Download Geometry in the plane

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Geometry in
the plane
4.1 Kick off with CAS
4.2 Review of basic geometry
4.3 Congruence and similarity
4.4 Geometric constructions
4.5 Polygons
4.6 Circle geometry
4.7 Tangents, chords and circles
4.8 Review
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4.1 Kick off with CAS
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PA
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To come
Please refer to the Resources tab in the Prelims section of your eBookPLUs for a comprehensive
step-by-step guide on how to use your CAS technology.
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4.2 Review of basic geometry
Plane geometry is the study of flat two-dimensional shapes. A point is a location
with no dimensions; that is, it is a concept that helps us determine the location of real
objects. A point is often labelled with a capital letter.
Lines and angles
PA
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FS
A line segment joins two points. It is a basic postulate (or principle) that only one
straight line segment can join two given points.
In the figure at right, there is exactly one straight line segment
A
B
that can join points A and B.
If we extend the line segment to ‘infinity’ in both
B
A
directions, we have what is properly called a line.
If we extend the line segment to ‘infinity’ in one direction only, we have a ray.
In the figure at right we have a ray extending from point A.
A
B
Often the terms line segment, line and ray are used
interchangeably. The context should help you determine
which of the three kinds of ‘line’ is really being used.
B
A
Three points can determine an angle. In Figure 1, the angle,
indicated by the arc of a circle, is named ∠BAC or ∠CAB. The
C
vertex of the angle is at point A and is placed in the middle of the
Figure 1
name, so we do not write ∠ABC or ∠BCA in this example.
D
A
C
B
Figure 2
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D
A straight angle is formed when the three points of the previous
example are all in a line. In Figure 2, the straight angle is ACB, with
the vertex at C. Note that another line segment can be drawn to point
D. By convention, a straight angle equals 180 degrees and is written
as ∠ACB = 180­°. Therefore, it can be seen that ∠ACD + ∠DCB =
∠ACB = 180°.
The two angles ∠ACD and ∠DCB are called supplementary
angles.
If the point D is moved so that ∠ACD = ∠DCB, then we have
created a right angle. In Figure 3, both ∠ACD and ∠DCB are right
angles. From the previous equation, because ∠ACD = ∠DCB, we
can rewrite
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to
A
C
B
Figure 3
∠ACD + ∠DCB = ∠ACB = 180°
∠ACD + ∠ACD = 180°
2(∠ACD) = 180°
∠ACD = 90°
So a right angle equals 90°. Therefore, we
can say that AB is perpendicular to CD.
∴ AB ⟂ CD
140 MATHS QUEST 11 SPECIALIST MATHEMATICS VCE Units 1 and 2
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WorKED
EXAMPLE
1
Find the value of x in the given diagram.
32°
A
tHiNK
D
E
x
C
B
WritE
1 Write an equation involving the required angle.
∠ACB is a straight angle.
∠BCD + x + ∠ECA = 180°
90° + x + 32° = 180°
2 Replace angles with known values.
FS
∠BCD = 90° (right angle)
122° + x = 180°
3 Solve for the missing angle.
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122° − 122° + x = 180° − 122°
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x = 58°
Parallel lines and angles
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PA
G
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It is a basic postulate of geometry that parallel lines never meet. When parallel lines
are intersected by a transversal, special angles are formed.
Consider the pair of parallel lines in the figure at right.
Transversal
Note the use of arrows to indicate that the lines are parallel
D
E
C || E. Now consider the two angles ∠ABC and ∠BDE,
B
marked with arcs. Because the parallel lines (E and C)
C
A
never meet, we could easily move one on top of the other
D
without affecting the angles. Thus it stands to reason that the
E
angles are equal. These equal angles are called
B
C
corresponding angles.
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EC
Now, consider the two crossing lines shown in the figure at
right. Because ∠EBD and ∠ABC are both straight angles
E
(180°), we can write:
∠EBC + ∠CBD = ∠CBD + ∠ABD
C
D
B
A
∠EBC = ∠ABD
(the two angles indicated with arcs). These angles are called vertically opposite angles.
Next, consider the parallel lines at right, and the two
angles ∠CBD and ∠EDB (indicated with arcs). They are
E
D
positioned ‘inside’ the parallel lines, on the same side
B
(to the right) of the transversal. These angles are called
A
C
co-interior angles and sum to 180°.
Finally, consider the pair of parallel lines on the right and the two angles ∠EDB and ∠FBD.
From the earlier result about parallel lines we know that
E
∠EDB = ∠CBA, and from the result about crossing
D
lines, that ∠CBA = ∠FBD. Therefore, ∠EDB = ∠FBD
B
C
F
(the two angles indicated with arcs). These are called
A
alternate angles.
From the above results, we are in a position to perform our first ‘proof’.
Topic 4 GEoMETry In THE PLAnE
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2
Prove that the sum of the interior angles of any triangle is
180 degrees. In other words, prove that a + b + c = 180°.
a
c
DrAW/WritE
1 Construct a line parallel to one of the sides.
b
a
2 Label some additional angles (d and e) that will
c
d
be needed.
b
e
a
3 The upper parallel line is also a straight angle.
c
d + b + e = 180°
FS
tHiNK
b
d=a
4 But we have some alternate angles.
e=c
a + b + c = 180°
… QED
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5 Substitute into the equation in step 3.
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WorKED
EXAMPLE
Angles in polygons
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As described in Chapter 2, QED is Latin for quod erat demonstrandum (‘as has been
demonstrated’), but could just as well be thought of as ‘quite easily done’! Remember
that, when proving something, you are allowed to use other facts that have already
been established, without having to prove them each time.
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A polygon is a closed figure made up of three or more straight line segments.
This figure is not a polygon
because it has a curved line.
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This figure is not a polygon
because it is not closed.
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This figure is a polygon because it is closed and has
four (more than two) line segments, all straight.
Now that we know that the sum of angles in a triangle is 180°, we can find the sum of
angles in any polygon by dividing it into triangles as shown.
F q
A
p
m n o
r
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s
x
t
u
D
B
w
v C
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m + q + r = 180°
n + s + t = 180°
o + u + v = 180°
p + w + x = 180°
The sum of the angles in a hexagon
= 4 × 180° (since it contains 4 triangles)
= 720°
MATHS QUEST 11 SPECIALIST MATHEMATICS VCE Units 1 and 2
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The general formula used to determine the sum of the interior angles in any polygon is:
(n − 2) × 180°, where n = number of sides
Exercise 4.2 Review of basic geometry
PRactise
1
WE1
Find the values of the pronumerals in the following figures.
a
b
c
x y
45° x x
34°
122°
128°
x
Prove the following theorem: ‘The value of the exterior angle
of a triangle equals the sum of the two interior opposite angles’.
In other words, in referring to the figure at right, d = a + b.
WE2
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38°
35°
E
130°
x°
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2 Find the value of x.
b
a
c d
4 ΔABC shown at right is isosceles. AB is produced to D, and BE is drawn through
D
B
E
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B, parallel to the base AC. Prove that BE bisects ∠CBD.
Interactivity
int-0997
Bisecting a line segment
C
5 Angles that add up to 180 degrees are called:
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CONSOLIDATE
A
A parallel
C complementary
Bisosceles
Dsupplementary
E right-angled
6 Which of the following statements is false?
A An angle is always formed when two lines meet.
BThe sum of angles in a right-angled triangle is 180 degrees.
C A line segment extended to infinity in both directions is called a ray.
DParallel lines never meet.
E Reflex angles are bigger than acute angles.
7 Prove that the co-interior (or allied) angles marked
with arcs in the figure at right are supplementary. In other
F
words, prove that ∠EDB + ∠DBC = 180°.
E
D
B
C
A
Topic 4 Geometry in the plane c04GeomentryInThePane.indd 143
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8 From the result in Worked example 2 and question 7,
101°
E
D
show that there are only two different angle values: all
B
angles in that figure must take on one or the other of
F
these values. Furthermore, show that if you know only
one of these angles, then all other angles can be
determined. Demonstrate this using the value in the figure at right.
