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C H A P T E R
12
MODULE 2
PL
What are the properties of parallel lines?
E
Geometry
What are the basic properties of triangles?
What are the basic properties of regular polygons?
How do we use and apply similarity and Pythagoras’ theorem in two dimensions?
How do we explore ratios of areas of similar figures?
How do we explore ratios of volumes of similar solids?
12.1
Properties of parallel lines – a review
SA
M
Angles 4 and 6 are called alternate angles.
l3
Angles 5 and 3 are called alternate angles.
Angles 2 and 6 are called corresponding angles.
1 2
l1
4 3
Angles 1 and 5 are called corresponding angles.
l2
Angles 4 and 8 are called corresponding angles.
56
Angles 3 and 7 are called corresponding angles.
8
7
Angles 3 and 6 are called cointerior angles.
Lines l1 and l2 are cut by a transversal l3 .
Angles 4 and 5 are called cointerior angles.
Angles 1 and 3 are called vertically opposite angles and are of equal magnitude.
Other pairs of vertically opposite angles are 2 and 4, 5 and 7, and 6 and 8.
Angles 1 and 2 are supplementary, i.e. their magnitudes add to 180◦ .
l3
When lines l1 and l2 are parallel, corresponding angles
are of equal magnitude, alternate angles are of equal
1 2
magnitude and cointerior angles are supplementary.
4 3
l1
Converse results also hold:
If corresponding angles are equal then
5 6
l1 is parallel to l2 .
8 7
l2
If alternate angles are equal then l1 is parallel to l2 .
If cointerior angles are supplementary then l1 is parallel to l2 .
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361
Chapter 12 — Geometry
Example 1
Parallel line properties
Find the values of the pronumerals.
Solution
(corresponding)
(alternate with a)
(cointerior with d)
(corresponding with b)
(vertically opposite e)
e° 65°
d° c°
b° a°
E
a = 65
d = 65
b = 115
e = 115
c = 115
There are lots of ways of finding these values. One sequence of reasoning has been used here.
Parallel line properties
PL
Example 2
Find the values of the pronumerals.
Solution
(2x – 50)°
2x − 50 = x + 10 (alternate angles)
∴ 2x − x = 50 + 10
∴
x = 60
Parallel line properties
M
Example 3
(x + 10)°
Find the values of the pronumerals.
Solution
SA
x + 100 = 2x + 80
(alternate angles)
∴ 100 − 80 = 2x − x
∴ x = 20
Also x + 100 + y + 60 = 180 (cointerior)
and x = 20
∴ y + 180 = 180 or y = 0
(x + 100)°
(2x + 80)° (y + 60)°
Exercise 12A
Questions 1 to 5 apply to the following diagram
l1
d a
b
c
l2
h e
g f
l3
1 Angles d and b are:
A alternate
D supplementary
B cointerior
E vertically opposite
C corresponding
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Essential Further Mathematics – Module 2 Geometry and trigonometry
2 Angles d and a are:
A alternate
D supplementary
B cointerior
E vertically opposite
C corresponding
B cointerior
E vertically opposite
C corresponding
3 Angles c and h are:
A alternate
D supplementary
A alternate
D supplementary
E
4 Angles b and f are:
B cointerior
E vertically opposite
C corresponding
B cointerior
E vertically opposite
C corresponding
5 Angles c and e are:
PL
A alternate
D supplementary
6 Find the values of the pronumerals in each of the following:
a
x°
z°
z°
40°
z° x°
y° 80°
(2x – 40)°
120°
y° x°
(x + 40)°
SA
y° 50°
(2z)°
f
e
d
c
x° y°
M
y° 70°
b
12.2
Properties of triangles–a review
a ◦ , b◦ and c◦ are the magnitudes of the interior angles of the triangle ABC.
B
d ◦ is the magnitude of an exterior angle at C.
The sum of the magnitudes of the interior angles of
b°
a triangle is equal to 180◦: a ◦ + b◦ + c◦ = 180◦ .
b◦ + a ◦ = d ◦ . The magnitude of an exterior angle is
a°
equal to the sum of the magnitudes of the two
A
opposite interior angles.
B
A triangle is said to be equilateral if all its sides
are of the same length: AB = BC = CA.
The angles of an equilateral triangle are all of
magnitude 60◦ .
c°
d°
C
60°
60°
A
60°
C
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363
Chapter 12 — Geometry
The bisector of each of the angles of an equilateral
triangle meets the opposite side at right angles and
passes through the midpoint of that side.
B
O
C
A
A triangle is said to be isosceles if it has two sides
of equal length. If a triangle is isosceles, the angles
opposite each of the equal sides are equal.
E
B
A
The sum of the magnitudes of the exterior angles
of a triangle is equal to 360◦: e◦ + d ◦ + f ◦ = 360◦
A triangle is said to be a right-angled triangle if it
has one angle of magnitude 90◦ .
B e°
f°
PL
Example 4
C
C
A
d°
Angle sum of a triangle
Find the values of the pronumerals.
Solution
M
20◦ + 22◦ + x◦ = 180◦ (sum angles = 180◦ )
∴ 42◦ + x◦ = 180◦ or x = 138
138◦ + y ◦ = 180◦ (sum angles = 180◦ )
A
∴ y = 42
B
22°
x° y°
C
20°
SA
Or, to find x:
Two of the angles sum to 42◦ and therefore the third angle is 138◦ . To find y ◦ . The two angles
sum to 180◦ . Therefore the second is 42◦
Example 5
Angle sum of an isosceles triangle
Find the values of the pronumerals.
Solution
100◦ + 2x ◦ = 80◦ (sum angles = 180◦ )
∴ 2x ◦ = 80◦ or x = 40
Or, observe the two unknown numbers are the same and
must sum to 80◦ , therefore each of them has size 40◦ .
