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Geometry 1
TERMINOLOGY
Altitude: Height. Any line segment from a vertex to the
opposite side of a polygon that is perpendicular to that side
Polygon: General term for a many sided plane figure. A
closed plane (two dimensional) figure with straight sides
Congruent triangles: Identical triangles that are the same
shape and size. Corresponding sides and angles are
equal. The symbol is /
Quadrilateral: A four-sided closed figure such as a square,
rectangle, trapezium etc.
Interval: Part of a line including the endpoints
Similar triangles: Triangles that are the same shape but
different sizes. The symbol is zy
Median: A line segment that joins a vertex to the
opposite side of a triangle that bisects that side
Vertex: The point where three planes meet. The corner of
a figure
Perpendicular: A line that is at right angles to another
line. The symbol is =
Vertically opposite angles: Angles that are formed
opposite each other when two lines intersect
Chapter 4 Geometry 1
INTRODUCTION
GEOMETRY IS USED IN many areas, including surveying, building and graphics.
These fields all require a knowledge of angles, parallel lines and so on, and
how to measure them. In this chapter, you will study angles, parallel lines,
triangles, types of quadrilaterals and general polygons.
Many exercises in this chapter on geometry need you to prove something
or give reasons for your answers. The solutions to geometry proofs only give
one method, but other methods are also acceptable.
DID YOU KNOW?
Geometry means measurement of the earth and comes from Greek. Geometry was used in ancient
civilisations such as Babylonia. However, it was the Greeks who formalised the study of geometry,
in the period between 500 BC and AD 300.
Notation
In order to show reasons for exercises, you must know how to name figures
correctly.
•B
The point is called B.
The interval (part of a line) is called AB or BA.
If AB and CD are parallel lines, we write AB < CD.
This angle is named +BAC or +CAB. It can sometimes be named +A.
^
Angles can also be written as BAC or BAC.
This triangle is named 3ABC.
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To name a quadrilateral,
go around it: for example,
BCDA is correct, but ACBD
is not.
Producing a line is the same
as extending it.
This quadrilateral is called ABCD.
Line AB is produced to C.
+ABD and +DBC are
equal.
DB bisects +ABC.
AM is a median of D ABC.
AP is an altitude of D ABC.
Types of Angles
Acute angle
0c1 xc1 90c
Chapter 4 Geometry 1
Right angle
A right angle is 90c.
Complementary angles are angles whose sum is 90c.
Obtuse angle
90c1 xc1180c
Straight angle
A straight angle is 180c.
Supplementary angles are angles whose sum is 180c.
Reflex angle
180c1 xc1 360c
Angle of revolution
An angle of revolution is 360c.
Vertically opposite angles
+AEC and +DEB are called vertically opposite angles. +AED and +CEB are
also vertically opposite angles.
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Vertically opposite angles are equal.
That is, +AEC = +DEB and +AED = +CEB.
Proof
Let +AEC = xc
Then +AED = 180c - xc
(+CED straight angle, 180c)
Now +DEB = 180c - (180c - xc)
(+AEB straight angle, 180c)
= xc
Also +CEB = 180c - xc
(+CED straight angle, 180c)
`
+AEC = +DEB and +AED =+CEB
EXAMPLES
Find the values of all pronumerals, giving reasons.
1.
Solution
x + 154 = 180
(+ABC is a straight angle, 180c)
x + 154 - 154 = 180 - 154
`
x = 26
2.
Solution
2x + 142 + 90 = 360
(angle of revolution, 360c )
2x + 232 = 360
2x + 232 - 232 = 360 - 232
2x = 128
2x
128
=
2
2
x = 64
Chapter 4 Geometry 1
3.
Solution
y + 2y + 30 = 90
(right angle, 90c)
3y + 30 = 90
3y + 30 - 30 = 90 - 30
3y = 60
3y
60
=
3
3
y = 20
4.
Solution
x + 50 = 165
x + 50 - 50 = 165 - 50
x = 115
y = 180 - 165
= 15
w = 15
(+WZX and +YZV vertically opposite)
(+XZY straight angle, 180c)
(+WZY and +XZV vertically opposite)
5.
CONTINUED
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Solution
a = 90
b + 53 + 90 = 180
b + 143 = 180
b + 143 - 143 = 180 - 143
b = 37
d = 37
c = 53
(vertically opposite angles)
(straight angle, 180c)
(vertically opposite angles)
(similarly)
6. Find the supplement of 57c 12l.
Solution
Supplementary angles add up to 180c.
So the supplement of 57c 12l is
180c - 57c 12l = 122c 48l.
7. Prove that AB and CD are straight lines.
A
D
(x + 30)c
C
(6x + 10)c
(2x
2 + 10)c
E
(5x + 30)c
B
Solution
6x + 10 + x + 30 + 5x + 30 + 2x + 10 = 360
^ angle of revolution h
14x + 80 - 80 = 360 - 80
14x = 280
14x
280
=
14
14
x = 20
+AEC = (20 + 30)c
= 50c
+DEB = (2 # 20 + 10)c
= 50c
These are equal vertically opposite angles.
` AB and CD are straight lines
Chapter 4 Geometry 1
4.1 Exercises
1.
Find values of all pronumerals,
giving reasons.
(a)
yc
(i)
133c
(b)
(j)
(c)
2.
Find the supplement of
(a) 59c
(b) 107c 31l
(c) 45c 12l
3.
Find the complement of
(a) 48c
(b) 34c 23l
(c) 16c 57l
4.