C
A
9 Find the values of the unknown angles in the following figures.
b
y
2x
c
y
47°
66°
55°
z
y z
31°
x
x
c
z
x
3x
30°
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y
x + 3°
A
11 Show that the sum of the interior angles of the pentagon at right
equals 540 degrees. In other words, show that: ∠AED + ∠EDC +
∠DCB + ∠CBA + ∠BAE = 540°
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20°
x
40°
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b
75°
y
E
a
O
10 Find the values of the unknown angles in the following figures.
x
FS
a x
B
C
E
D
x
x − 22°
x + 15°
x − 20°
x + 10° x + 5°
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EC
12 Find the value of x in the figure below.
13 Find the value of the pronumerals in the figure at right.
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E
C
O
60°
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A
y°
x°
B
70°
F
G
14 Using the figure at right, find the value of x, giving reasons for your answer.
E
F
D 95°
70°
x°
C
B
A
144 MATHS QUEST 11 SPECIALIST MATHEMATICS VCE Units 1 and 2
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MASTER
15 Prove that the diagonals of a square form a right angle; that is,
A
B
that angle a in the figure below equals 90 degrees.
a
D
C
16 a
Copy and complete the following table relating the sum of the interior angles to
the number of sides of a polygon.
b Can you establish a general formula for the sum in terms of n, the number
of sides?
4
Sum of interior angles 180°
6
7
8
540°
Congruence and similarity
9
10
20
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4.3
5
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3
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Number of sides (n)
Congruence
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Congruent figures are shapes that have the exact same shape and size. If ΔABC and
ΔDEF are congruent, we write ΔABC ≊ ΔDEF.
Congruent triangles have all angles and sides equal. There are four tests to prove
triangles are congruent: SSS, SAS, AAS and RHS.
Similarity
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Similar figures are shapes that have the same shape but different size. If triangles
ΔABC and ΔDEF are similar, we write ΔABC ~ ΔDEF.
There are three tests to prove triangles are similar.
The following triangles will be used to prove the tests for similar triangles.
G
A
EC
80°
R
9
R
B
40°
8
6
D
60°
4.5 80° 3
40° 60°
E
F H
4
C 80°
9
40°
6
8
60°
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Test 1
Triangles are said to be similar if all the corresponding angles are equal (abbreviated
to AAA). That is, the three angles in one of the triangles are equal to the three angles
in the other triangle.
ΔABC ~ ΔDEF, because ∠A = ∠D
ΔDEF ~ ΔGHI, because ∠D = ∠G
∠B = ∠E
∠E = ∠H
∠C = ∠F
∠F = ∠I
Test 2
Triangles are said to be similar if the ratios between the corresponding side lengths
are equal (abbreviated to SSS).
ΔABC ~ ΔDEF, because the ratio of corresponding side lengths is 2:
9
BC 8
AC 6
AB
= =2
=
= 2,
= = 2, and
EF 4
DE 4.5
DF 3
Topic 4 Geometry in the plane c04GeomentryInThePane.indd 145
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ΔDEF ~ ΔGHI, because the ratio of corresponding side lengths is 12:
DE 4.5 1 DF 3 1
EF 4 1
= =
=
= ,
= = , and
HI 8 2
GH
9
2 GI 6 2
tEst 3
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D
Triangles are said to be similar if two of
their corresponding sides are in the same
ratio (ratio value does not equate to 1)
18
and the angle between these two sides
A
(the included angle) is the same in both
6
60°
60°
triangles (abbreviated to SAS).
B 8
C E
F
24
ΔABC ~ ΔDEF, because the
included angle
corresponding sides have the same ratio
value that does not equate to 1 and the included angle for both triangles is the same:
ED 18
EF 24
=
= 3 and
=
= 3; ∠B = ∠E
BC
8
AB
6
3
Compare each of the following triangles with ΔABC and state whether they
are similar, congruent or there is not enough information given for a decision
to be made. Justify your answers.
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WorKED
EXAMPLE
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Sometimes it is hard to decide whether the two triangles are similar or not, because
they are not oriented the same way. In such cases it is helpful to re-draw the
triangles so that the sides and angles that we think might be corresponding are in the
same order.
a ΔDEF
b ΔGHI
d ΔMNO
EC
c ΔJKL
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R
6.6 110°
30°
B
10
30°
40°
110°
D
40°
110°
H 4.5 I
10
110°
20
WritE
a 1 Re-draw ΔDEF so that its angles correspond
a
B
110°
A
6.6
6
M
6.6
110° K
40°
O
L
tHiNK
to those of ΔABC.
146
C
40°
J
7.5
F
6
G
20
C
U
N
A
E
C E
30° 10 40°
D
110°
30°
20
40°
F
MATHS QUEST 11 SPECIALIST MATHEMATICS VCE Units 1 and 2
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∠A = ∠F
2 Compare the corresponding angles and write
down what you observe.
∠B = ∠D
∠C = ∠E
All corresponding angles are equal.
FE 20
=
=2
CB 10
2 ≠ 1, so the triangles are not congruent.
3 Compare the side measurements given to test
for congruency.
b
A
sides correspond.
6.6 110°
B
30°
C G
40°
10
AC
6
=
HI
4.5
110° 4.5
7.5 40°
I
E
2 Compare the side measurements.
H
6
PR
O
b 1 Re-draw ΔABC and ΔGHI so that angles/
O
and specify the test upon which your
conclusion is based.
FS
ΔABC ~ ΔDEF because all corresponding
angles are equal (AAA).
4 State whether the triangles are similar
PA
G
= 1.333 and
10
BC
=
GI
7.5
= 1.333
∠C = ∠I = 40°
TE
D
3 Check whether the angles formed by
EC
the sides that have measurements are in
the same ratio (that is the included
angles).
ΔABC and ΔGHI are similar because two
pairs of corresponding sides are in the
same ratio that doesn’t equate to 1 and the
included angles are equal (SAS).
R
R
4 State and justify your conclusion.
O
c 1 Re-draw ΔABC and ΔJKL so the angles/
c
6.6
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C
sides correspond.
2 Check if the corresponding angles are equal.
3 As side measurements are given, check for
congruency.
4 State your conclusion and justify.
B
30°
K
6.6 110°
A
110° 6
10
40°
C
J
10
40°
L
∠J = 180° − 150° = 30°
∠J = ∠B, ∠A = ∠K, ∠C = ∠L
BC 10
AB 6.6
=
= 1 and
=
=1
JL
10
KJ 6.6
Initially, the two triangles appeared to be
similar because of AAA; however, as the
side ratio equates to 1, these triangles are
congruent.
Topic 4 Geometry in the plane c04GeomentryInThePane.indd 147
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d 1 Re-draw ΔABC and ΔMNO so the angles/
d
M
sides correspond.
110°
A
6.6
110° 6
CN
B
30° 10 40°
can be performed since we only know the
measurements of one side and one angle in
the ΔMNO. State this in writing.
a the corresponding angles are equal
b the corresponding side ratios are equal (to a value
WritE
a 1 Draw the triangles separately with angles
a
A
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D
2 State the corresponding pairs of angles which
3 Write your conclusion.
EC
are equal in size. Specify the reason.
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N
C
O
R
corresponding sides.
R
b 1 Evaluate the ratio between the
A
B
7.5
b°
PA
G
marked and side measurements shown.
7.5
E
other than 1).
tHiNK
3.75 D
B
4.8
O
Show that ΔABC ~ ΔADE, because:
PR
O
4
FS
Unable to determine whether the triangles are
similar, as ΔMNO does not provide enough
information to test AAA, SSS or SAS.
2 None of the three tests (AAA, SSS or SAS)
WorKED
EXAMPLE
O
20
a°
10
b°
5
E
D
4.8
c° C 11.25
A
C
10
7.2
a°
15
7.2
c° E
∠A = ∠A (shared)
∠B = ∠D (corresponding)
∠C = ∠E (corresponding)
∴ ΔABC ~ ΔADE using AAA test.
b AD = 11.25
AB
7.5
= 1.5
AE 15
=
AC 10
= 1.5
DE 7.2
=
BC 4.8
= 1.5
2 Write your conclusion.
148
The corresponding side lengths of the
triangles are in the same ratio. (The ratio is
not equal to 1.)
∴ ΔABC ~ ΔADE (SSS test).
MATHS QUEST 11 SPECIALIST MATHEMATICS VCE Units 1 and 2
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WorKED
EXAMPLE
5
Find the value of the pronumeral x.