B
100°
A
x°
x°
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364
Essential Further Mathematics – Module 2 Geometry and trigonometry
Exercise 12B
1 Find the values of the pronumerals in each of the following:
a
b
B
80°
50°
x°
A
c
Q
y°
x°
30°
A
C
d
B
70°
y°
Y
50°
30°
C
y°
X
P
e
g B
x°
D
PL
y°
C
y°
b° c°
A
w°
x°
a° 75°
40°
C
y° E
A
Properties of regular polygons—a review
M
12.3
Z
R
f
B
z°
40°
E
x°
x°
x°
Equilateral triangle
Square
Regular pentagon
Regular hexagon
Regular octagon
SA
A regular polygon has all sides of equal length and all angles of equal magnitude.
A polygon with n sides can be divided into n triangles. The first three polygons below are
regular polygons.
O
O
O
The angle sum of the interior angles of an n-sided convex polygon is given by the
formula:
S = [180(n − 2)]◦ = (180n − 360)◦
The result holds for any convex polygon. Convex means that a line you draw from any
vertex to another vertex lies inside the polygon.
The magnitude of each of the interior angles of an n-sided polygon is given by:
x=
(180n − 360)◦
n
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365
Chapter 12 — Geometry
The angle bisectors of a regular polygon meet at a point O.
For a regular polygon, a circle can be drawn with centre O on which all the vertices lie.
O
O
E
O
Example 6
PL
The sum of the angles at O of a regular polygon is 360◦ .
The sum of the exterior angles of a regular polygon is 360◦ .
Angle properties of an octagon
The diagram opposite shows a regular octagon.
a Show that x = 45.
b Find the size of angle y.
Solution
SA
M
a 8x ◦ = 360◦ (sum angles at O = 360◦ )
360◦
∴ x◦ =
= 45◦
8
∴ x = 45
b y ◦ + y ◦ + 45◦ = 180◦ (OBC isosceles)
∴ 2y ◦ = 135 or y = 67.5
Example 7
A
H
O
G
B
x°
y°
C
F
E
D
Angle sum of an octagon
Find the sum of the interior angles of an 8-sided convex polygon (octagon).
Solution
1 Use the rule for the sum of the interior
angles of an n-sided polygon:
S = (180n − 360)◦
2 In this example, n = 8. Substitute and
evaluate.
3 Write down your answer.
S = (180n − 360)◦
n =8
∴ S = (180 × 8 − 360)◦ = 1080◦
The sum of the interior angles is 1080◦ .
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366
Essential Further Mathematics – Module 2 Geometry and trigonometry
Exercise 12C
1 Name each of the following regular polygons.
a
e
d
c
b
B
C
E
2 ABCD is a square. BD and AC are diagonals which meet at O.
a Find the size of each of the angles at O.
b What type of triangle is triangle:
i BAD
ii AOB
D
A
PL
3 ABCDE is a regular pentagon.
O
a Find the value of:
i x
ii y
b Find the sum of the interior angles of the regular
pentagon ABCDE.
A
B
y°
x° O
C
D
E
4 ABCDEF is a regular hexagon.
Find the value of:
b y
E
M
a x
5 State the sum of the interior angles of:
a a 7-sided regular polygon
b a hexagon
F
A
y°
x°
O
B
D
C
c an octagon
SA
6 The angle sum of a regular polygon is 1260◦ . How many sides
does the polygon have?
7 A circle is circumscribed about a hexagon ABCDEF.
a Find the area of the circle if OA = 2 cm.
b Find the area of the shaded region.
8 The diagram shows a tessellation of regular hexagons
and equilateral triangles.
State the values of a and b and use these to explain
the existence of the tessellation.
A
B
F
O
C
E
D
b°
a°
9 If the magnitude of each angle of a regular polygon is 135◦ , how many sides does the
polygon have?
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367
Chapter 12 — Geometry
Pythagoras’ theorem
B
Pythagoras’ theorem
Pythagoras’ theorem states that for a right-angled
triangle ABC with side lengths a, b and c, as shown
in the diagram, a 2 + b2 = c2 .
c
A
a
C
b
E
Pythagoras’ theorem can be illustrated by the diagram shown
here. The sum of the areas of the two smaller squares is
equal to the area of the square on the longest side
(hypotenuse).
area = c2 cm2
area =
b2 cm2
m
b cm
PL
cc
a cm
area = a2 cm2
There are many different proofs of Pythagoras’
theorem. A proof due to the 20th President of the
United States, James A. Garfield, is produced using
the following diagram.
Y
Area of trapezium WXYZ = 12 (a + b)(a + b)
Z
Area of EYZ + area of EWX + area of EWZ
c
M
b
a
a
X
c
b
= 12 ab + 12 c2 + 12 ab
= ab + 12 c2
a2
b2
1
∴
+ ab +
= ab + c2
2
2
2
2
2
2
∴
a +b =c
E
W
SA
12.4
Pythagorean triads
A triple of natural numbers (a, b, c) is called a
Pythagorean triad if c2 = a 2 + b2 .
The table presents the first six such ‘primitive’
triples. The adjective ‘primitive’ indicates that the
highest common factor of the three numbers is 1.
Example 8
a
b
c
3
4
5
5
12
13
7
24
25
8
15
17
9
40
41
Pythagoras’ theorem
Find the value, correct to two decimal places, of
the unknown length for the triangle opposite.
x cm
6.1 cm
cm Evans, Lipson
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Essential Further Mathematics – Module 2 Geometry and trigonometry
Solution
Example 9
The length is 8.08 cm correct to two
decimal places.
PL
2 Write down your answer for the length correct
to two decimal places.