Find the (i) complement and
(ii) supplement of
(a) 43c
(b) 81c
(c) 27c
(d) 55c
(e) 38c
(f) 74c 53l
(g) 42c 24l
(h) 17c 39l
(i) 63c 49l
(j) 51c 9l
5.
(a) Evaluate x.
(b) Find the complement of x.
(c) Find the supplement of x.
(d)
(e)
(f)
(g)
(h)
(2x + 30)c
142c
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6.
Find the values of all
pronumerals, giving reasons for
each step of your working.
8.
(a)
Prove that CD bisects +AFE.
9.
Prove that AC is a straight line.
D
C
(b)
(3x + 70)c
(110 - 3x)c
B
(c)
A
10. Show that +AED is a right angle.
A
(d)
B
(50 - 8y)c
(e)
C
(5y - 20)c
E
(f)
7.
Prove that AC and DE are straight
lines.
(3y + 60)c
D
Chapter 4 Geometry 1
149
Parallel Lines
When a transversal cuts two lines, it forms pairs of angles. When the two
lines are parallel, these pairs of angles have special properties.
Alternate angles
Alternate angles form
a Z shape. Can you
find another set of
alternate angles?
If the lines are parallel, then alternate angles are equal.
Corresponding angles
Corresponding angles form
an F shape. There are 4 pairs
of corresponding angles. Can
you find them?
If the lines are parallel, then corresponding angles are equal.
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Cointerior angles
Cointerior angles form
a U shape. Can you find
another pair?
If the lines are parallel, cointerior angles are supplementary (i.e. their sum
is 180c).
Tests for parallel lines
If alternate angles are equal, then the lines are parallel.
If +AEF = +EFD,
then AB < CD.
If corresponding angles are equal, then the lines are parallel.
If +BEF = +DFG,
then AB < CD.
If cointerior angles are supplementary, then the lines are parallel.
If +BEF + +DFE = 180c,
then AB < CD.
Chapter 4 Geometry 1
If 2 lines are both parallel to a third line, then the 3 lines are parallel to
each other. That is, if AB < CD and EF < CD, then AB < EF.
EXAMPLES
1. Find the value of y, giving reasons for each step of your working.
Solution
+AGF = 180c - 125c
= 55c
(+FGH is a straight angle)
`
(+AGF, +CFE corresponding angles, AB < CD)
y = 55c
2. Prove EF < GH.
Solution
+CBF = 180c - 120c (+ABC is a straight angle)
= 60c
` +CBF = +HCD = 60c
But +CBF and +HCD are corresponding angles
` EF < GH
Can you prove this
in a different way?
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Think about the reasons for
each step of your calculations.
4.2 Exercises
1.
Find values of all pronumerals.
(a)
(h)
(i)
(b)
(j)
(c)
2.
Prove AB < CD.
(a)
(d)
(b)
(e)
(c)
A
(f)
(g)
B
104c
C 76c
D
E
Chapter 4 Geometry 1
A
(d)
(e)
B
138c
B
52c
E
C
C
E
128c
D
23c
F
115c
G
H
F
Types of Triangles
Names of triangles
A scalene triangle has no two sides or angles equal.
A right (or right-angled) triangle contains a right angle.
The side opposite the right angle (the longest side) is called the
hypotenuse.
An isosceles triangle has two equal sides.
The angles (called the base angles) opposite the equal sides in an
isosceles triangle are equal.
An equilateral triangle has three equal sides and angles.
A
D
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All the angles are acute in an acute-angled triangle.
An obtuse-angled triangle contains an obtuse angle.
Angle sum of a triangle
The sum of the interior angles in any triangle is 180c,
that is, a + b + c = 180
Proof
Let +YXZ = ac, +XYZ = bc and +YZX = cc
Draw line AB < YZ
Then +BXZ = cc
(+BXZ, +XZY alternate angles, AB < YZ)
+AXY = bc
(similarly)
+YXZ + +AXY + +BXZ = 180c
(+AXB is a straight angle)
`
a + b + c = 180
Chapter 4 Geometry 1
Class Investigation
1.
2.
3.
4.
5.
Could you prove the base angles in an isosceles triangle are equal?
Can there be more than one obtuse angle in a triangle?
Could you prove that each angle in an equilateral triangle is 60c?
Can a right-angled triangle be an obtuse-angled triangle?
Can you find an isosceles triangle with a right angle in it?
Exterior angle of a triangle
The exterior angle in any triangle is equal to the sum of the two opposite
interior angles. That is,
x+y=z
Proof
Let +ABC = xc , +BAC = yc and +ACD = zc
Draw line CE < AB
zc = +ACE + +ECD
+ECD = xc
+ACE = yc
`
z=x+y
(+ECD,+ABC corresponding angles, AB < CE)
(+ACE,+BAC alternate angles, AB < CE)
EXAMPLES
Find the values of all pronumerals, giving reasons for each step.
1.
CONTINUED
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Solution
x + 53 + 82 = 180
(angle sum of D 180c)
x + 135 = 180
x + 135 - 135 = 180 - 135
x = 45
2.
Solution
+A = +C = x
x + x + 48 = 180
2x + 48 = 180
2x + 48 - 48 = 180 - 48
2x = 132
132
2x
=
2
2
x = 66
(base angles of isosceles D)
(angle sum in a D 180c)
3.
Solution
y + 35 = 141
(exterior angle of D)
y + 35 - 35 = 141 - 35
`
y = 106
This example can be done using the interior sum of angles.