D
x
B
7.5
10
A
15
tHiNK
C
E
WritE
∠A = ∠A (shared)
1 Establish whether the triangles ABC and
ADE are similar by applying the AAA test.
FS
∠B = ∠D (corresponding)
corresponding sides are in the same ratio.
Find this ratio using the corresponding pair of
sides whose lengths are known.
= 1.5
3 The missing length AD in ΔADE corresponds AD = 1.5
PA
G
E
AB
to the side AB in ΔABC. Use this
information to form the ratio equation.
PR
O
AE 15
=
AC 10
2 Since the triangles are similar, their
O
∠C = ∠E (corresponding)
∴ ΔABC – ΔADE (AAA test)
4 Identify the values of AD and AB.
AD = x, AB = 7.5
x
= 1.5
7.5
5 Substitute the values of AD and AB into the
x = 1.5 × 7.5
TE
D
equation and solve for x.
= 11.25
1
For each of the following, compare ΔABC and ΔEFD and state whether
they are similar, congruent, or there is not enough information given for a decision
to be made. Justify your answers.
WE3
R
R
PrACtisE
EC
ExErCisE 4.3 Congruence and similarity
U
N
C
O
a
b
A
D 120°
5.5
6 4 E 20°
8.25
40°
C
B 120°
20°
F
B
6 cm
40°
25°
3 cm
A
115° 40°
E
C
3 cm
D
6 cm
F
2 For each of the following, compare ΔABC and ΔEFD and state whether they are
similar, congruent, or there is not enough information given for a decision to be
made. Justify your answers.
a
B
3.375
A
7.875
2.5
5.625
E
b
D
3.5
1.5
F
D
12
A
C
8
6
B
40°
6
E
110°
30°
F
C
Topic 4 GEoMETry In THE PLAnE
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3
Show that ΔABC is similar to ΔDBE because:
a the corresponding angles are equal
b the corresponding side ratios are equal (to a value
other than 1).
B
WE4
6 cm
D
4.2 cm
A
4 Show that ΔABC is similar to ΔDBE because:
E
4 cm
3.5 cm
C
6.8 cm
B
a the corresponding angles are equal
b the corresponding side ratios are equal (to a
10 cm
PR
O
A
E 5 cm
8 cm
O
D
3 cm
FS
6 cm
value other than 1).
C
12 cm
Find the values of the pronumerals (to 1 decimal place).
a
b
37 cm
9.6 cm
x
*
6.2 cm
e
15
6
B
TE
D
d A
y
x
PA
G
*
12.4
9.6
31 cm
y
c A
2.4
E
WE5
B
C 5 D
x
4
E
EC
y
C
1.7 m
x
D
4m
y
C
x
5
B
18
D
12
E
A
3.4 m
5
5 cm
E
6 Find the values of the pronumerals (to 1 decimal place).
a
R
A
R
30°
O
2 cm
C
x°
22°37' A
b
B 3 cm
C
y°
z
z
36
U
N
D 9 cm E
C
x
y°
D 15
c
B
d
15
r=8
x
y
9
9
24
x
y
r=4
150 MATHS QUEST 11 SPECIALIST MATHEMATICS VCE Units 1 and 2
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CONSOLIDATE
7 In each of the following diagrams, find and re-draw two triangles that are similar.
Give reasons for your answer.
a A
b
c
A
B
D
d
C
D
C
E
D
e
E
B
E
B
E
C
A
D
A
D
A
C
B
FS
B
C
E 70°
E
9 The magnitude of angle a° is:
PR
O
B180° − 30°
D180° − 60°
A 30°
C 60°
O
8 The magnitude of angle a° is:
PA
G
A 64°
C 52°
E 104°
B128
D154°
30°
a°
64°
a°
A 24
C 16
B22
D15
16
b
R
EC
E 9.6
TE
D
10 In the given diagram, the length of side b is closest to:
Questions 11 and 12 refer to the following
information.
A young tennis player’s serve is shown in the
diagram at right. Assume the ball travels in a
straight line.
20
R
1.1 m
O
C
U
N
12
Not to scale
x
y
0.9 m
5m
10 m
11 The height of the ball just as it is hit, x, is closest to:
A 3.6 m
D1.8 m
B2.7 m
E 1.6 m
C 2.5 m
12 The height of the player, y, as shown is closest to:
A 190 cm
D160 cm
B180 cm
E 150 cm
C 170 cm
Topic 4 Geometry in the plane c04GeomentryInThePane.indd 151
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13 Calculate the size of the missing angles. Justify your answer.
a°
24°
b°
68°
d°
32°
b°
c°
b°
a°
d
a°
e
c°
d°
e°
78°
a°
x°
f
99°
y°
c°
c°
b°
d°
b°
e°
d°
PR
O
a°
e°
O
c
b
c°
FS
a
63°
32°
58°
14 Find the length of the bridge, AB, needed to span the river, using similar triangles
PA
G
B
E
as shown (to 1 decimal point).
TE
D
(Not to scale.
All measurements are
in metres.)
A
12. 5 m
2. 5 m
4. 3 m
EC
15 A triangle with sides 12 cm, 24 cm and 32 cm is similar to a smaller triangle
U
N
C
O
R
R
that has a longest side measurement of 8 cm. Draw a diagram to represent this
situation and then calculate the perimeter of the smaller triangle.
16 It just happens that you always carry a wooden rod 1.5 m in length and a tape
measure. For each of the following situations, draw a diagram showing the two
similar triangles and calculate the height of the vertical object.
a The clearance sign is missing from a low bridge over a road. The sun has
created an image of the bridge and
its opening on the ground. The
image of the opening is 1.44 m in
length. The semi-trailer you are
driving is 2.5 m high. You step out
of the truck and place your trusty
rod on the ground, producing a
90 cm shadow. Will you need to
find an alternative route or can
you proceed?
b The measuring tape for the pole
vault height is missing. How
high did Australian National
152 MATHS QUEST 11 SPECIALIST MATHEMATICS VCE Units 1 and 2
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O
17 A curved section of a fun-ride is to be supported
by similar triangles.
a If the dimensions of the first and last
triangle are as shown, calculate the
13.75
dimensions of the triangle in between.
b The construction people decide to make the
1.5 x y
track even steeper by extending the track
2.2
with one more similar triangle. What are the
z
6.25
1
dimensions of the next triangle?
18 At your 18th birthday party you want to show
slides of you and your friends growing up. The diagram
at right shows the set-up of the projector lens, slide and
projector screen.
9.375
TE
D
PA
G
E
PR
O
Master
FS
Champion Elizabeth Parnov pole vault if the pole vault creates a shadow of
2.4 m on the ground, while your wooden rod creates a shadow of 80 cm?
c Does the fire brigade need to bring an extension ladder longer than 15 m to
rescue people from the top of a high-rise building if the shadow created by the
building is 24 m and the shadow created by your pole is 2 m?
d Slam-dunking a basketball ring can be dangerous, people have been injured by
the ring and backboard collapsing. You have always wanted to know how high
the basketball players had to jump to touch the top of the ring. Your wooden
rod held next to the basketball pole created a shadow half the length of the
basketball pole’s 1 m shadow.
If the ring is 45 cm from the top of the board, how far 0 from the ground do the
players have to reach to touch the ring?
EC
If: a a slide is 5 cm × 5 cm
b the distance from A to B is 10 cm
c the projector screen is 1.5 m × 1.5 m
Slide
B
Lens
Projector
screen
Geometric constructions
U
N
C
O
4.4
R
R
how far horizontally from the screen do you need to
place the projector’s lens so the image just covers the
entire screen?
A
A traditional part of any study of geometry includes the skill of producing
constructions. In this section, you are to use only a straight edge (you can use a ruler,
but no measuring!) and a pair of compasses.
Using our straight edge and compasses, we are able to construct a variety of
geometric figures. We shall now look at constructions that show us how to bisect
lines, bisect angles, and draw special angles (for example, 60°).
The only new definition required here is bisection. You can bisect a line by dividing it
in half, or bisect an angle so that its measure is halved.
Bisecting lines
In the following examples, constructions drawn with a ruler and a pair of compasses
are shown in grey.