2
2
x2 = 5.3
+ 6.1 (Pythagoras)
∴ x = 5.32 + 6.12 = 8.080 . . .
E
1 Using Pythagoras’ theorem, write down an
expression for x in terms of the two other
sides of the right-angled triangle. Solve for x.
Pythagoras’ theorem
Find the value, correct to two decimal places, of
the unknown length for the triangle opposite.
Solution
5.6 cm
y cm
M
1 Using Pythagoras’ theorem, write down an
expression for y in terms of the two other
sides of the right-angled triangle. Solve for y.
2 Write down your answer for the length correct
to two decimal places.
5.62 + y 2 = 8.62
∴ y 2 =8.62 − 5.62
∴y =
(Pythagoras)
8.62 − 5.62 = 6.526 . . .
The length is 6.53 cm correct to two
decimal places.
Pythagoras’ theorem
SA
Example 10
8.6 cm
The diagonal of a soccer ground is 130 m and the long side of the ground measures 100 m.
Find the length of the short side correct to the nearest cm.
Solution
1 Draw a diagram. Let x be the length of the
short side.
2 Using Pythagoras’ theorem, write down an
expression for x in terms of the two other sides
of the right-angled triangle. Solve for x.
4 Write down your answer correct to the
nearest cm.
130 m
xm
100 m
Let x m be the length of the shorter
side.
x 2 + 1002 = 1302 (Pythagoras)
∴ x 2 =1302 − 1002 = 6900
∴x =
1302 − 1002 = 83.066 . . .
Correct to the nearest centimetre, the
length of the short side is 83.07 m.
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369
Chapter 12 — Geometry
Exercise 12D
1 Find the length of the ‘unknown’ side for each of the following:
b
a
6 cm
c
10 cm
5 cm
3 cm
10 cm
11 cm
C
9 cm
7 cm
33 cm
B
15 cm
12 cm
PL
44 cm
f
E
e
d
A
2 In each of the following find the value of x correct to two decimal places.
b
a
4.8 cm
2.8 cm
d
c
x cm
x cm
6.2 cm
3.2 cm
9.8 cm
x cm
x cm
3.5 cm
M
5.2 cm
4 cm
3 cm
3 Find the value of x for each of the following (x > 0). Give your answers correct
to 2 decimal places.
a x 2 = 62 + 42
b 52 + x 2 = 92
c 4.62 + 6.12 = x 2
4 In triangle VWX, there is a right angle at X.VX = 2.4 cm and XW = 4.6 cm. Find VW.
A
SA
5 Find AD, the height of the triangle.
32 cm
32 cm
C
D
20 cm
B
6 An 18 m ladder is 7 m away from the bottom of a vertical wall. How far up the wall does it
reach?
7 Find the length of the diagonal of a rectangle with dimensions 40 m × 9 m.
8 Triangle ABC is isosceles. Find the length of CB.
A
8
14
C
B Lipson
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370
Essential Further Mathematics – Module 2 Geometry and trigonometry
9 In a circle of centre O, a chord AB is of length 4 cm.
The radius of the circle is 14 cm. Find the distance
of the chord from O.
O
x cm
B
A
10 Find the value of x.
18 cm
E
x cm
x cm
X
PL
11 How high is the kite above the ground?
0m
17
Y
90 m
12 A square has an area of 169 cm2 . What is the length of the diagonal?
M
13 Find the area of a square with a diagonal of length:
√
a 8 2 cm
b 8 cm
20 cm
14 Find the length of AB.
SA
15 ABCD is a square of side length 2 cm.
If CA = CE, find the length of DE.
C
D
12 cm
8 cm
A
B
B
A
C
D
E
16 The midpoints of a square of side length 2 cm
are joined to form a new square. Find the area
of the new square.
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371
Chapter 12 — Geometry
Similar figures
Similarity
In this section we informally define two objects to be similar if they have the same shape
but not the same size.
Examples
Any two circles are similar to each other.
3 cm
4 cm
E
C2
Any two squares are similar to each other.
3 cm
S1
4 cm
S2
PL
C1
It is not true that any rectangle is similar to any other rectangle. For example, rectangle 1 is not
similar to rectangle 2.
4 cm
R1
8 cm
R2
1 cm
1 cm
A rectangle similar to R1 is:
8 cm
2 cm
M
R3
So, for
to be similar, their corresponding sides must be in the same
two rectangles
8
4
ratio
=
.
2
1
Similar triangles
Two triangles are similar if one of the following conditions holds:
corresponding angles in the triangles are equal
SA
12.5
A
45°
B'
B
100°
100°
35°
C
A'
45°
35°
C'
corresponding sides are in the same ratio
AB
BC AC =
=
=k
AB
BC
AC
k is the scale factor.
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Essential Further Mathematics – Module 2 Geometry and trigonometry
two pairs of corresponding sides have the same ratio and the included angles are equal
B
B'
A
45°
C
A'
45°
C'
Example 11
Similar triangles
PL
Find the value of length of side AC in ABC
correct to two decimal places.
E
AB
AC
= AB
AC
If triangle ABC is similar to triangle XYZ this can be written symbolically as ABC ∼ XYZ.
The triangles are named so that angles of equal magnitude hold the same position, i.e. A
corresponds to X, B corresponds to Y, C corresponds to Z.
5 cm 20°
A
Solution
SA
M
1 Triangle ABC is similar to triangle AB C : two
pairs of corresponding
sides have the same
6.25
5
=
and included angles (20◦ ).
ratio
3
3.75
2 For similar triangles, the ratios of corresponding
AC
AB
sides are equal, for example, = .
AC
AB
Use this fact to write down an expression
involving x. Solve for x.
3 Write down your answer correct to two decimal
places.