+BCA = 180c - 141c
= 39c
y + 39 + 35 = 180
y + 74 = 180
y + 74 - 74 = 180 - 74
`
y = 106
(+BCD is a straight angle 180c)
(angle sum of D 180c)
Chapter 4 Geometry 1
Think of the reasons
for each step of your
calculations.
4.3 Exercises
1.
Find the values of all
pronumerals.
(a)
(h)
(b)
(i)
(j)
(c)
(d)
(k)
(e)
(f)
(g)
157
2.
Show that each angle in an
equilateral triangle is 60c.
3.
Find +ACB in terms of x.
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4.
5.
6.
Prove AB < ED.
(d)
8.
Prove D IJL is equilateral and
D JKL is isosceles.
9.
In triangle BCD below, BC = BD.
Prove AB ED.
Show D ABC is isosceles.
Line CE bisects +BCD. Find the
value of y, giving reasons.
A
B
C
46c
E
88c
D
7.
Evaluate all pronumerals, giving
reasons for your working.
(a)
10. Prove that MN QP .
32c
M
(b)
75c
O
73c
Q
(c)
P
N
Chapter 4 Geometry 1
Congruent Triangles
Two triangles are congruent if they are the same shape and size. All pairs of
corresponding sides and angles are equal.
For example:
We write D ABC / D XYZ.
Tests
To prove that two triangles are congruent, we only need to prove that certain
combinations of sides or angles are equal.
Two triangles are congruent if
• SSS: all three pairs of corresponding sides are equal
• SAS: two pairs of corresponding sides and their included angles are
equal
• AAS: two pairs of angles and one pair of corresponding sides are equal
• RHS: both have a right angle, their hypotenuses are equal and one
other pair of corresponding sides are equal
EXAMPLES
1. Prove that DOTS / DOQP where O is the centre of the circle.
CONTINUED
The included angle
is the angle between
the 2 sides.
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Solution
S:
A:
S:
OS = OQ
+TOS = +QOP
OT = OP
`
by SAS, DOTS / DOQP
(equal radii)
(vertically opposite angles)
(equal radii)
2. Which two triangles are congruent?
Solution
To find corresponding sides, look at each side in relation to the angles.
For example, one set of corresponding sides is AB, DF, GH and JL.
D ABC / D JKL (by SAS)
3. Show that triangles ABC and DEC are congruent. Hence prove that
AB = ED.
Solution
A: +BAC = +CDE
A: +ABC = +CED
S:
AC = CD
(alternate angles, AB < ED)
(similarly)
(given)
` by AAS, D ABC / D DEC
`
AB = ED
(corresponding sides in congruent D s)
Chapter 4 Geometry 1
4.4 Exercises
1.
Are these triangles congruent?
If they are, prove that they are
congruent.
(a)
2.
Prove that these triangles are
congruent.
(a)
B
(b)
Y
4.7
m
110c
2.3
4.7
m
m
Z
110
c
C
A
2
.3 m
(b)
X
(c)
(c)
(d)
(d)
(e)
(e)
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3.
A
Prove that
(a) Δ ABD is congruent to Δ ACD
(b) AB bisects BC, given D ABC is
isosceles with AB = AC.
D
B
4.
Prove that triangles ABD and CDB
are congruent. Hence prove that
AD = BC.
C
(a) Prove that TABC and TADC
are congruent.
(b) Show that +ABC = +ADC.
7.
The centre of a circle is O and AC
is perpendicular to OB.
A
5.
In the circle below, O is the centre
of the circle.
A
O
D
B
O
C
B
C
(a) Prove that TOAB and TOCD
are congruent.
(b) Show that AB = CD.
6.
(a) Show that TOAB and TOBC
are congruent.
(b) Prove that +ABC = 90c.
8.
ABCF is a trapezium with
AF = BC and FE = CD. AE and BD
are perpendicular to FC.
In the kite ABCD, AB = AD and
BC = DC.
F
A
B
E
D
C
(a) Show that TAFE and TBCD
are congruent.
(b) Prove that +AFE = +BCD.
Chapter 4 Geometry 1
9.
The circle below has centre O and
OB bisects chord AC.
10. ABCD is a rectangle as shown
below.
A
B
D
C
C
O
B
A
(a) Prove that TOAB is congruent
to TOBC.
(b) Prove that OB is perpendicular
to AC.
(a) Prove that TADC is
congruent to TBCD.
(b) Show that diagonals AC and
BD are equal.
Investigation
The triangle is used in many
structures, for example trestle
tables, stepladders and roofs.
Find out how many different ways
the triangle is used in the building
industry. Visit a building site, or
interview a carpenter. Write a
report on what you find.
Similar Triangles
Triangles, for example ABC and XYZ, are similar if they are the same shape but
different sizes.
As in the example, all three pairs of corresponding angles are equal.
All three pairs of corresponding sides are in proportion (in the same ratio).
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We write: D ABC <; D XYZ
D XYZ is three times larger than D ABC.
6
XY
= =3
AB
2
XZ
12
=
=3
4
AC
15
YZ
=
=3
5
BC
XY
XZ
YZ
`
=
=
AB
AC
BC
This shows that all 3 pairs
of sides are in proportion.
Application
Similar figures are used in many areas, including maps, scale drawings, models
and enlargements.
EXAMPLE
1. Find the values of x and y in similar triangles CBA and XYZ.
Solution
First check which sides correspond to one another (by looking at their
relationships to the angles).
