Calculate
to one decimal place, the distance from A to B
direct.
ii. From B he walks 5km north and then 4km west
to C, calculate to one decimal place:
a. the distance from B to C,
b. the distance from A to C.
7. Two points A and C are on the same level
ground as the foot of a pole B. The distance
between A and C is 16m and A and C are on the
same side of the vertical pole. The distances from
the top of the pole, D, to A and C are 40m and
50m respectively. Find :
i. the distance between the foot of the pole, B, and
the point A,
ii. the height, BD of the pole.
A Quadrilateral
It is a plane figure bounded by four straight lines.
Examples of quadrilaterals are Rectangles,
Squares, Rhombuses, Kites, Trapeziums and
Parallelograms. The interior angles of a
quadrilateral sum up to 3600
Types and Properties of Quadrilaterals
Rectangle : It is a quadrilateral with opposite pair
of sides equal. Also:
4. A kite is flying so that it is 55 feet high, and its
above the point 75feet from the flyer. How long
is the string of the flyer?
5. A right – angled triangle with sides A, B and C
with respective sides a, b, c has the following
D
C
A
B
I. Each interior angle is a right angle.
II. Each pair of opposite sides is parallel.
Baffour – Ba Series, Core Maths for Schools and Colleges
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III. It has two diagonals that are not symmetrical.
IV. It fits its outline in two ways.
IV. Area, A = L × B
A=
×
D
C
h
A
Square: It is a plane figure ( special rectangle)
with four sides equal. In the figure
below,
=
=
=
Also:
D
C
A
B
I. Each interior angle is a right angle.
II. Each pair of opposite sides is parallel.
III. It has two diagonals that are symmetrical.
IV. It fits its outline in four ways.
V. Area, A = L × L
A=
×
Parallelogram : It is any quadrilateral with both
pair of opposite sides equal and parallel. Also:
B
I. It fits it outline in one way.
II. Area, A = (
h.
Rhombus: It is a parallelogram with four sides
equal. Also:
C
B
A
D
I. It consist of two congruent isosceles triangle,
base to base.
II. It fits it outline in four ways.
III. Its diagonals are axes of symmetry.
III. Its diagonals bisect at right angles.
IV. Area = × (Product of diagonals)
A= ×(
h
b
I. Opposite angles are equal.
II. Diagonals bisect each other.
III. It has half turn of symmetry.
IV. It can be formed by two pairs of parallel lines.
V. Area = base × perpendicular height
A = bh
Trapezium: It is any quadrilateral with one pair
of opposite sides parallel. Also;
Kite: It consists of two isosceles triangles with
equal bases. Also;
I. It fits its outline in two ways.
II. One diagonal is an axis of symmetry.
III. Each pair of adjacent sides is equal.
Baffour – Ba Series, Core Maths for Schools and Colleges
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IV. Its diagonals bisect at right angles.
Unknown Angle of a Quadrilateral
The sum of interior angles of a quadrilateral is
3600.
To calculate the value of an unknown angle in a
quadrialateral;
I. Represent the unknown angle by any preferred
variable.
II. Sum up the angles and equate to 3600.
III. Workout for the value of the variable.
Worked Examples
1. In the figure below,
find the value of the
1000
angle marked y
Exercises 7.17
A. Find the values of the unknown angles
1.
a
a
2a
2.
126
1150
x
y
a
x
0
600
a
Solution
For all parallelograms, opposite angles are equal.
Therefore, from the diagram, a = 1200 and 2b =
600
2b = 600
b = 300
⇒a = 1200 and b = 300
106
0
3.
x
4.
1100
y
700
5.
2. Find the values of the letters in the diagram
below;
2b
0
650
Solution
x + 1100 + 1000 + 1150 = 3600
x = 3600 – 1100 – 1000 – 1150
x = 3600 – 3250
x = 350
1200
680
x
1100
115
2a
870
x
940
700
B. 1. The interior angles of a quadrilateral are 5x0,
3x0, 300 and 7x0. Find the value of x and the
value of each interior angle of the quadrilateral.
2. A parallelogram has an interior angle of 950.
Find the values of the other three interior angles.
3. Find the value of x and the value of each angle
of a quadrilateral whose values are given as x,
x, x and x.
4. In the diagram below, ABCD is a kite.
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Line of Symmetry of Some Figures:
Rectangle, Square, Equilateral triangle, kite…
1. Square
b
a
1150
L1
L4
L3
300
L3
L2
Calculate the value of the angles a and b.