Example 12
B'
B
x cm
6.25 cm
20°
3.75 cm
3 cm
C
A'
3.013 cm
C'
Triangles similar
∴
5
x
=
3.013
6.25
5
∴x =
× 3.013
6.25
= 2.4104
The length of side AC is 2.41 cm,
correct to two decimal places.
Similar triangles
Find the value of length of side AB in ABC.
m
6c
B
m
xc
A
Solution
1 Triangle ABC is similar to triangle AXY
(corresponding angles are equal).
X
3 cm
C
Y
2.5 cm
Triangles similar
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373
Chapter 12 — Geometry
3
x
=
x +6
5.5
∴ 5.5x = 3(x + 6)
= 3x + 18
2 For similar triangles, the ratios of corresponding
AB
AC
sides are equal (for example,
=
).
AX
AY
Use this fact to write down an expression
involving x. Solve for x. Note that,
if AB = x then AX = x + 6.
∴ 2.5x = 18 or x =
3 Write down your answer.
The length of side AB is 7.2 cm.
∴
E
18
= 7.2
2.5
Exercise 12E
1 Find the value of x for each of the following pairs of similar triangles.
a
A
PL
A'
4 cm 82° 5 cm
56°
B
C
b
9 cm
B'
135°
18°
18°
C
X
d
x cm
13 cm
Y
Z
x cm
5 cm Y
B
M
c
C'
135°
A
10 cm
56°
C
6 cm
B
x cm
82°
D
10 cm
°
C
14 cm
×
13 cm
B
X
8 cm
12 cm
C
e
SA
°
A
10 cm
f
A
6 cm
°
B
°
×
E
P
6 cm
B
R
g
C
x cm
12 cm x cm
8 cm
Q
x cm
h
°
A
A
A
12 cm
P
x cm
B
E
2 cm
°
D
4 cm
i
B
16 cm
Q
j
A
x cm
8 cm
C
2 cm
x cm
P
Q
1.5
P
D
2 cm
A
3 cm
E
x cm
A
C
2 cm
6 cm
B
cm B
8 cm
C
10 cm
C
Q
2 cm
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Essential Further Mathematics – Module 2 Geometry and trigonometry
2 Given that AD = 14, ED = 12, BC = 15 and EB = 4,
find AC, AE and AB.
D
C
15
14
12
A
4
B
E
°
33 m
°
0.24 m
E
3 A tree casts a shadow of 33 m and at the same time a stick 30 cm long casts a shadow
24 cm long. How high is the tree?
0.3 m
4 A 20 metre high neon sign is supported by a 40 m steel
cable as shown. An ant crawls along the cable starting at A.
How high is the ant when it is 15 m from A?
m
PL
40
20 m
A
5 A hill has a gradient of 1 in 20, i.e. for every 20 m horizontally there is a 1 m increase in
height. If you go 300 m horizontally, how high up will you be?
M
6 A man stands at A and looks at point Y across the river.
He gets a friend to place a stone at X so that A, X and Y
are collinear. He then measures AB, BX and XC to be
15 m, 30 m and 45 m respectively. Find CY, the
distance across the river.
45 m
C
A
15 m
X
30 m
B
river
Y
7 Find the height, h m, of a tree that casts a shadow 32 m long at the same time that a vertical
straight stick 2 m long casts a shadow 6.2 m long.
8 A plank is placed straight up stairs that are 20 cm wide
and 12 cm deep. Find x, where x cm is the width of the
widest rectangular box of height 8 cm that can be placed
on a stair under the plank.
SA
nk
pla
8 cm
x cm 12 cm
20 cm
9 The sloping edge of a technical drawing table is 1 m
from front to back. Calculate the height above the ground
of a point A, which is 30 cm from the front edge.
1m
30
cm
A
92 cm
80 cm
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375
Chapter 12 — Geometry
10 Two similar rods 1.3 m long have to be hinged together
to support a table 1.5 m wide. The rods have been fixed
to the floor 0.8 m apart. Find the position of the hinge
by finding the value of x.
1.5 m
xm
(1.3 – x) m
E
0.8 m
11 A man whose eye is 1.7 m from the ground when standing 3.5 m in front of a wall 3 m high
can just see the top of a tower that is 100 m away from the wall. Find the height of the tower.
PL
12 A man is 8 m up a 10 m ladder, the top of which leans against a vertical wall and touches it
at a height of 9 m above the ground. Find the height of the man above the ground.
13 A spotlight is at a height of 0.6 m above ground level.
A vertical post 1.1 m high stands 3 m away, and 5 m
further away there is a vertical wall. How high up the
wall does the shadow reach?
wall
Volumes and surface areas
M
Volume of a prism
A prism is a solid which has a constant cross-section. Examples are cubes, cylinders,
rectangular prisms and triangular prisms
SA
12.6
spotlight
0.6 m 1.1 m
vertical post
The volume of a prism can be found by using its cross-sectional area.
volume = area of cross-section × height (or length)
V = A×h
Answers will be in cubic units, i.e. mm3 , cm3 , m3 etc.
Example 13
Volume of a cylinder
Find the volume of this cylinder which has radius 3 cm
and height 4 cm correct to two decimal places.
3 cm
4 cm
Solution
1 Find the cross-sectional area of the prism.
Area of cross-section = πr 2 = π × 32
= 28.27 cm 2
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376
Essential Further Mathematics – Module 2 Geometry and trigonometry
2 Multiply by the height
3 Make sure that accuracy is given to the
correct number of decimal places
Volume = 28.27 × 4
= 113.10 cm 3 (correct to two
decimal places)
The formulas for determining the volumes of some ‘standard’ prisms are given here.