YZ and BA, XZ and CA, and XY and CB are corresponding sides.
`
XZ
XY
=
CA
CB
y
5.4
=
4.9
3.6
3.6y = 4.9 # 5.4
Chapter 4 Geometry 1
165
4 . 9 # 5 .4
3 .6
= 7.35
XY
=
CB
5 .4
=
3.6
= 2 . 3 # 5 .4
2 . 3 # 5 .4
=
3 .6
= 3.45
y=
YZ
BA
x
2 .3
3 .6x
x
Tests
There are three tests for similar triangles.
Two triangles are similar if:
• three pairs of corresponding angles are equal
• three pairs of corresponding sides are in proportion
• two pairs of sides are in proportion and their included angles
are equal
EXAMPLES
1.
(a) Prove that triangles ABC and ADE are similar.
(b) Hence find the value of y, to 1 decimal place.
Solution
(a) +A is common
+ABC = +ADE
+ACB = +AED
` D ABC <; D ADE
(b)
(corresponding angles, BC < DE)
(similarly)
(3 pairs of angles equal)
CONTINUED
If 2 pairs of angles are
equal then the third
pair must also be equal.
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AE = 2.4 + 1.9
= 4 .3
DE
AE
=
BC
AC
y
4 .3
=
3 .7
2.4
2 .4 y = 3 . 7 # 4. 3
3 .7 # 4 .3
y=
2 .4
= 6 .6
2. Prove D XYZ <; DWVZ.
Solution
XZ
ZV
YZ
ZW
XZ
`
ZV
+XZY
3
15
=
7
35
3
6
=
=
7
14
YZ
=
ZW
= +WZV
=
(vertically opposite angles)
` since two pairs of sides are in proportion and their included angles are
equal the triangles are similar
Ratio of intercepts
The following result comes from similar triangles.
When two (or more) transversals cut a series of parallel lines, the
ratios of their intercepts are equal.
That is, AB : BC = DE : EF
AB
DE
or
=
EF
BC
Chapter 4 Geometry 1
Proof
Draw DG and EH parallel to AC.
Then
Also
`
`
DG = AB
EH = BC
DG
AB
=
EH
BC
+GDE = +HEF
+DEG = +EFH
+DGE = +EHF
(opposite sides of a parallelogram)
(similarly)
(1)
(corresponding +s, DG < EH)
(corresponding +s, BE < CF)
(angle sum of Ds)
So D DGE <; D EHF
DG
DE
=
`
EH
EF
From (1) and (2):
(2)
AB
DE
=
EF
BC
EXAMPLES
1. Find the value of x, to 3 significant figures.
Solution
x
1.5
=
^ ratios of intercepts on parallel lines h
8.9
9.3
9.3x = 8.9 # 1.5
8.9 # 1.5
x=
9.3
= 1.44
CONTINUED
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2. Evaluate x and y, to 1 decimal place.
Solution
Use either similar triangles or ratios of intercepts to find x. You must use
similar triangles to find y.
Why?
x
2.7
=
5.8
3. 4
2.7 # 5.8
x=
3.4
= 4.6
y
2.7 + 3.4
=
7.1
3.4
6.1 # 7.1
y=
3.4
= 12.7
These ratios come
from intercepts on
parallel lines.
These ratios come from
similar triangles.
4.5 Exercises
1.
Find the value of all pronumerals,
to 1 decimal place where
appropriate.
(a)
(c)
(d)
(e)
(b)
Chapter 4 Geometry 1
(f)
46 c
xc
11
8.9
25.7
9.1
1.3
5c
c
1.82
E
14.3
19
c
4.
52c
A
4.2
B
4.9
5.88
yc
6.86
F
The diagram shows two
concentric circles with centre O.
(a) Prove that DOAB <; D OCD.
(b) If radius OC = 5.9 cm and
radius OB = 8.3 cm, and the
length of CD = 3.7 cm, find the
length of AB, correct to 2 decimal
places.
7.
(a) Prove that D ABC <; D ADE.
(b) Find the values of x and y,
correct to 2 decimal places.
8.
ABCD is a parallelogram, with
CD produced to E. Prove that
D ABF <; DCEB.
Evaluate a and b to 2 decimal
places.
Show that D ABC and DCDE are
similar.
EF bisects +GFD. Show that
D DEF and D FGE are similar.
C
6.
(g)
3.
D
87c
y
46
a
2.
Show that D ABC and D DEF are
similar. Hence find the value of y.
5.
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9.
Show that D AED <; D ABC. Find
the value of m.
10. Prove that D ABC and D ACD are
similar. Hence evaluate x and y.
(e)
12. Show that
AB
AF
(a)
=
BC
FG
AB
AF
(b)
=
AC
AG
BD
DF
(c)
=
CE
EG
11. Find the values of all
pronumerals, to 1 decimal place.
(a)
13. Evaluate a and b correct to
1 decimal place.
(b)
14. Find the value of y to 2
significant figures.
(c)
(d)
15. Evaluate x and y correct to
2 decimal places.
Chapter 4 Geometry 1
Pythagoras’ Theorem
DID YOU KNOW?
The triangle with sides in the
proportion 3:4:5 was known to be
right angled as far back as ancient
Egyptian times. Egyptian surveyors
used to measure right angles by
stretching out a rope with knots tied
in it at regular intervals.
They used the rope for forming
right angles while building and
dividing fields into rectangular plots.