Symmetrical Objects and Line of Symmetry
Symmetrical objects are objects that can be
folded or divided into two equal halves such that
one-half fits exactly on the other half. For
example, it is possible to draw a line (mirror)
through a rectangle and an isosceles triangle to
divide them into two halves such that one-half fits
exactly on the other half. In these cases, the
rectangle and the isosceles triangle are said to be
Symmetrical.
Consider the figures below;
A
D
A
Fig. I
M Fig. II
2. Rectangle
L1
L2
L2
L1
Line of Symmetry of a rectangle = 2
3. Equilateral triangle
L1
L3
L2
L2
L3
L1
Line of symmetry of an equilateral triangle is 3.
B
4. Kite.
L1
C
L1
B
M
P
D
L4
L1
Line of symmetry of a square = 4
A line that divides a figure into two equal halves
is called a line of symmetry. It is also called a
line of fold.
P
L2
Line of symmetry of a kite = 1
C
In both figures, PM is the line of symmetry. This
implies that in fig. I, APMD = PBCM and in fig.
II, ABMP = PMDC
Exercises 7.18
1. Name any five figures that are symmetrical
2. Draw each of the following and indicate the
lines of symmetry;
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i. Rectangle
iii. Parallelogram
v. Circle
ii. Rhombus
iv. Scalene triangle
vi. Isosceles
3. Copy and complete the table below:
Object
Square
Rectangle
Isosceles triangle
Rhombus
Parallelogram
Kite
Circle
Equilateral
triangle
Line of Symmetry
5 sides
4 sides
1
2
1
3
2
2 × 1800 = 3600
3 × 1800 = 5400
.
Polygons
A polygon is any plane figure bounded by
three or more straight lines. The following
are names of polygons of three to ten sides.
Number of sides
3
4
5
6
7
8
9
10
Polygon by drawing diagonal lines, making
sure that no line crosses another.
III. Multiply the number of triangles in
thepolygon by 1800 (the sum of interior angles of
a triangle) to obtain the sum of interior angles, ,
of the polygon.
2
3
1
3
4 × 1800 = 7200
4
5 × 1800 =9000
8 sides
9 sides
1
2
5
2
4
Name
Triangle
Quadrilateral
Pentagon
Hexagon
Heptagon
Octagon
Nonagon
Decagon
Sum of Interior Angles of a Polygon
Investigations
Reminder: The sum of interior angles of a
triangle is 1800.
7 sides
6 sides
1
1
3
6
3
5
2
7
4
4
5
6 × 1800 = 10800
6
7 × 1800 = 12600
Summary
To investigate the sum of interior angles of a
polygon given the number of sides :
I. Draw the polygon.
II. Find the total number of triangles in the
Baffour – Ba Series, Core Maths for Schools and Colleges
Number of
sides
3
4
5
6
7
8
Number of
Triangles
(3 – 2) = 1
(4− 2) = 2
(5 − 2) = 3
(6 – 2) = 4
(7 − 2) = 5
(8 − 2) = 6
Sum of Interior
Angles ( )
1×1800 = 1800
2×1800 = 3600
3×1800 = 5400
4×1800 =7200
5×1800 = 9400
6×1800=10200
Page 221
9
10
n
(9 − 2) = 7 7×1800 =12600
(10 − 2) = 8 8×1800 =14400
(n − 2)
(n − 2)×1800
From the table, it can be seen that the number
of triangles is two less than the number of sides
of the polygon. The conclusion is that;
For any polygon with n sides, the sum of
0
interior angles, = (
Worked Examples
Find the sum of interior angles of a polygon
with thefollowing sides;
1. 15
2. 12
3. 20
Substitute n = 20
⇒ Sum of interior angles
= ( 20 – 2) 1800
= 18 × 1800
= 3,2400
Exercises 7.19
A. Find the sum of interior angles of a polygon
with the following sides:
(1) 14
(2) 22
(3) 42
(4) 35
(5) 17
(6) 50
B. Find the value of the unknown angles;
1.
a
Solution
1. 15 sides
0
60
0
110
0
70
Solution
Sum of interior angles = ( n – 2) 1800
Substitute n = 15
⇒ Sum of interior angles,
= ( 15 – 2) 1800
= 13 × 1800
= 2,3400
2.
b
0
0
32
60
3.
76
0
125
0
c
2. 12 sides
Solution
Sum of interior angles = ( n – 2) 1800
Substitute n = 12
⇒ Sum of interior angles,
= ( 12 – 2) 1800
= 10 × 1800
= 18000
3. 20 sides
Solution
Sum of interior angles = ( n – 2) 1800
4.