Solid
Cylinder (radius r cm, height h cm)
Formula
V = r 2 h
E
r cm
h cm
Cube (all edges x cm)
PL
V = x3
Rectangular prism (length l cm,
width w cm, height h cm)
V = lwh
h cm
w cm
l cm
Triangular prism
h cm
The triangular prism shown has a right-angled
triangle base but the following formula holds
for all triangular prisms
l cm
M
b cm
V = 12 bhl
Volume of a pyramid
The formula for finding the volume of a right pyramid
can be stated as:
SA
Volume of pyramid =
× base area
× perpendicular height
1
3
h
For the square pyramid shown:
V = 13 x 2 h
x
The term right in this context means that the apex of the pyramid is directly over the centre of
the base.
Example 14
Volume of a pyramid
Find the volume of this hexagonal pyramid with a base area of 40 cm2
and a height of 20 cm. Give the answer correct to one decimal place.
Solution
20 cm
V =
1
3
× A ×h
=
1
3
× 40 × 20
3
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377
Chapter 12 — Geometry
Example 15
Volume of a pyramid
Find the volume of this square pyramid with a square base
with each edge 10 cm and a height of 27 cm.
27 cm
Solution
1
V = 3x2 h
1
3
× 10 × 10 × 27
= 900 cm3
Volume of a cone
10 cm
10 cm
E
=
Volume of cone =
V =
PL
The formula for finding the volume of a cone can be stated as:
1
× base
3
1
r 2 h
3
area × height
Volume of a sphere
The formula for the volume of a sphere is:
V = 43 r 3
r
M
where r is the radius of the sphere.
Example 16
Volume of a sphere
Find the volume of this sphere.
Solution
SA
4
Volume of sphere = πr 3
3
4
= × π × 43
3
4 cm
= 268.08 cm3 (2 decimal places)
Composite shapes
Using the shapes above, new shapes can be made. The volumes of these can be found by
summing the volumes of the component solids.
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378
Essential Further Mathematics – Module 2 Geometry and trigonometry
Example 17
Volume of a composite shape
Solution
1 Use the formula V = r 2 h to find the
volume of the cylinder
The volume of the cylinder
= π × 52 × 5 = 125π mm3
The volume of the hemisphere
2
250
= π × 53 =
π mm3
3
3
PL
2 Find the volume of hemisphere using
the formula that the volume of a
hemisphere = 12 43 r 3 = 23 r 3 .
E
A hemisphere is placed on top of a cylinder to form a capsule.
The radius of both the hemisphere and the cylinder is 5 mm.
The height of the cylinder is also 5 mm. What is the volume of the
composite solid in cubic millimeters, correct to two decimal places?
Therefore the volume of the composite =
250
625 π
125π +
π=
= 654.498 . . .
3
3
The volume of the composite = 654.50 mm3
(correct to two decimal places)
3 Add the two together.
4 Write down your answer.
M
Surface area of three-dimensional shapes
The surface area of a solid can be found by calculating and totalling the area of each of its
surfaces. The net of the cylinder in the diagram demonstrates how this can be done.
l
l
SA
r
Total area = 2r + 2rl
= 2r (r + l)
2
r
A = πr2
r
A = 2πrl
A = πr2
2πr
Here are some more formulas for the surface areas of some solids. The derivation of these is
left to the reader.
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379
Chapter 12 — Geometry
Solid
Cylinder (radius r cm, height h cm)
Formula
r cm
S = 2r 2 + 2r h
h cm
Cube (all edges x cm)
E
S = 6x 2
Rectangular prism (length l cm,
width w cm, height h cm)
S = 2(lw + lh + w h)
h cm
l cm
PL
w cm
Triangular prism
√
S = bh + bl + hl + l b2 + h 2
h cm
b cm
Surface area of a right square pyramid
M
Example 18
l cm
Find the surface of the right square pyramid shown if the
square base has each edge 10 cm in length and the isosceles
triangles each have height 15 cm.
SA
Solution
1 Draw the net of the pyramid.
2 First determine the area of the square.
3 Determine the area of one of the
isosceles triangles.
4 Find the sum of the areas of the four
triangles and add to the area of the
square.
Area of the square = 102 = 100 cm2
The area of one triangle
1
= × 10 × 15 = 75 cm2
2
The surface area of the solid
= 100 + 4 × 75 = 100 + 300
= 400 cm2
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380
Essential Further Mathematics – Module 2 Geometry and trigonometry
Exercise 12F
1 Find the volume in cm3 of each of the following shapes, correct to two decimal places.
b
radius 6.3 cm and height 2.1 cm
c
dimensions 2.1 cm, 8.3 cm and 12.2 cm
d
PL
E
a
area of cross section = 2.8 cm2
height = 6.2 cm
radius 2.3 cm and length 4.8 cm
2 Find, correct to two decimal places, the surface area and
volume of the solid shown given that the cross section is
a right angled isosceles triangle.
4 cm
4 cm
12 cm
M
3 The box shown has dimensions length: 13 cm,
width: 4 cm and height 3 cm.
a Find the surface area of the box in square centimetres (cm2 ).
b Find the volume of the box in cubic centimetres (cm3 ).
4 Find the volumes, to two decimal places, of spheres with:
a radius = 4 mm
b diameter = 23 cm c radius = 3.8 m d diameter = 15 cm
SA
5 Find the volume, to two decimal places, of hemispheres with:
a radius = 12 cm
b diameter = 32 mm c radius = 16 mm
d radius = 15 cm
6 Calculate the volume of a right pyramid with a rectangular base 18 m by 15 m. The vertex
of the pyramid is 20 m perpendicularly above the centre of the base.
7 Each side of the square base of one of the great Egyptian pyramids is 275 m long. Calculate
the volume of the pyramid, to the nearest cubic metre, if it has a perpendicular height of
175 m.
V
8 Find the surface area and volume of the right square pyramid shown.
12 cm
B
The length of each edge of the square base is 10 cm and the
height of the pyramid is 12 cm.