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WorKED
EXAMPLE
6
Use a ruler and a pair of compasses to bisect a line, AB.
tHiNK
DrAW
1 a Draw a line AB.
b Place the compass point at A, with any
radius more than half the length of the line.
c Draw a circle.
B
FS
A
2 With the same radius as in step 1, repeat for point B.
O
3 The two circles will intersect at two points. Join these
PR
O
points with a straight line. This line bisects AB.
Use a ruler and a pair of compasses to construct a line parallel to a given
line.
C
7
U
N
WorKED
EXAMPLE
O
R
R
EC
TE
D
PA
G
E
With this construction, not only have you divided the original line (AB) in half, but
the line that you drew in step 3 is perpendicular to AB. Therefore, this line is called
the perpendicular bisector of AB.
Why does it work? It is useful to examine the geometry of the construction to help
understand why it provides the correct result.
The figure below shows the essential part of the
C
construction. Points C and D are the intersections of the
two circles.
A
E
B
Join point A to C and B to C to create two triangles, as
D
shown in the next figure.
Because the two circles had the same radius, AC = BC. The two triangles also share
the common side CE. By symmetry, ∠ACE = ∠BCE and ∠CAE = ∠CBE.
Therefore, since two of the angles are equal, the third angles are also equal; that is,
∠AEC = ∠BEC.
C
From all this it is clear that the two triangles are identical.
Therefore, AE = EB and we have effectively bisected AB
(divided it into two equal parts).
A
E
B
tHiNK
DrAW
1 Pick any two points, A and B, on the given line.
2 From point A, draw a circle of any radius more than
half the distance from A to B.
3 With the same radius, repeat step 2 at point B.
A
B
4 Join the highest points (or lowest points) of the circles
with a straight line. This line will be parallel to AB.
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Note: The radius is the distance between the compass points, and this determines the
distance between the parallel lines.
Bisecting angles
Another important construction is that employed to bisect angles.
WorKED
EXAMPLE
Use a ruler and a pair of compasses to bisect any angle.
8
tHiNK
DrAW
FS
1 With any radius, and with the compass point at the vertex V,
O
draw an arc of a circle that crosses both arms of the angle. The
crossings are labelled A and B.
B
PR
O
V
A
2 With any radius, and with the compass point at A, draw an arc
V
B
A
PA
G
E
inside the angle. This arc should be long enough so that the line
representing half the angle would cross it.
3 With the same radius, repeat step 2, putting the point of the
compass at B. The two arcs will cross at point C.
4 Join the vertex V to C. This line bisects the angle, namely:
TE
D
∠AVB
2
V
C
B
A
EC
∠AVC =
R
R
When drawn carefully, this construction is an accurate way of halving an angle; it
is even more accurate than with a protractor. For example, if the original angle was
68.3°, it would be difficult with a protractor to obtain an angle of 34.15°.
U
N
C
O
Constructing angles
WorKED
EXAMPLE
9
The last group of constructions involves the ‘special angles’ of 30°, 45°, 60° and 90°.
You have already seen how to construct a 90° angle (see page xxx); bisecting this will
produce a 45° angle.
Use a ruler and compasses to construct a 60° angle.
tHiNK
DrAW
1 Draw a line as the base of the angle. Select a point for the
vertex, A; put the compass point there and draw a circle of any
radius, crossing the line at B.
A
B
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2 With the same radius as in step 1, put the compass point at
B and draw an arc, crossing the other circle
at two places, C and D.
D
B
A
C
3 Join A to C. ∠CAB = 60°.
D
B
O
C
FS
A
PR
O
There are several other 60° angles in the construction above. Can you find them?
(Hint: Use the symmetry of the construction.) A 30° angle can be constructed by
bisecting a 60° angle.
Exercise 4.4 Geometric constructions
Using a ruler and compasses, construct perpendicular
lines AC and BD to the line AB.
WE6
E
1
C
A
PA
G
PRACTISE
2 Draw two line segments of different lengths and construct
B
D
the perpendicular bisectors of each. Check your accuracy by
measuring the two parts of each line segment.
3
Use a ruler and a pair of compasses to construct two parallel lines which are
3 cm apart.
WE7
TE
D
Interactivities
int-0998
Bisecting an angle
int‐0999
Circumcentre
4 Draw two parallel lines exactly 20 cm apart. (Hint: How can you do this if the
compasses won’t open wide enough?)
WE8
Draw an obtuse angle, and use a ruler and a pair of compasses to bisect it.
EC
5
6 Draw a reflex angle, and use a ruler and a pair of compasses to bisect it.
WE9
Construct a 45° angle using a ruler and compasses only.
R
7
R
8 Construct a 30° angle using a ruler and compasses only.
O
9 Construct a right angle using a straight edge and a pair of compasses.
10 Draw a line AB. Use a straight edge and compasses to construct a line
U
N
C
CONSOLIDATE
156 parallel to AB.
11 Use the following set of instructions to copy the angle ABC.
Step 1. With your compass point at B, trace an arc cutting AB and BC at D and E
respectively.
Step 2. Draw a line similar to line BC on another sheet of paper.
Step 3. With the same radius as in step 1, draw an arc on the new line.
Step 4. Use your compass to ‘measure’ the distance from D to E.
A
Step 5. Use this radius to draw an arc on the new line, putting
the compass point where the first arc cut the line.
Step 6. Join the vertex of the new line with the point where
B
C
the two arcs cross.
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12 Draw an acute angle ABC and construct a copy of this angle.
13 Devise a method of constructing a perpendicular through a point on the line.
A
14 Prove that ∠ABC = ∠DBC.
A
PR
O
O
D
FS
C
B
15 Given an angle ABC with BD the bisector of this angle, prove that the point D is
the same distance from both arms of the angle.
A
E
E
PA
G
D
B
C
F
16 Given an isosceles triangle, ΔABC, with BD the bisector of ∠ABC, prove that BD
B
A
D
C
R
EC
TE
D
also bisects the base of the triangle, AC.
17 a
Consider the isosceles triangle in the figure at right.
A
R
O
C
U
N
Master
Construct perpendicular bisectors of each of the three
lines AB, BC, and CA.
b What do you notice about these bisectors?
B
C
18 a Repeat question 17 for any scalene triangle.
b What do you notice?
19 A pair of circular pulleys of the same radius is connected by a closed band of
rubber. The distance between the pulleys is equal to their diameter. Make a
straight‐edge‐and‐compass construction of this system.
20 Three circular pulleys of the same radius are arranged as follows: Pulley A is
directly above Pulley B, at a distance equal to twice the diameter. Pulley C is to
the right of Pulley B at a distance equal to three times the diameter. The rubber
band connecting them runs on the ‘outside’ of the system. Make a straight‐edge‐
and‐compass construction of this system.
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4.5
Polygons
As we have seen, a polygon is a closed figure with straight sides. In this section, we
look at the following aspects of polygons:
1. triangle constructions
2. quadrilaterals
3. regular polygons
4. star polygons.
Triangle constructions
10
Construct the perpendicular bisectors and median bisectors of each side of
any triangle and investigate their properties.
PR
O
WorKED
EXAMPLE
O
FS
Let us investigate the properties of the perpendicular bisectors of each side of
any triangle.
PA
G
E
Tutorial
eles‐1542
Worked example 10
tHiNK
DrAW
1 Draw any triangle and add circles centred at each vertex. The
EC
TE
D
radius should be large enough so that perpendicular bisectors
can be drawn.
2 Use the construction circles to draw perpendicular bisectors.
C
O
R
R
The black lines join pairs of intersecting arcs. It should be
clear that the bisectors all meet at a point. This point is called
the circumcentre.
U
N
3 Use this point as a centre and draw a circle that just touches
each vertex. To do this, put the compass point at the point
where the bisectors meet, and put the pencil point at any
vertex. You should observe that the resultant circumcircle (or
outcircle) just touches each vertex.
4 Furthermore, from step 2, we can determine the midpoint
of each side from the perpendicular bisectors. Note the use
of short bars to indicate the bisection of the sides at points
P, Q and R.
158
Q
R
P
MATHS QUEST 11 SPECIALIST MATHEMATICS VCE Units 1 and 2
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5 Join each midpoint to its opposite vertex. These lines
also meet at a single point, called the centroid, which has
applications in physics and engineering, as it is effectively the
point of symmetry of the triangle.