0
0
0
0
Finding the Number of Sides of a Polygon
Given the sum of interior of a polygon as;
θ = (n – 2) × 1800, the number of sides, n, of the
polygon can be found by making n the subject of
the formula as shown:
Baffour – Ba Series, Core Maths for Schools and Colleges
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In θ = (n – 2) × 1800
θ = 1800n – 3600
θ + 3600 = 1800 n
n=
, where
is the sum of interior <s
Worked Examples
1. The sum of interior angles of a polygon is 9000.
Find the number of sides of the polygon
Solution
= 9000n = ?
n=
=
=
= 7 sides
2. Find the number of sides of a polygon with
whose sum of interior angles is 1,9800.
The sum of the pair of interior and exterior
angles are supplementary (equal to 1800) because
they are angles formed on a straight line.
Therefore, from the diagram above;
a + b = 1800 and c + d = 1800
It is conclusive that for all polygons;
1. Interior angle exterior angle 1800
2. Exterior angle 1800 – interior angle
3. Interior angle 1800 – Exterior angle
Worked Examples
1. What is the value of
the angles marked
and in the figure
below?
1100
0
55
Solution
Solution
= 19800n = ?
n=
0
0
(
– 55 = 1250
0
0
(
0
0
y = 180 – 110
y = 700
0
=
=
= 13 sides
Exercises 7.20
A. Find the number of sides of a polygon
with the following sums of interior angles:
1. 10800
2. 14400
3. 28800
4. 36000
5. 30600
6. 48600
0
Exercises 7.21
A. List the pair of exterior and interior angles
in the polygon below;
f
Interior and Exterior Angles of a Polygon
Consider the polygon below;
e
gh
a b
m
n
k
p
c d
Angles a and b are pair of inerior angles and
exterior angles respectively, likewise angles c
and d.
B. In the hexagon below, determine the size
of the interior and exterior angles.
Baffour – Ba Series, Core Maths for Schools and Colleges
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Interior angle + exterior angle 1800
⇒Interior angle 1800 – Exterior angle
1800 – 300
= 1500
1120
0
125
0
92
y
z
0
86
0
x 75
A Regular Polygon
A regular polygon is a polygon that has allthe
sides and angles equal. Examples are equilateral
triangle, rhombus and a square.
2. A regular polygon has 9 sides. Find:
i. each interior angle of the polygon,
ii. the Sum of the interior angles of the polygon.
Solution
i. For a regular polygon each interior angle,
=
For all regular polygons of n sides, it implies that
there are n interior angles and n exterior angles.
If the sum of all interior angles of a regular
polygon is = (n – 2 ) × 1800, then each interior
angle,
=
From
=
(
(
n=
, the number of sides
(for regular polygons)
Worked Examples
1. A regular polygon has 12 sides. What is the
size of each interior angle?
Solution
Method 1
For a regular polygon, each interior angle
=
B. 1. The diagram below shows part of a regular
10 – sided polygon. Workout the size of the angle
marked x.
x
=
= 1500
Method 2
Because it is a regular polygon, all the 12 sides
are equal. n =12
Exterior angle,
= 1400
=
Exercises 7.22
A. Find the value of each interior angle of a
regular polygon with the following sides;
1. 13 2. 14
3. 17
4. 19
5. 21
(
(
, but n = 9
(
ii. Sum of interior angles,
= (n – 2 ) × 1800 , but n = 9
= (9 – 2 ) × 1800
= 7 × 1800
= 1,2600
=
.
Substitute n =12
=
(
2. The diagram below shows a regular polygon
and a regular hexagon. Calculate the size of the
angle marked x.
= 300
Baffour – Ba Series, Core Maths for Schools and Colleges
x
Page 224
Sum of Exterior Angles of a Polygon
Consider the diagrams below:
e
a
i. the value of angle x
ii. the value of angle exterior angle 3x
iii. the value of the interior angle of 3x
a
b
e
c
d
Fig. I
c
d
b
Fig. II
When all the exterior angles are cut and moved to
a fixed point, they form a complete
turn of 3600 as shown below;
e a
b
d
c
a + b + c + d + e = 3600
In general, for every polygon (regular or
irregular), the sum of exterior angles is equal to
3600. ⇒: = 3600
Solution
i. Sum of exterior < s of a polygon = 3600
720 + 3x0 + 520 + 360 + 500 = 3600
3x = 360 – 72 – 52 – 36 – 50
3x = 1500
x = 500
ii. The value of angle 3x = 3 × 500 = 150
iii. Let the interior angle of 3x be y
3x + y = 1800, but 3x = 1500
1500 + y = 1800
y = 1800 – 1500
y = 300
3. The exterior angles of a polygon are 2x0, (x
– 20)0, x0, (3x + 10)0, (x + 15)0 and (2x + 5)0.