A
C
O
10 cm
X
D
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381
Chapter 12 — Geometry
9 The diagram shows a capsule, which consists of two hemispheres,
each of radius 2 cm, and a cylinder length 5 cm and radius 2 cm.
The surface area of a sphere is given by the formula S = 4r 2 and
the surface area of the curved section of a cylinder is given by the
formula S = 2r h.
Find the surface area and volume of the capsule. Give your answers
correct to two decimal places.
5 cm
E
10 Find:
a the surface area
b the volume
of the object shown.
2 cm
3m
4m
2m
10 m
PL
5m
11 The diagram opposite shows a right pyramid on a cube.
Each edge of the cube is 14 cm.
The height of the pyramid is 24 cm.
Find:
a the volume of the solid
b the surface area of the solid
M
24 cm
14 cm
SA
12 Find:
a the surface
b the volume
of the solid shown opposite.
7 cm
4 cm
10 cm
5 cm
20 cm
4 cm
13 The solid opposite consists of a half cylinder on
a rectangular prism. Find, correct to two decimal
places:
a the surface area
b the volume
10 cm
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12.7
Essential Further Mathematics – Module 2 Geometry and trigonometry
Areas, volumes and similarity
Areas
Some examples of similar shapes and the ratio of their areas are considered in the following.
Similar circles
4
radius circle 2
=
radius circle 1
3
× 42
42
4 2
Ratio of areas =
= 2 =
= k2
× 32
3
3
Scale factor = k =
Area = × 32
3 cm
Area = 3 × 2
= 6 cm2
length rectangle 2
6
= =2
length rectangle 1
3
24
Ratio of areas =
= 4 = (2)2 = k 2
6
5 cm
3 cm
4 cm
Area = × 4 × 3
= 6 cm2
SA
1
2
4 cm
Scale factor = k =
Scale factor = k =
M
Similar triangles
Area = × 42
PL
Similar rectangles
2 cm
4 cm
E
3 cm
Ratio of areas =
height triangle 2
9
= =3
height triangle 1
3
54
= 9 = (3)2 = k 2
6
6 cm
Area = 6 × 4
= 24 cm2
15 cm
9 cm
12 cm
Area = 12 × 12 × 9
= 54 cm2
A similar pattern emerges for other shapes. Scaling the linear dimension of a shape by a factor
of k scales the area by a factor of k 2 .
Scaling areas
If two shapes are similar and the scale factor is k, then the area of the similar
shape = k2 × area of the original shape.
Example 19
Using area scale factors with similarity
The two triangles shown are similar.
The base of the smaller triangle has a length of 10 cm.
Its area is 40 cm2 .
The base of the larger triangle has a length of 25 cm.
Determine its area.
40 cm2
10 cm
25 cm
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383
Chapter 12 — Geometry
Solution
2 Write down the area of the small triangle.
3 Area of larger triangle = k 2 × area of
smaller triangle.
25
= 2.5
10
Area of small triangle = 40 cm2
∴ Area of larger triangle = 2.52 × 40
= 250
Substitute the appropriate values and
evaluate.
4 Write down your answer.
The area of the larger triangle is 250 cm2 .
Example 20
Scale factors and area
Solution
12 cm
60 cm
PL
The two hearts shown are similar shapes.
The width of the larger heart is 60 cm.
Its area is 100 cm2 .
The width of the smaller heart is 12 cm.
Determine its area.
k=
E
1 Determine the scale factor k.
M
1 Determine the scale factor k. Note we are
scaling down.
2 Write down the area of the larger heart.
3 Area of smaller heart = k 2 × area of
larger heart.
Substitute the appropriate values and
evaluate.
4 Write down your answer.
Area = 100 cm2
k=
12
= 0.2
60
Area of larger heart = 100 cm2
∴ Area of smaller heart = 0.22 × 100
=4
The area of the smaller heart is 4 cm2 .
SA
Volumes
Two solids are considered to be similar if they have the same shape and the ratio of their
corresponding linear dimensions is equal.
Some examples of similar volume and the ratio of their areas are considered in the
following.
Similar spheres
3 cm
Volume = 43 × 33
= 36 cm3
Scale factor = k =
radius sphere 2
4
=
radius sphere 1
3
256
256
Ratio of volumes = 3
=
36 108
64
4 3
=
= k3
=
27
3
4 cm
Volume = 43 × 43
256
=
cm3
3
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384
Essential Further Mathematics – Module 2 Geometry and trigonometry
Similar cubes
4 cm
2 cm
Scale factor = k =
2 cm
2 cm
64
=8
8
3
= (2) = k 3
4 cm
4 cm
Ratio of volumes =
Volume = 4 × 4 × 4
= 64 cm3
E
Volume = 2 × 2 × 2
= 8 cm3
side length 2
4
= =2
side length 1
2
Similar cylinders
1 cm
3 cm
radius 2
3
Scale factor = k =
= =3
radius 1
1
6 cm
PL
2 cm
54
Volume = × 12 × 2 Ratio of volumes =
= 27 = (3)3 = k 3
2
Volume = × 32 × 6
= 2 cm3
= 54 cm3
A similar pattern emerges for other solids. Scaling the linear dimension of a solid by a factor
of k scales the volume by a factor of k 3 .
M
Scaling volumes
If two solids are similar and the scale factor is k, then the volume of the similar
solid = k3 × volume of the original solid.
Example 21
Similar solids
SA
The two cuboids shown are similar solids.
The height of the larger cuboid is 6 cm.
Its volume is 120 cm3 .
The height of the smaller cuboid is 1.5 cm.
Determine its volume.
1.5 cm
6 cm
volume = 120 cm3
Solution
1 Determine the scale factor k. Note we are
scaling down.