Q
R
P
11
Construct the incentre of the triangle shown at right by
hand.
PR
O
WorKED
EXAMPLE
O
FS
Imagine that the triangle in the previous example is made out of a thick piece of
cardboard. The centroid is the point where you could place a finger and ‘balance’ the
triangle. The reason for this is that the median bisectors in step 5 are the three axes of
symmetry of the triangle, each axis dividing the triangle into two equal areas. Since
they all meet, the centroid is the point of symmetry.
The incentre is the intersection of the angle bisectors of a triangle.
E
tHiNK
PA
G
A
DrAW
1 a Construct the angle bisectors by drawing arcs centred at
B
A
C
TE
D
each vertex (A, B and C).
b From the intersection of these arcs and the sides of the
triangles, draw intersecting arcs between pairs of sides. In
the figure at right this has been done to vertex A only, to
keep the drawing uncluttered.
C
B
2 Complete the construction of the angle bisectors and observe
B
R
EC
that they meet at a point — the incentre.
A
C
3 By placing the compass point at the incentre and carefully
R
B
U
N
C
O
drawing a circle, it is possible to construct the incircle, which
just touches each side of the triangle.
C
A
Quadrilaterals
There are many names for the various kinds of quadrilaterals, and it is useful to look
at their definitions in terms of their features.
Square
The square has:
1. four equal sides
2. four right angles.
A square can be constructed easily from a circle by finding the
perpendicular bisector of the diameter.
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rectangle
The rectangle has:
1. opposite sides equal
2. four right angles.
A rectangle can be constructed from a circle by using the
construction shown to draw two lines parallel to a diameter. The
rectangle is formed from the intersection of the parallel lines with the circle.
O
FS
Parallelogram
A parallelogram has:
1. opposite sides equal in length
2. opposite sides parallel
3. opposite angles equal.
PR
O
rhombus
A rhombus has:
1. all four sides equal in length
2. opposite sides parallel
3. opposite angles equal.
PA
G
E
Trapezium
A trapezium (or trapezoid) has one pair of opposite sides parallel.
12
Construct a rhombus using compasses and a straight edge.
EC
WorKED
EXAMPLE
TE
D
other
All other four-sided figures are generally just called quadrilaterals,
even though the above figures are also quadrilaterals.
Obviously there is some overlap between these definitions; for
example, a square is a kind of rectangle, just as a rhombus is a kind of parallelogram.
DrAW
R
tHiNK
1 a Construct a pair of parallel lines.
R
b From any point on one line (A), draw an arc of radius equal
U
N
C
O
to the length of the side of the rhombus. This arc cuts the
first line at B and the second line at C.
A
B
C
2 Using the same radius as in step 1, put the compass point at C
and draw an arc cutting the same line at D.
A
C
3 Join points A to C and B to D to form the rhombus ABDC.
A
D
B
C
D
It is easy to extend this kind of construction to other types of quadrilaterals.
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regular polygons
A regular polygon is one with each side the same length and with each interior
angle the same size. For triangles, the regular polygon is the equilateral triangle; for
quadrilaterals, it is the square. Worked example 9 should give you an idea of how
to construct an equilateral triangle; Worked example 12 should help you construct a
square. The construction of a regular hexagon (6 sides) is particularly easy.
13
Construct a regular hexagon using a straight edge and compasses.
tHiNK
DrAW
1 Draw a circle whose radius is equal to the length of one side of
PR
O
O
the hexagon.
FS
WorKED
EXAMPLE
A
2 Using the same radius as in step 1, and with the compass point
PA
G
E
anywhere on the circle (A), draw an arc that cuts the circle
at point B.
3 Repeat step 2 by:
putting the compass at B and cutting the circle at C
putting the compass at C and cutting the circle at D
putting the compass at D and cutting the circle at E
putting the compass at E and cutting the circle at F.
TE
D
a
b
c
d
A
B
E
F
D
A
C
B
EC
4 Join points A–B–C–D–E–F to form a hexagon.
U
N
C
O
R
R
Several regular polygons, such as those with 7, 9, 11 and 13 sides, cannot be drawn
using just a straight edge and ruler. By dividing up the circle’s 360 degrees, it is
possible to ∘work out the angle ‘between’ each side. For example, the hexagon’s 6 sides
must be 360
= 60° apart. Because construction of a 60° angle is easy, the hexagon can
6
be constructed. This does
not occur with regular polygons such as the heptagon, where
360 ∘
the angle would be 7 = 51.428 …°. In cases like this, a protractor may be used.
Star polygons
The last step of Worked example 13 was to join the points, in order, to form the
regular polygon. What would happen if you joined points by ‘skipping’ others?
Consider the figure at right. Point A has been joined to E and then to C and back to A
again. A similar pattern has been used starting at point B. In other words, one point
was ‘skipped’ for each line. In this case, the polygon formed
E
F
(after removing the central lines) is the only one possible with 6
points: the ‘Star of David’. These types of polygon are called star
polygons because of the obvious resemblance.
D
Star polygons can be constructed with the aid of a protractor,
as polygons that are 7-sided, 9-sided and so on have no exact
C
construction method.
B
A
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WorKED
EXAMPLE
14
Construct all the star polygons from a regular nonagon (9 sides), using a
straight edge, compasses and protractor.
tHiNK
DrAW
1 a Draw a circle of any radius, and draw a line from the centre to
B
the circumference (point A). Because there are to be 9 sides, the
interior angle will be 360°
= 40°.
9
b Using a protractor, measure a 40° angle from A to cross the
circle at B.
40°
remaining 7 angles.
C
FS
2 Measure a 40° angle from B to cross the circle at C. Repeat for the
A
D
B
40° 40°
40°
40° A
40°
40°
40° 40°
J
40°
F
PR
O
O
E
G
3 Instead of joining A–B–C–D–E–F–G–H–J–A, try skipping one point,
PA
G
E
joining: A–C–E–G–J–B–D–F–H–A.
TE
D
EC
A
F
J
G
H
C
D
B
A
E
J
F
G
R
H
C
D
B–E–H–B and C–F–J–C.
R
B
E
5 Repeat steps 3 and 4, but skip two points, joining A–D–G–A and
O
C
D
4 Remove the construction lines to form the star polygon.
B
E
A
J
F
G
U
N
C
H
6 Repeat steps 3 and 4, but skip three points, joining
C
D
A–E–J–D–H–C–G–B–F–A.
H
B
E
A
J
F
G
H
7 Repeat steps 3 and 4, but skip four points, joining A–F–B–G–C–H–
D–J–E–A. As this is the same figure as in step 6 with the order of the
points reversed, there are no more different star polygons to be drawn.
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Note: The greater the number of points that are skipped, the sharper the points of the
star polygon become. Furthermore, if we skip too many points (as in step 7) the same
star polygon is produced; thus, there is a limited number of star polygons for a given
regular polygon.
Exercise 4.5 Polygons
PRACTISE
1
Construct a scalene triangle with one angle greater than 90 degrees.
Investigate the properties of perpendicular bisectors and side length bisectors.
WE10
2 Which of the basic constructions of the previous sections might be used to
Construct the following triangle. Start with a base side of 6 cm. From the lefthand end point of the line draw a circle of radius 7 cm, and from the right-hand end
point draw a circle of 5 cm. Join the two end points to the place where the two arcs
meet, above the line, to form the triangle. Construct the incentre of this triangle.
WE11
PR
O
O
3
FS
construct an isosceles triangle?
4 Construct a scalene triangle and determine the incentre. What did you observe
about the properties of the incentre?
WE12
Construct a parallelogram using a straight edge and compasses.
E
5
PA
G
6 Construct a square of side length 5 cm using only a ruler and a pair of compasses.
Begin the construction of a regular dodecagon (12-sided polygon) using the
following steps.
a Construct a regular hexagon.
b Join the vertices of the hexagon to the centre of the circle first used in the
construction. What should the final three steps be?
8 Construct all the star polygons from the regular dodecagon drawn in question 7.
WE13
9
The stars on the Australian flag are
7-pointed (the points represent the 6 states and
the territories). There is no compass-and-straightedge construction, so use a protractor to help you
construct the 7-pointed star polygon.