It was Pythagoras (572–495 BC)
who actually discovered the
relationship between the sides of the
right-angled triangle. He was able to
generalise the rule to all right-angled triangles.
Pythagoras was a Greek mathematician,
philosopher and mystic. He founded the Pythagorean
School, where mathematics, science and philosophy
were studied. The school developed a brotherhood and
performed secret rituals. He and his followers believed
that the whole universe was based on numbers.
Pythagoras was murdered when he was 77, and the
brotherhood was disbanded.
The square on the hypotenuse in any right-angled triangle is equal to the
sum of the squares on the other two sides.
That is,
c2 = a2 + b2
or
c=
a2 + b2
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Proof
Draw CD perpendicular to AB
Let AD = x, DB = y
Then x + y = c
In D ADC and D ABC,
+A is common
+ADC = +ACB = 90c
`
D ADC <; D ABC (equal corresponding +s)
AC
AD
=
AB
AC
x
b
c =b
b 2 = xc
D BDC <; D ABC
Similarly,
BC
DB
=
AB
BC
y
a
a= c
a 2 = yc
Now
a 2 + b 2 = yc + xc
= c ^y + xh
= c ]c g
= c2
If c 2 = a 2 + b 2, then D ABC must be right angled
EXAMPLES
1. Find the value of x, correct to 2 decimal places.
Solution
c2 = a2 + b2
x2 = 72 + 42
= 49 + 16
= 65
Chapter 4 Geometry 1
173
x = 65
= 8.06 to 2 decimal places
2. Find the exact value of y.
Solution
c2 = a2 + b2
82 = y2 + 42
64 = y 2 + 16
48 = y 2
`
Leave the answer in
surd form for the exact
answer.
y = 48
= 16 # 3
=4 3
3. Find the length of the diagonal in a square with sides 6 cm. Answer to
1 decimal place.
Solution
6 cm
6 cm
c =a +b
= 62 + 62
= 72
2
2
2
c = 72
= 8 .5
So the length of the diagonal is 8.5 cm.
CONTINUED
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Maths In Focus Mathematics Extension 1 Preliminary Course
4. A triangle has sides 5.1 cm, 6.8 cm and 8.5 cm. Prove that the triangle
is right angled.
Solution
5.1 cm
8.5 cm
6.8 cm
Let c = 8.5 (largest side) and a and b the other two smaller sides.
a 2 + b 2 = 5 . 1 2 + 6. 8 2
= 72.25
c 2 = 8. 5 2
= 72.25
` c2 = a2 + b2
So the triangle is right angled.
4.6 Exercises
1.
Find the value of all pronumerals,
correct to 1 decimal place.
(a)
2.
Find the exact value of all
pronumerals.
(a)
(b)
(b)
(c)
(c)
(d)
(d)
Chapter 4 Geometry 1
3.
Find the slant height s of a
cone with diameter 6.8 m and
perpendicular height 5.2 m, to
1 decimal place.
4.
Find the length of CE, correct
to 1 decimal place, in this
rectangular pyramid. AB = 8.6 cm
and CF = 15.9 cm.
5.
Prove that D ABC is a right-angled
triangle.
6.
7.
Show that AC = 2 BC.
8.
(a) Find the length of diagonal
AC in the figure.
(b) Hence, or otherwise, prove
that AC is perpendicular to DC.
9.
Find the length of side AB in
terms of b.
Show that D XYZ is a right-angled
isosceles triangle.
X
XY
in
YZ
terms of x and y in D XYZ.
10. Find the exact ratio of
2
1
Y
1
Z
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11. Show that the distance squared
between A and B is given by
d 2 = 13t 2 - 180t + 625.
16. A ramp is 4.5 m long and 1.3 m
high. How far along the ground
does the ramp go? Answer correct
to one decimal place.
12. An 850 mm by 1200 mm gate
is to have a diagonal timber
brace to give it strength. To what
length should the timber be cut,
to the nearest mm?
4.5 m
1.3 m
17. The diagonal of a television
screen is 72 cm. If the screen is
58 cm high, how wide is it?
18. A property has one side 1.3 km
and another 1.1 km as shown
with a straight road diagonally
through the middle of the
property. If the road is 1.5 km
long, show that the property is
not rectangular.
13. A rectangular park has a length of
620 m and a width of 287 m. If I
walk diagonally across the park,
how far do I walk?
14. The triangular garden bed below
is to have a border around it.
How many metres of border are
needed, to 1 decimal place?
1.5 km
1.3 km
1.1 km
15. What is the longest length of
stick that will fit into the box
below, to 1 decimal place?
19. Jodie buys a ladder 2 m long and
wants to take it home in the boot
of her car. If the boot is 1.2 m by
0.7 m, will the ladder fit?
Chapter 4 Geometry 1
20. A chord AB in a circle with
centre O and radius 6 cm has a
perpendicular line OC as shown
4 cm long.
(a) By finding the lengths of AC
and BC, show that OC bisects the
chord.
(b) By proving congruent
triangles, show that OC bisects
the chord.
O
A
4 cm
6 cm
C
B
Types of Quadrilaterals
A quadrilateral is any four-sided figure
In any quadrilateral the sum of the interior angles is 360c
Proof
Draw in diagonal AC
+ADC + +DCA + +CAD = 180c
(angle sum of D)
+ABC + +BCA + +CAB = 180c
(similarly)
` +ADC + +DCA + +CAD + +ABC + +BCA + +CAB = 360c
That is,
+ADC + +DCB + +CBA + +BAD = 360c
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Maths In Focus Mathematics Extension 1 Preliminary Course
EXAMPLE
Find the value of i.