2 Write down the volume of the larger cuboid.
3 Volume smaller cuboid = k 3 × volume
larger cuboid.
k=
1.5
= 0.25
6
Volume larger cuboid = 120 cm3
Volume smaller cuboid = 0.253 × 120
= 1.875
Substitute the appropriate values and evaluate.
4 Write down your answer.
The volume of the smaller cuboid is
1.875 cm3 .
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385
Chapter 12 — Geometry
Example 22
Similar solids
V'
V
The two square pyramids shown are similar
with a base dimensions 4 and 5 cm respectively.
The height of the first pyramid is 9 cm and
its volume is 48 cm3 . Find the height and
volume of the second pyramid.
A
9 cm
C
B
4 cm
O
A'
D
D'
k=
5
= 1.25
4
PL
Height
2 Write down the height of Pyramid 1.
3 Height Pyramid 2 = k× height Pyramid 1.
Substitute the appropriate values and evaluate.
4 Write down your answer.
Pyramid 2
E
1 Determine the scale factor k. Use the base
measurements.
5 cm
O'
Pyramid 1
Solution
C'
B'
Height 1 = 9 cm
∴ Height 2 = 1.25 × 9 = 11.25
The height of Pyramid 2 is 11.25 cm.
6 Write down your answer.
The volume of Pyramid 2 is 93.75 cm3 .
M
Volume
5 Volume Pyramid 2 = k 3 × volume Pyramid 1
Substitute the appropriate values and evaluate.
Volume 1 = 48 cm3
∴ Volume 2 = 1.253 × 48 = 93.75
Exercise 12G
SA
1 Triangle ABC is similar to triangle XYZ.
The length scale factor k = 1.2. The area
of triangle ABC is 6 cm2 . Find the area
of triangle XYZ.
2 The two rectangles are similar. The area
of rectangle ABCD is 20 cm2 . Find the
area of rectangle AB C D .
B
A
B
3 cm
A
Y
×1.2
C X
C
Z
B'
5 cm
D
A'
3 The two shapes shown are similar. The length
scale factor is 32 . The area of the shape to the
right is 30 cm2 . What is the area of the shape
to the left?
4 Triangle ABC is similar to triangle XYZ.
YZ
ZX
XY
=
=
= 2.1.
AB
BC
CA
2
C'
TheUniversity
area of triangle
XYZ is 20Sample
cm . Find
area of triangle •ABC.
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386
Essential Further Mathematics – Module 2 Geometry and trigonometry
5 Triangles ABC and AB C are equilateral triangles.
a Find the length of BF.
b Find a.
area of triangle AB C .
c Find the ratio
area of triangle ABC
B'
B
2 cm
A
2 cm
F
2 cm
C
a cm
2 cm
A'
a cm
C'
F'
a cm
E
6 The areas of two similar triangles are 16 and 25. What is the ratio of a pair of
corresponding sides?
7 The areas of two similar triangles are 144 and 81. If the base of the larger triangle is 30,
what is the corresponding base of the smaller triangle?
PL
8 These two rectangular prisms are similar. The length scale
factor is 1.8. The volume of the first solid is 20 cm3 .
What is the volume of the second solid?
9 Two cones are similar. The ratio of volumes is 8 : 125. Find the ratio of the:
i heights
ii lengths of sloping edges iii areas of bases.
5 cm
10 cm
M
10 A cone has water poured into it as shown. Find the ratio
of the volume of empty space in the cone to volume
of water.
11 Consider two similar cuboids that have edges where lengths are in the ratio 1 : 4.
a Find the ratio of the surface area of the two cuboids.
b Find the ratio of the volumes.
SA
12 An inverted right circular cone of capacity 100 m3 is filled with
water to half its depth. Find the volume of water.
13 The ratio of the radii of two spheres is 2 : 5. Find the ratio of:
i the surface areas
ii the volumes
14 Two right circular cones are as shown. Find:
a the ratio of the heights of the cones
b the ratio of the surface areas
c the ratio of the volumes
15 cm
45 cm
10 cm
30 cm
15 The ratio of the volumes of two cubes is 1 : 27. Find:
a the ratio of the surface areas of the cubes
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387
Chapter 12 — Geometry
Review
Key ideas and chapter summary
Alternate, corresponding,
cointerior and vertically
opposite angles
Angles 4 and 6 are examples of alternate angles.
Angles 2 and 6 are examples of corresponding angles.
Angles 3 and 6 are examples of cointerior angles.
l3
Angles 1 and 3 are examples
2
1
l1
of vertically opposite angles
4 3
l2
and are of equal magnitude.
5 6
7
When lines l1 and l2 are parallel corresponding angles
are of equal magnitude, alternate angles are of equal
magnitude and cointerior angles are supplementary.
PL
Angles associated with parallel
lines crossed by a transversal
line
E
8
Corresponding
Corresponding
Alternate
Alternate
Converse results also hold:
If corresponding angles are equal then l1 is parallel to l2 .
If alternate angles are equal then l1 is parallel to l2 .
If cointerior angles are supplementary then l1 is parallel to l2 .
The sum of the magnitudes of the interior angles of a
triangle is equal to 180◦: a ◦ + b◦ + c◦ = 180◦ .
Equilateral triangle
A triangle is said to be equilateral if
all of its sides are of the same length.
The angles of an equilateral triangle
are all of magnitude 60◦ .
SA
M
Angle sum of triangle
Isosceles triangle
Polygon
10 cm
10 cm
10 cm
A triangle is said to be isosceles if
it has two sides of equal length.
5 cm
5 cm
If a triangle is isosceles the angles
opposite each of the equal sides are equal.