R
EC
WE14
TE
D
7
C
O
R
10 Construct all the different star polygons possible
U
N
CONSOLIDATE
from an octagon. What is an easy compass-andstraight-edge construction of the octagon?
11 Construct a right-angled triangle (one with an angle of 90 degrees). Investigate the
properties of the circumcentre and centroid.
12 Devise a method of constructing a parallelogram (with unequal sides) similar to
that of Worked example 12.
13 State whether each of the following is true or false.
a An irregular quadrilateral cannot have a 90° angle.
b A rectangle is a quadrilateral because it has two pairs of parallel sides.
c A rhombus is a parallelogram because it has two pairs of parallel sides.
d Not every parallelogram has opposite sides equal in length.
e Not every square is a parallelogram.
14 One angle in a rhombus is 40°. Find all other angles.
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15 A quadrilateral has each angle 10° greater than the previous one, except for the
smallest angle. What size is the smallest angle?
16 For the polygon shown, find:
a°
a the sum of the interior angles
b the value of the pronumeral, a.
55°
240°
170°
260°
17 ABCD is a quadrilateral with AF parallel to BC.
FS
55°
B
PR
O
O
The diagonal BD intersects AF at E. Prove that
∠AED = ∠BCD + ∠BDC.
E
A
y°
D
B
x° x°
D
C
19 ABCD is a quadrilateral with right angles at B and D
B
C
TE
D
PA
G
E
AC, bisects ∠ABC. Prove that ΔABC is isosceles.
C
F
A
18 In ΔABC, the perpendicular line BD, from B to side
x°
A
D
and diagonal AC, as shown. If AB = CD, prove that
ABCD is a rectangle.
20 ABCD is a parallelogram and AF = EC. Prove that
R
EC
AE = CF.
D
F
B
E
C
R
21 An ancient method of getting a right angle is to use the Pythagorean triple: a
triangle with sides of 3, 4 and 5 units. Devise a method of constructing such a
triangle with compasses and a straight edge only.
C
O
Master
A
U
N
22 The construction of a regular pentagon is quite difficult. Use the following
164 instructions to help in the compass-and-straight-edge construction. The various
points defined in the steps are shown at right.
D
Step 1. Draw a circle of any radius, and mark the centre C
and the diameter AB.
B
A
Step 2. Find the perpendicular bisector of AB meeting the
F C
M
circle at D and E.
Step 3. Find the midpoint (M) of CB, and draw an arc with
E
radius MD cutting AB at F.
The side length of the pentagon is equal to the distance DF. Using this
information, complete the construction of the pentagon. (Hint: This step is similar
to the construction of a hexagon.)
MATHS QUEST 11 SPECIALIST MATHEMATICS VCE Units 1 and 2
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4.6 Circle geometry
Until now, we have avoided the most ‘perfect’ form in geometry, the circle. There are
several important properties and theorems of circles that form part of the classical
study of geometry.
Review of circle definitions
E
B
O
FS
A circle is formed by drawing the curved line that is at a given
D
distance from a single point (O). This distance is called the radius.
The line itself is called the circumference. A line from a point on
A
the circumference (A), through the centre (O), to another point (B)
on the circumference is called a diameter. A similar line (DE) that
does not pass through the centre is called a chord. OC is a radius.
Mathematically, it is easy to see that the length of AB is twice that of
OC. In other words, diameter = 2 × radius.
The length of any chord is less than that of the diameter
D
(DE < AB). A chord divides the circle into two regions. In the
figure at right, the shaded area is called the minor segment. The
A
rest of the circle’s area is called the major segment. A minor
segment’s area is always less than half the area of the circle. The
part of the circumference joining points D and E is called an arc.
The shorter arc (in the shaded region) is the minor arc; the longer
one is the major arc.
A chord can define (or subtend) certain angles. The angle
D
formed by joining the chord to the centre (∠DOE) is called the
angle subtended by the chord at the centre. The angle formed
by joining the chord to a third point on the circumference
(∠DFE) is called the angle subtended by the chord at the
circumference.
Obviously for a given chord (DE) there is only one angle
subtended at the centre, but there are different angles subtended
at the circumference as the point F is moved around the circle.
A tangent line is one that just touches the circumference at a
single point (C) and is always perpendicular to the radius (OC).
There is a close connection between this tangent and the one
defined in trigonometry.
E
B
O
E
B
O
F
O
U
N
C
O
R
R
EC
TE
D
PA
G
E
PR
O
O
C
C
Circle theorems
Circle theorem 1
Consider the figure at right. A chord AB is drawn and a third point
on the circumference (C) defines the angle ACB (marked angle y
in the figure). From the same chord, lines are drawn to the centre
(O), defining the angle x. It can be proven that x = 2y.
Circle theorem 1: The angle subtended by the chord at
the centre is twice the value of the angle subtended by the
same chord at the circumference.
A
B
x
O
y
C
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Circle theorem 2
Now, imagine moving the point C in the figure anywhere along the
circumference to, say, points C1 or C2. In each case ‘new’ angles
(y1, y2) are subtended by the same chord.
But from circle theorem 1: x = 2y 2 and x = 2y1, therefore y2 = y1.
This is true as long as C1 and C2 stay on the same side of AB as C
(that is, in the same segment as C).
A
B
x
O
y2 y y1
C2 C
C1
C
Find the values of the angles x, y and z.
y
D
Tutorial
eles‐1543
Worked example 15
x
62°
O
E
PA
G
circumference and the centre respectively. The angle at the centre
(62°) is twice the angle at the circumference (x).
WritE
TE
D
∠ADB = ∠ACB
x =y
31° = y
y = 31°
2 The angles x and y are subtended by the same chord at the
EC
circumference, so they are equal.
A
∠AOB = 2∠ADB
62° = 2x
x = 31°
1 The angles x and 62° are subtended by the same chord at the
x =z
31° = z
z = 31°
3 Triangle ADO is isosceles, and angles on the base of an isosceles
O
R
R
triangle are equal, so x = z.
Find the values of the angles x, y and z.
A
C
16
B
z
tHiNK
O
46°
Tutorial
eles‐1544
Worked example 16
tHiNK
1 Use circle theorem 1 to find x.
z
x
U
N
WorKED
EXAMPLE
O
15
PR
O
WorKED
EXAMPLE
FS
Circle theorem 2: Angles subtended at the circumference, in the same segment, by the same chord are equal.
B
y
D
C
WritE
x = ∠AOB = 2∠ACB
(circle theorem 1)
x = 2(46°)
= 92°
166
MATHS QUEST 11 SPECIALIST MATHEMATICS VCE Units 1 and 2
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2 Use circle theorem 2 to find y.
y = ∠ADB = ∠ACB
(circle theorem 2)
y = 46°
3 Use properties of isosceles triangle AOB
x + z + ∠OAB
x+z+z
x + 2z
2z
= 180°
= 180°
= 180°
= 180° − 92°
= 88°
z = 44°
FS
to find z.
PA
G
E
PR
O
O
Circle theorem 3
Now, imagine moving the chord AB to the points A1 and B1 as shown in Figure 1.
Notice that the values of both angles x and y increase.
What would happen if A1 and B1 were
A
B
A
B
moved even further, so that the chord
A1
B1
x
x
became a diameter? This is shown in
B1
A1
Figure 2. Now, x is a straight angle
O
O
(180°), and from circle theorem 1, x = 2y.
y
y
Therefore, y = 90°. This special case
C
C
of the theorem can be stated as circle
Figure 2
Figure 1
theorem 3.
TE
D
Circle theorem 3: The angle subtended by a diameter is a right angle (90°).
R
EC
Circle theorem 4
The figure at right is called a cyclic quadrilateral, because all four
vertices touch the circumference and it encloses the centre. It can
be proven that the sum of opposite angles is always 180°; that is,
a + c = 180° and b + d = 180°.
A
D
d
a
O
c C
b
B
17
D
Find the values of x and y.
89°
U
N
C
WorKED
EXAMPLE
O
R
Circle theorem 4: Opposite angles in a cyclic quadrilateral add up to 180 degrees.
tHiNK
Ax
O 103° C
y
B
WritE
1 ABCD is a cyclic quadrilateral, and opposite
angles in such a shape add up to 180°.