Solution
i + 120 + 56 + 90 = 360 ^ angle sum of quadrilateral h
i + 266 = 360
i = 94
Parallelogram
A parallelogram is a quadrilateral with opposite sides parallel
PROPERTIES
•
•
•
•
These properties can
all be proven.
opposite sides of a parallelogram are equal
opposite angles of a parallelogram are equal
diagonals in a parallelogram bisect each other
each diagonal bisects the parallelogram into two
congruent triangles
TESTS
A quadrilateral is a parallelogram if:
• both pairs of opposite sides are equal
• both pairs of opposite angles are equal
• one pair of sides is both equal and parallel
• the diagonals bisect each other
Chapter 4 Geometry 1
179
Rectangle
A rectangle is a parallelogram with one angle a right angle
If one angle is a right
angle, then you can
prove all angles are
right angles.
PROPERTIES
• the same as for a parallelogram, and also
• diagonals are equal
TEST
A quadrilateral is a rectangle if its diagonals are equal
Application
Builders use the property of equal diagonals to check if a rectangle is accurate.
For example, a timber frame may look rectangular, but may be slightly slanting.
Checking the diagonals makes sure that a building does not end up like the
Leaning Tower of Pisa!
Rhombus
A rhombus is a parallelogram with a pair of adjacent sides equal
PROPERTIES
• the same as for parallelogram, and also
• diagonals bisect at right angles
• diagonals bisect the angles of the rhombus
It can be proved that
all sides are equal.
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Maths In Focus Mathematics Extension 1 Preliminary Course
TESTS
A quadrilateral is a rhombus if:
• all sides are equal
• diagonals bisect each other at right angles
Square
A square is a rectangle with a pair of adjacent sides equal
PROPERTIES
• the same as for rectangle, and also
• diagonals are perpendicular
• diagonals make angles of 45c with the sides
Trapezium
A trapezium is a quadrilateral with one pair of sides parallel
Kite
A kite is a quadrilateral with two pairs of adjacent sides equal
Chapter 4 Geometry 1
EXAMPLES
1. Find the values of i, x and y, giving reasons.
Solution
i = 83c
x = 6.7 cm
y = 2.3 cm
(opposite +s in < gram)
(opposite sides in < gram)
(opposite sides in < gram)
2. Find the length of AB in square ABCD as a surd in its simplest form if
BD = 6 cm.
Solution
Let AB = x
Since ABCD is a square, AB = AD = x (adjacent sides equal)
Also, +A = 90c
(by definition)
By Pythagoras’ theorem:
c2 = a2 + b2
62 = x2 + x2
36 = 2x 2
18 = x 2
` x = 18
= 3 2 cm
CONTINUED
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3.
(a) Two equal circles have centres O and P respectively. Prove that OAPB
is a rhombus.
(b) Hence, or otherwise, show that AB is the perpendicular bisector
of OP.
Solution
OA = OB
(equal radii)
PA = PB
(similarly)
Since the circles are equal, OA = OB = PA = PB
` since all sides are equal, OAPB is a rhombus
(b) The diagonals in any rhombus are perpendicular bisectors.
Since OAPB is a rhombus, with diagonals AB and OP, AB is the
perpendicular bisector of OP.
(a)
4.7 Exercises
1.
Find the value of all pronumerals,
giving reasons.
(a)
(e)
(f)
(b)
(g)
(c)
(d)
Chapter 4 Geometry 1
2.
Given AB = AE, prove CD is
perpendicular to AD.
(c)
(d)
3.
(a) Show that +C = xc and
+B = +D = (180 - x)c.
(b) Hence show that the sum of
angles of ABCD is 360c.
(e)
(f)
4.
5.
Find the value of a and b.
7
3x
x+
6
y
6.
In the figure, BD bisects
+ADC. Prove BD also bisects
+ABC.
7.
(a)
Prove that each figure is a
parallelogram.
(a)
(b)
(b)
Find the values of all
pronumerals, giving reasons.
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Maths In Focus Mathematics Extension 1 Preliminary Course
(c)
(d)
(e)
(d)
8.
Evaluate all pronumerals.
(a)
9.
The diagonals of a rhombus
are 8 cm and 10 cm long. Find
the length of the sides of the
rhombus.
10. ABCD is a rectangle with
+EBC = 59c . Find +ECB, +EDC
and +ADE.
(b)
(c)
11. The diagonals of a square are
8 cm long. Find the exact length
of the side of the square.
12. In the rhombus, +ECB = 33c.
Find the value of x and y.
ABCD is a kite
Polygons
A polygon is a closed plane figure with straight sides
A regular polygon has all sides and all interior angles equal
Chapter 4 Geometry 1
EXAMPLES
3-sided
(equilateral
triangle)
4-sided
(square)
5-sided
(pentagon)
6-sided
(hexagon)
8-sided
(octagon)
10-sided
(decagon)
DID YOU KNOW?
Carl Gauss (1777–1855) was a famous German mathematician, physicist and astronomer. When
he was 19 years old, he showed that a 17-sided polygon could be constructed using a ruler and
compasses. This was a major achievement in geometry.
Gauss made a huge contribution to the study of mathematics and science, including
correctly calculating where the magnetic south pole is and designing a lens to correct
astigmatism.
He was the director of the Göttingen Observatory for 40 years. It is said that he did not
become a professor of mathematics because he did not like teaching.