Find the value of x
Worked Examples
1. In the figure below, determine the value of a
880
a
0
36
0
73
0
80
Not drawn to
scale
Solution
Exterior <s of a polygon sum up to 3600
a0 + 880 + 730 + 800 + 360 = 3600
a0 = 3600 – 880 – 730 – 800 – 360
a0 = 830
2. The exterior angles of a pentagon are 720,
3x0, 520, 360, and 500 . Determine;
Solution
Exterior angles of a polygon = 3600
⇒2x0 + (x – 20)0 + x0 + (3x + 10)0 + (x + 15)0
+ (2x + 5)0 = 3600
2x0 + x0 – 200 + x0 + 3x0 + 100 + x0 + 150 + 2x0
+ 50 = 3600
2x0 + x0 + x0 + 3x0 + x0 + 2x0 – 200 + 100 + 150
+ 50 = 3600
10x0 + 100 = 3600
10x0 = 3600 – 100
10x0 = 350
x = 350
Exercises 7.23
A. Find the value of the angles marked x
Baffour – Ba Series, Core Maths for Schools and Colleges
Page 225
1.
i. Find its exterior angle.
ii. How many sides has the polygon?
0
Solution
i. Interior angle exterior angle 1800
Exterior angle
1800 – interior angle
But interior angle
1720
Exterior angle
1800 – 1720 80
30
0
50
=
2.
ii. sum of exterior angles
0
60
3600
But the exterior angle is 80
⇒80
3.
0
130
+ 40
Therefore n =
0
= 45 sides
3. Each interior angle of a regular polygon is
1600, how many sides has it?
0
120
0
70
Finding the Number of Sides, n, of a Regular
Polygon Given its Exterior Angle,
The exterior angle, , of a regular polygon is
related to the number of sides, n, by the formula:
. Therefore, knowing the value of each
exterior angle
, the number of sides, n
is
calculated by the formula; n
Solution
Interior angle + Exterior angle 1800
Exterior angle
1800 – Interior angle
But interior angle
1600
Exterior angle 1800 – 1600 200
Exterior angle
Substitute exterior angle = 200
⇒200
n=
Worked Examples
1. If each exterior angle of a regular polygon is
300, find the number of sides of the polygon.
= 18 sides
4. Find the size of the interior angle of a regular
polygon with 5 sides.
Solution
= 300 and n = ?
Solution
n=
Substitute n = 5
=
= 12 sides
Exterior angle
Exterior angle
2. A regular polygon has an interior angle of 1720.
,
Exterior angle
Baffour – Ba Series, Core Maths for Schools and Colleges
,
720,
Interior angle
1800
Page 226
⇒Interior angle
1800 – 720
= 1080
1800 – exterior angle
Area of ∆
= (14)(14) sin 600
= 84.87 cm
2
14
14
0
60
14
0
5. Four interior angles of a hexagon are 130 ,
1600, 1120 and 800. If the remaining angles are
equal, find the size of each of them.
Solution
Let each of the remaining angles be x
⇒sum of interior angles
= 1300 + 1600 + 1120 + 800 + x + x = 7200
2x = 7200 – 1300 – 1600 – 1120 – 800
2x = 2380
x = 1190
Therefore each remaining angle is 1190
6. A regular polygon is inscribed in a circle of
radius 14cm. If each interior angle is 1200,
find:
i. the value of each exterior angle,
ii. the area of the polygon.
Solution
Let the exterior angle be A
Interior angle + Exterior angle = 1800
1200 + A = 1800
A = 1800 – 1200
A = 600
ii. From the diagram,
Area of the polygon
6 × 84.87 = 509.2 cm2
Solved Past Question
Three interior angles of a polygon are 1600
each. If the other interior angles are 1200 each,
find the number of sides of the polygon.
Solution
Let the interior angles of the polygon be n
3(1600) + (n – 3)1200 = (n – 2) 1800
4800 + 1200n – 3600 = 1800 n – 3600
1200n – 1800n = -3600 + 3600 – 4800
-600 n = - 4800
n=
= 8 sides
Exercises 7.24
A. 1. The exterior angle of a regular polygon is
300, find the number of sides of the polygon.
2. The interior angle of a regular polygon is twice
its exterior angle. Find the number of sides of the
polygon.
3. A regular polygon has 12 sides. Find:
(i) its exterior and interior angles,
(ii) the sum of the interior angles.
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