2 Angles x and 103° are opposite angles.
3 Angles y and 89° are opposite angles.
x + 103° = 180°
x = 77°
y + 89° = 180°
y = 91°
Topic 4 GEoMETry In THE PLAnE
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Exercise 4.6 Circle geometry
Find the values of the angles x, y and z in the figure at right.
y
z
2 The angle subtended by a chord at the centre is _____________ the
88°
angle subtended by the same chord at the _____________.
O
x
The above sentence can be completed using the words:
A half, circumference
Btwice, diameter
C twice, circumference
Dequal, circumference
E twice, edge
3 WE16 Find the values of the angles x, y and z in Figure 1 below.
4 Using the result from circle theorem 2, prove the additional result that x = z in
Figure 2 below.
WE15
A
x 80° z
B
z
x
PR
O
34°
E
O
y
y
O
D
FS
1
O
PRACTISE
y
C
E
Figure 1 Figure 2
Find the values of x and y in Figure 3 below.
6 In Figure 4 below, find the value of angle y in terms of angle x. (Hint: Use circle
theorems 2 and 4 combined.)
WE17
PA
G
5
54°
TE
D
y
x
O
x
O
Figure 3 Figure 4
7 Find the value of the angle x in Figure 5 below.
R
CONSOLIDATE
EC
84°
y
8 In question 7, if angle ∠OAC = 20°, then ∠OBC equals:
U
N
C
O
R
A 37°
B57°
C 59°
D94°
E 131°
9 In figure 6 (below), the angle (or angles) that is (are) half of ∠AOC is (are):
A ∠ABC only
C ∠ABC and ∠ADC
E ∠ACD only
B∠ADC only
D∠ADC and ∠ACD
10 The angle subtended at the circumference by a diameter is ________ degrees.
11 Find the values of x and y in Figure 7 below.
B
B
A
A
x
O
168 50°
x
O
O
37º C
C
Figure 5
D
Figure 6
30°
y
Figure 7
MATHS QUEST 11 SPECIALIST MATHEMATICS VCE Units 1 and 2
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12 In the following five figures, identify the circle theorem that helps you find the
value of the angle labelled x. Then find the actual value of x.
a
111°
b 102°
x
51°
O
c 81°
x
O
x
e O
41°
22°
x
51°
O 110°
O
O
x
81°
FS
d
120°
PR
O
13 Find the values of the angles x, y and z in Figure 8 below.
14 Prove the general case of the result of question 13, namely that: x = y = z
15 In Figure 9 below, chords AB and CD are parallel. Find the values of the angles
A
B
z
PA
G
y
40°
O
A
B
O
C
z
42°
x
TE
D
D
E
x, y and z.
D
y
x
C
Figure 8 Figure 9
16 Prove the general case of the result of question 15, namely that: x = y = z
17 The proof of circle theorem 1 relied on a result about the external angle of any
EC
triangle, as shown in the figure at right. The result is that d = a + b. Prove
this result.
b
R
R
MASTER
a
c d
U
N
C
O
18 The proof of circle theorem 1 relies on a particular construction
whereby a line joining the vertices of the two angles in question
are joined and extended towards the circumference.
This is the line COD in the figure at right. Imagine the
point C being moved so that it was much closer to B as in
the figure. Clearly, it is not possible to form the line COD,
so how can we prove the theorem in this case?
Step 1. Draw a line from C to O (see the figure at right).
Step 2. Using the result about equal angles in isosceles
triangles, find some equal angles. There are three isosceles
triangles in the diagram.
To get you started, ∠EBC = y + ∠ECO. (Why?)
Complete the proof, namely that x = 2y.
D
A
B
x
y
O
E
C
y
C
A
z
x
O
E
B
y
C
C
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4.7 Tangents, chords and circles
There are many geometric problems that use the four circle theorems to help in their
solution. To begin, we need to recall the definition of a tangent to a circle as a line
that just touches the circumference and forms a right angle to the radius at the point
of contact.
18
Construct, with a straight edge and compasses, a tangent to a circle at any
point by hand. (This task is relatively easy, relying on the earlier construction
of a perpendicular bisector.)
tHiNK
FS
WorKED
EXAMPLE
DrAW
circumference, extend the radius outward.
b Using the same radius as the circle (OA), draw an arc
crossing the extended line at B. The result is that A is
the midpoint of OB.
B
A
E
O
PA
G
2 Construct the perpendicular bisector of the line OB.
PR
O
O
1 a Given the circle centred at O, and any point A on the
You will need a larger radius for the compasses than
in step 1. Draw arcs above and below the line OB by
first placing the compass point at B, then at O. The arcs
should cross at C and D.
C
B
TE
D
A
D
O
EC
3 Join the points C and D. Because the line CD is
C
perpendicular to OB, it is also perpendicular to OA.
Because it just touches the circle at A, it must also be
the tangent.
R
B
O
R
A
D
U
N
C
O
The alternate segment theorem
Consider the figures shown below. Line BC is a tangent to the circle at the point A.
E
D
B
170
D
O
A
C
B
O
F
A
C
MATHS QUEST 11 SPECIALIST MATHEMATICS VCE Units 1 and 2
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A line is drawn from A to anywhere on the circle, point D.
The angle ∠BAD defines a segment (the shaded area).
The unshaded part of the circle is called the alternate segment to ∠BAD.
Now consider angles subtended by the chord AD in the alternate segment, such as the
angles marked in red and blue.
The alternate segment theorem states that these are equal to the angle that made the
segment, namely:
∠BAD = ∠AED and ∠BAD = ∠AFD
FS
Proof of the alternate segment theorem
We are required to prove that ∠BAD = ∠AFD.
D
B
O
PR
O
O
G
F
A
C
R
EC
TE
D
PA
G
E
1. Construct the diameter from A through O, meeting the circle at G. This can be
done simply with a straight edge, as both points A and O are known. Join G to the
points D and F.
2. Use circle theorem 3 to find some right angles. This refers to the property that
angles subtended at the circumference by a diameter are right angles.
∠BAG = ∠CAG = 90°(property of tangents)
∠GFA = 90°(circle theorem 3)
∠GDA = 90°(circle theorem 3)
3. Consider triangle GDA. We know that ∠GDA = 90°. Solve.
∠GDA + ∠DAG + ∠AGD = 180°
90° + ∠DAG + ∠AGD = 180°
∠DAG + ∠AGD = 90°
U
N
C
O
R
4. a. ∠BAG is also a right angle.
∠BAG = ∠BAD + ∠DAG = 90°
b. Equate the two results.
∠DAG + ∠AGD = ∠BAD + ∠DAG
c. Cancel the equal angles on both sides.
∠AGD = ∠BAD
5. Now consider the fact that both triangles DAG and DAF are subtended from the
same chord (DA).
∠AGD = ∠AFD (circle theorem 2)
Equate the two equations.
∠AFD = ∠BAD QED
Chords and tangents in circles
We shall now consider three situations: where chords intersect inside a circle, where
they meet outside a circle and where one of the chords is a tangent. First, let us
consider the case of two chords that meet inside a circle.
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Chords meeting inside a circle
WorKED
EXAMPLE
19
Two chords, AB and CD, meet at a 50° angle at point E.
Find the values of the angles x, y and z.
C
44° x
50°
E
B
A y
z
D
2 a Construct a line joining B to D.
b Now the angle y and the 44° angle (∠DCB)
are subtended at the circumference by the
same chord (BD).
C
44° x
50°
E
A
PA
G
D
y = 44°
3 a Now, construct a line joining C to A.
TE
D
b This time the angles x and z are subtended
by the same chord (AC).
c Apply circle theorem 2.
EC
A
B
y
z
c Apply circle theorem 2.
O
a triangle.
x + 50° + 44° = 180°
x = 180° − 50° − 44°
= 86°
PR
O
1 Find the value of x using the sum of angles in
FS
WritE/DrAW
E
tHiNK
C
44° x
50°
E
(circle theorem 2)
B
y
z
D
z = x = 86°
(circle theorem 2)
O
R
R
As a consequence of this example, the two triangles AED and CEB have all three
angles equal and are therefore similar triangles. This can be stated as a theorem:
Two chords that meet inside a circle form two similar triangles.