A polygon is a closed geometric shape with sides which are
segments of straight lines. Examples are:
3 sides:
Triangle
4 sides:
Quadrilateral
5 sides:
Pentagon
6 sides:
Hexagon
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Essential Further Mathematics – Module 2 Geometry and trigonometry
A polygon is said to be convex if any diagonal lies
inside the polygon.
Regular polygon
A regular polygon has all sides of equal length and
all angles of equal magnitude.
Sum of the interior angles
The angle sum of the interior angles of an n-sided
polygon is given by the formula: S = (180n − 360)◦ .
Pythagoras’ theorem
Pythagoras’ theorem states that for a right-angled
triangle ABC with side lengths a, b and c,
a 2 + b2 = c2 , where c is the longest side.
E
Convex polygon
We informally define two objects to be similar if they
have the same shape but not the same size.
Conditions for similarity of triangles r Corresponding angles in the triangles are equal.
r Corresponding sides are in the same ratio.
PL
Similar figures
AB B C AC =
=
=k
AB
BC
AC
where k is the scale factor
r Two pairs of corresponding sides have the same
ratio and the included angles are equal.
Volumes of solids
r cm
M
Cylinder:
V = r 2 h
h cm
x
Cube:
V = x3
SA
Review
388
Surface area of solids
h cm
Rectangular prism:
V = lw h
Right-angled triangular prism
V = 12 bhl
w cm
l cm
h cm
b cm
l cm
r cm
Cylinder:
S = 2r 2 + 2r h
h cm
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389
Chapter 12 — Geometry
Cube:
S = 6x 2
h cm
Rectangular prism:
S = 2 (lw + lh + w h)
w cm
E
l cm
Right-angled triangular prism
√
S = bh + bl + hl + l b2 + h 2
Scaling, areas and volumes
h cm
b cm
l cm
r If two shapes are similar and the scale factor is k,
PL
then the area of the similar shape = k 2 × area of
the original shape.
k =
3
2
k2 =
9
4
r If two solids are similar and the scale factor
k=
3
2
k3 =
27
8
SA
M
is k, then the volume of the similar
solid = k 3 × volume of the original solid.
Skills check
Having completed this chapter you should be able to:
apply the properties of parallel lines and triangles and regular polygons to find the
size of an angle given suitable information
find the size of each interior angle of a regular polygon with a given number of sides
use the definition of objects such as triangles, quadrilaterals, squares, pentagons,
hexagons, equilateral triangles, isosceles triangles to determine angles
recognise when two objects are similar
determine unknown lengths and angles through use of similar triangles
find surface areas and volumes of solids
use Pythagoras’ theorem to find unknown lengths in right-angled triangles
use similarity of two- and three-dimensional shapes to determine areas and volumes
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Essential Further Mathematics – Module 2 Geometry and trigonometry
Multiple-choice questions
Questions 1 to 3 relate to the diagram
1 Angle PRS =
B 40◦
C 50◦
D 60◦
A 20◦
E 110◦
2 Angle RPS =
B 40◦
A 20◦
E 110◦
C 50◦
D 60◦
P
120°
70°
E
3 Given that PS bisects angle QPR, the size of angle PQS is:
B 35◦
C 40◦
D 50◦
E 60◦
A 20◦
R
S
Q
PL
Questions 4 to 6 relate to the diagram
Lines m and l are parallel and cut by a transversal q.
4 The value of x is:
A 65
B 125
5 The value of y is:
A 65
B 125
6 The value of z is:
A 65
B 125
C 62.5
D 60
C 62.5
D 60
E 55
m
z° y°
l
q
C 62.5
D 60
E 55
C
M
6 cm
A
8 The triangle ABC has a right angle at A. The length of
side BC to the nearest cm is:
A 10
B 14
C 9
D 12
E 11
C
7 cm
Z
A 3 cm B
X.
10 YZ is parallel to Y Z and Y Y =
The area of triangle XYZ is 60cm . The area of triangle X Y Z is:
B 30 cm2
C 15 cm2
A 20 cm2
80
20
cm2
cm2 E
D
3
3
B
9 cm
9 Triangles ABC and XYZ are similar isosceles triangles.
The length of XY is:
C
A 4 cm
B 5 cm
C 4.2 cm
D 8.5 cm
E 7.2 cm
5 cm
5 cm
B
8 cm
A
125°
x°
E 55
7 The triangle ABC has a right angle at A. The length
of side BC in cm is:
A 10
B 14
C 9
D 9. 8
E 11
SA
Review
390
1
Y
2
2
12 cm
12 cm
X
Y
X
Y'
Z'
Y
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Chapter 12 — Geometry
1.2 cm
C 2.16
1.8 m
xm
18 m
E
12 A regular convex polygon has 12 sides. The magnitude of each of its interior angles
is:
B 45◦
C 60◦
D 150◦
E 120◦
A 30◦
Z
C
7 cm
7 cm
A
B
10 cm
14 Two similar cylinders are shown. The ratio of the volume
of the smaller cylinder to the larger cylinder is:
15 cm
A 1:4
B 1:16
C 1:64
10 cm
D 15:60
E 1:3
10 cm
Y
X
4 cm
PL
13 Triangles ABC and XYZ are similar isosceles triangles.
The length of XY correct to one decimal place is:
A 4.8 cm
B 5.7 cm
C 4.2 cm
D 8.5 cm
E 8.2 cm
60 cm
40 cm
15 Each interior angle of a regular polygon measures 135◦ . The number of sides the
polygon has is:
A 4
B 6
C 8
D 10
E 7
D
C
10 cm
A
B
SA
M
16 Each side length of a square is 10 cm. The length of the
diagonal is:
√
√
C 10 2
D 8
E 1.4
A 10
B 5 2
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11 The value of x is:
A 12
B 27
D 20.8
E 13.81