The sum of the interior angles of an n-sided polygon is given by
S = 180n - 360
or S = (n - 2) # 180c
Proof
Draw any n-sided polygon and divide it into n triangles as
shown. Then the total sum of angles is n # 180c or 180n.
But this sum includes all the angles at O. So the sum of
interior angles is 180n - 360c .
That is, S = 180n - 360
= ] n - 2 g #180c
The sum of the exterior angles of any polygon is 360c
Proof
Draw any n-sided polygon. Then the sum of both the
exterior and interior angles is n #180c.
Sum of exterior angles = n #180c - sum of interior angles
= 180n - ] 180n - 360c g
= 180n - 180n + 360c
= 360c
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Maths In Focus Mathematics Extension 1 Preliminary Course
EXAMPLES
1. Find the sum of the interior angles of a regular polygon with 15 sides.
How large is each angle?
Solution
n = 15
S = (n - 2)#180c
= (15 - 2)#180c
= 13#180c
= 2340c
Each angle has size 2340c' 15 = 156c.
2. Find the number of sides in a regular polygon whose interior angles
are 140c.
Solution
Let n be the number of sides
Then the sum of interior angles is 140n
But
S = (n - 2)#180c
So 140n = (n - 2)#180c
= 180n - 360
360 = 40n
9=n
So the polygon has 9 sides.
There are n sides and so n
angles, each 140c.
4.8 Exercises
1.
2.
Find the sum of the interior
angles of
(a) a pentagon
(b) a hexagon
(c) an octagon
(d) a decagon
(e) a 12-sided polygon
(f) an 18-sided polygon
Find the size of each interior
angle of a regular
(a) pentagon
(b) octagon
(c) 12-sided polygon
(d) 20-sided polygon
(e) 15-sided polygon
3.
Find the size of each exterior
angle of a regular
(a) hexagon
(b) decagon
(c) octagon
(d) 15-sided polygon
4.
Calculate the size of each
interior angle in a regular 7-sided
polygon, to the nearest minute.
5.
The sum of the interior angles of
a regular polygon is 1980c.
(a) How many sides has the
polygon?
(b) Find the size of each interior
angle, to the nearest minute.
Chapter 4 Geometry 1
6.
Find the number of sides of a
regular polygon whose interior
angles are 157c 30l.
7.
Find the sum of the interior
angles of a regular polygon whose
exterior angles are 18c.
8.
9.
A regular polygon has interior
angles of 156c. Find the sum of its
interior angles.
13. A regular octagon has a
quadrilateral ACEG inscribed as
shown.
B
D
H
Find the size of each interior
angle in a regular polygon if
the sum of the interior angles is
5220c.
10. Show that there is no regular
polygon with interior angles of
145c.
11. Find the number of sides of a
regular polygon with exterior
angles
(a) 40c
(b) 30c
(c) 45c
(d) 36c
(e) 12c
C
A
G
E
F
Show that ACEG is a square.
14. In the regular pentagon below,
show that EAC is an isosceles
triangle.
A
E
B
12. ABCDEF is a regular hexagon.
A
B
D
F
C
E
D
(a) Show that triangles AFE and
BCD are congruent.
(b) Show that AE and BD are
parallel.
C
15. (a) Find the size of each exterior
angle in a regular polygon with
side p.
(b) Hence show that each interior
180 (p - 2)
.
angle is
p
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Maths In Focus Mathematics Extension 1 Preliminary Course
Areas
Most areas of plane figures come from the area of a rectangle.
Rectangle
A = lb
Square
A square is a
special rectangle.
A = x2
Triangle
The area of a triangle
is half the area of a
rectangle.
A=
1
bh
2
Proof
h
b
Draw rectangle ABCD, where b = length and h = breadth.
Chapter 4 Geometry 1
189
area = bh
`
1
1
area AEFD and area DCEF = area EBCF
2
2
1
area DCDE =
area ABCD
2
1
That is, A = bh
2
Area D DEF =
`
Parallelogram
A = bh
Proof
In parallelogram ABCD, produce DC to E and draw BE perpendicular to CE.
Then ABEF is a rectangle.
Area ABEF = bh
In D ADF and D BCE,
+AFD = +BEC = 90c
AF = BE = h
(opposite sides of a rectangle)
AD = BC
(opposite sides of a parallelogram)
` by RHS, D ADF / D BCE
`
area D ADF = area D BCE
So
area ABCD = area ABEF
= bh
Rhombus
1
xy
2
(x and y are lengths of diagonals)
A=
The area of a
parallelogram is the
same as the area of
two triangles.
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Maths In Focus Mathematics Extension 1 Preliminary Course
Proof
Let AC = x and BD = y
By properties of a rhombus,
1
1
AE = EC = x and DE = EB = y
2
2
Also +AEB = 90c
Area D ABC =
=
Area D ADC =
=
` total area of rhombus =
=
1
1
x: y
2
2
1
xy
4
1
1
x: y
2
2
1
xy
4
1
1
xy + xy
4
4
1
xy
2
Trapezium
A=
Proof
Let
DE = x
Then DF = x + a
`
FC = b - ] x + a g
=b-x-a
1
h ( a + b)
2
Chapter 4 Geometry 1
Area trapezium = area D ADE + area rectangle ABFE + area D BFC
1
1
= xh + ah + (b - x - a) h
2
2
1
= h (x + 2a + b - x - a)
2
1
= h (a + b)
2
Circle
You will study the circle in
more detail in Chapter 9.
A = rr 2
EXAMPLES
1. Find the area of this trapezium.
Solution
1
h ( a + b)
2
1
= ( 4) ( 7 + 5)
2
= 2 # 12
A=
= 24 m 2
4.2 cm
8.9 cm
3.7 cm
2. Find the area of the shaded region in this figure.
12.1 cm
CONTINUED
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Maths In Focus Mathematics Extension 1 Preliminary Course
Solution
Area large rectangle = lb
= 8.9 #12.1
= 107.69 cm 2
Area small rectangle = lb
= 3 . 7 # 4 .2
`
= 15.54 cm 2
shaded area = 107.69 - 15.54
= 92.15 cm 2
3. A park with straight sides of length 126 m and width 54 m has semicircular ends as shown. Find its area, correct to 2 decimal places.
126 m
54 m
192
Solution
Area of 2 semi-circles = area of 1 circle
54
r =
2
= 27
A = rr 2
= r (27) 2
= 2290.22 m2
Area rectangle = 126 # 54
= 6804
Total area = 2290.22 + 6804
= 9094.22 m2
4.9 Exercises
1.
Find the area of each figure.
(a)
(b)
Chapter 4 Geometry 1
(c)
(b)
(c)
(d)
(e)
(d)
(e)
(f)
6
cm
2 cm
(g)
4.
2.
Find the area of a rhombus with
diagonals 2.3 m and 4.2 m.
3.
Find each shaded area.
(a)
Find the area of each figure.
(a)
(b)
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Maths In Focus Mathematics Extension 1 Preliminary Course
(c)
(d)
8.
The dimensions of a battleaxe
block of land are shown below.
(a) Find its area.
(b) A house in the district where
this land is can only take up 55%
of the land. How large (to the
nearest m 2) can the area of the
house be?
(c) If the house is to be a
rectangular shape with width
8.5 m, what will its length be?
9.
A rhombus has one diagonal
25 cm long and its area is
600 cm 2 . Find the length of
(a) its other diagonal and
(b) its side, to the nearest cm.
(e)
5.
Find the exact area of the figure.
6.
Find the area of this figure,
correct to 4 significant figures.
The arch is a semicircle.
7.
Jenny buys tiles for the floor of
her bathroom (shown top next
column) at $45.50 per m 2 . How
much do they cost altogether?
10. The width w of a rectangle is
a quarter the size of its length.
If the width is increased by 3
units while the length remains
constant, find the amount of
increase in its area in terms of w.
Chapter 4 Geometry 1
195
Test Yourself 4
1.
Find the values of all pronumerals
(a)
2.
Prove that AB and CD are parallel lines.
3.
Find the area of the figure, to 2 decimal
places.
4.
(a) Prove that triangles ABC and ADE are
similar.
(b) Evaluate x and y to 1 decimal place.
5.
Find the size of each interior angle in a
regular 20-sided polygon.
6.
Find the volume of a cylinder with radius
5.7 cm and height 10 cm, correct to
1 decimal place.
7.
Find the perimeter of the triangle below.
(b)
(c)
(d)
x
(e)
(O is the centre
of the circle.)
(f)
(g)
The perimeter
is the distance
around the outside
of the figure.
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Maths In Focus Mathematics Extension 1 Preliminary Course
8.
(a) Prove triangles ABC and ADC are
congruent in the kite below.
(b) Prove triangle AOB and COD are
congruent. (O is the centre of the circle.)
12. Triangle ABC is isosceles, and AD bisects
BC.
(a) Prove triangles ABD and ACD are
congruent.
(b) Prove AD and BC are perpendicular.
13. Triangle ABC is isosceles, with AB = AC.
Show that triangle ACD is isosceles.
9.
Find the area of the figure below.
14. Prove that opposite sides in any
parallelogram are equal.
10. Prove triangle ABC is right angled.
15. A rhombus has diagonals 6 cm and 8 cm.
(a) Find the area of the rhombus.
(b) Find the length of its side.
16. The interior angles in a regular polygon
are 140c . How many sides has the
polygon?
17. Prove AB and CD are parallel.
11. Prove
AF
AB
.
=
AG
AC
Chapter 4 Geometry 1
18. Find the area of the figure below.
6 cm
5 cm
8 cm
20. (a) Prove triangles ABC and DEF are
similar.
(b) Evaluate x to 1 decimal place.
2 cm
10 cm
19. Prove that z = x + y in the triangle
below.
Challenge Exercise 4
1.
Find the value of x.
4.
Given +BAD =+DBC, show that D ABD
and D BCD are similar and hence find d.
2.
Evaluate x, y and z.
5.
Prove that ABCD is a parallelogram.
AB = DC.
3.
Find the sum of the interior angles of a
regular 11-sided polygon. How large is
each exterior angle?
6.
Find the shaded area.
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Maths In Focus Mathematics Extension 1 Preliminary Course
7.
Prove that the diagonals in a square
make angles of 45c with the sides.
8.
Prove that the diagonals in a kite are
perpendicular.
9.
Prove that MN is parallel to XY.
12. Find the values of x and y correct to
1 decimal place.
13. Find the values of x and y, correct to
2 decimal places.
10. Evaluate x.
11. The letter Z is painted on a billboard.
(a) Find the area of the letter.
(b) Find the exact perimeter of the letter.
14. ABCD is a square and BD is produced to
1
E such that DE = BD.
2
(a) Show that ABCE is a kite.
2x
(b) Prove that DE =
units when
2
sides of the square are x units long.