U
N
C
Chords meeting outside a circle
Let us now consider the case where the two chords meet outside the circle.
WorKED
EXAMPLE
20
Two chords, PA and PB, meeting at P, intersect
the circumference of a circle at C and D. Find
the values of the angles x, y and z.
A
y
C
B
tHiNK
1 Use the sum of angles in a triangle to find x.
172
z
88°
x
28°
D
P
WritE
x + 88° + 28° = 180° (sum of angles in atriangle)
x = 180° − 88° − 28°
= 64°
MATHS QUEST 11 SPECIALIST MATHEMATICS VCE Units 1 and 2
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∠DCA + 88° = 180°(straight angle = 180°)
∠DCA = 92°
∠DCA + z = 180°(circle theorem 4)
92° + z = 180°
z = 88°
∠BDC + x = 180°(straight angle = 180°)
2 a Find z. Consider the straight angle ∠ACP.
b Consider the cyclic quadrilateral ABDC.
Use circle theorem 4 to find z.
(Note: z will always equal ∠DCP.)
3 a Consider the straight angle ∠BDP.
b Use circle theorem 4 to find y.
(Note: y will always equal x (∠CDP).)
∠BDC + 64° = 180°(from step 1)
∠BDC = 116°
FS
y + ∠BDC = 180°(circle theorem 4)
PR
O
y = 64°
O
y + 116° = 180°
A chord meeting a tangent
Finally, consider what happens when one of the chords becomes a tangent.
B
A chord (BD) meets a tangent (AD) at point D. The chord
crosses the circumference at C. Construct a pair of similar
triangles based on the diagram shown.
E
21
PA
G
WorKED
EXAMPLE
tHiNK
C
A
D
DrAW/WritE
1 Join point A to B and A to C and label
TE
D
B
EC
angles a, b, c, d, e, x and y.
To find similar triangles, attempt to find
equal angles.
R
2 Because AD is a tangent, and AB and
O
R
AC are chords, the alternate segment
theorem can apply.
3 There are two straight angles along lines
U
N
C
BD and AD.
4 Consider the angles in the small
triangle ACD and the largest
triangle BAD.
5 Use the results from steps 2 and 3
(b = e and c = a + b).
6 Because three angles are equal, the
e
x C
c
a
y
b d
A
D
y=x
b=e
(alternate segment theorem on AB)
(alternate segment theorem on AC)
x + c = 180°
(straight angle along AD)
y + a + b = 180°
(straight angle along BD)
x + c = y + a + b (equate two straight angles)
c=a+b
(result from step 2, x = y)
ACD angles: b, c, d
BAD angles: e, (a + b), d
= b, c, d
Hence, triangle ACD and triangle BAD are similar.
triangles are similar.
Topic 4 GEoMETry In THE PLAnE
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So, no matter what the geometry of the two chords (even where one is a tangent), a
pair of similar triangles is always formed.
22
A plank rests on two cylindrical rollers, as shown
in the diagram. The radii of the rollers are 40
cm and 50 cm and their distance apart on the
ground is 120 cm. Find the value of the angle x.
50 cm
40 cm
x
Tutorial
eles‐1545
Worked example 22
120 cm
DrAW/WritE
O
tHiNK
FS
WorKED
EXAMPLE
1 Redraw the diagram as a similar triangle problem. Label
PR
O
A
the triangle and transfer all the relevant information onto
the diagram.
C
40 cm
x
–
2
D
y
2 Write a similarity statement.
Note: With any similar figures, the corresponding angles are
equal and their corresponding sides are in equal ratio.
TE
D
3 To find missing side lengths, establish the scale factor by
ΔCED is similar to ΔAEB.
ΔCED − ΔAEB
Scale factor =
AB
CD
= 50
40
= 54
EC
calculating the ratio of two corresponding sides whose
lengths are given.
B
120 cm
PA
G
E
E
50 cm
= 1.25
R
4 Use the scale factor to find the length of side ED.
EB = 1.25 ED
R
a Write the relationship between EB and ED.
b Substitute the known values into the equation.
O
y + 120 = 1.25y
C
y − y + 120 = 1.25y − y
U
N
120 = 0.25y
c Transpose the equation to make y the subject.
0.25y = 120
d Divide both sides by 0.25.
0.25y
120
0.25
0.25
y = 480
5 Use triangle CED to find the magnitude of the
=
C
angle required.
E
174
x
–
2
480 cm
40 cm
D
MATHS QUEST 11 SPECIALIST MATHEMATICS VCE Units 1 and 2
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O
A
a Identify the appropriate ratio to use.
Note: The opposite and adjacent sides are known;
therefore, choose the tangent ratio.
tan (θ) =
b Substitute the known values into the ratio and evaluate.
40
x
tan a b =
2
480
x
1
tan a b =
2
12
x
1
= tan−1 a b
2
12
x
6
= 4.7636
2
d Evaluate for x and round the answer correct to
2 decimal places.
FS
x
c Transpose the equation to make the angle the subject,
2
using the inverse tangent function.
O
x = 2 × 4.76366
PR
O
= 9.5272°
= 9.53°
6 Answer the question.
The required angle is 9.53°.
Draw a circle of any radius. Construct a radius that points to the right
(that is, due east). Construct a tangent to the circle at the point where this radius
touches the circle.
2 Draw a circle of radius 3 cm and construct a tangent to the circle at any point,
using a straight edge and compasses.
3 WE19 In Figure 1 on the below, find the values of the angles x, y and z.
1
WE18
TE
D
PRACTISE
PA
G
E
Exercise 4.7 Tangents, chords and circles
4 Find the values of the angles x and y in Figure 2 on the below. (Hint: Use the
y
z
75°
R
R
EC
alternate segment theorem.)
x
O
O
C
U
N
A
y
20°
B
WE20
O
42°
62°
Figure 2
Figure 1
5
x
a Identify a pair of similar triangles in Figure 3 on the below.
b Find the values of the angles x and y.
c If AC = 10 cm, BC = 4 cm and EC = 11 cm, find ED.
6 Find the lengths of the sides labelled x and y, as shown in Figure 4 on the below.
A
B
x
y
E
B
70°
D
Figure 3
25°
C
A
3 2.5 E y
x
5
O
7
C
D
Figure 4
Topic 4 Geometry in the plane c04GeomentryInThePane.indd 175
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7
In Figure 5, chord SQ meets a tangent PQ at point Q. Construct a pair of
similar triangles based on the diagram shown.
WE21
8 Find two pairs of similar triangles in Figure 6.
B
S
E
A
R
O
Q
D
Figure 6
P
A plank rests on two cylindrical rollers,
as shown. The radii of the rollers are 10 cm
and 25 cm and their distance apart on the
ground is 90 cm. Find the value of the angle x.
FS
Figure 5
O
WE22
PR
O
9
PA
G
shown. The radii of the rollers are 18 cm and
24 cm. The plank makes an angle of 14° with
the horizontal ground. Find the distance, x cm,
between the rollers.
24 cm
TE
D
14°
18 cm
x
11 Line AB is a tangent to the circle, as shown in Figure 7, on the below. Find the
values of the angles labelled x and y.
Questions 12, 13 and 14 refer to Figure 8 on the below. The line MN is a tangent
to the circle and EA is a straight line. The circles have the same radius.
B y
D
x
A
U
N
E
O
C
O
R
R
EC
CONSOLIDATE
90 cm
E
10 A plank rests on two cylindrical rollers, as
25 cm
10 cm
x
DIGITAL DOC
doc‐9953
Investigation
Circle constructions
C
21°
F
O
M
G
C
B
Figure 8
Figure 7
A
N
12 Find 6 different right angles.
13 If ∠DAC = 20°, then ∠CFD and ∠FDG are respectively:
A 70° and 50°
D70° and 70°
B70° and 40°
E 50° and 50°
C 40° and 70°
BFGB
E none of the above
C EDA
14 A triangle similar to FDA is:
A FDG
DGDE
176 MATHS QUEST 11 SPECIALIST MATHEMATICS VCE Units 1 and 2
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15 In Figure 9 below, express x in terms of a and b.
16 Two tangent lines to a circle meet at an angle y, as shown in Figure 10 below.
Find the values of the angles x, y and z.
10°
x
z
a
O
O
x
b
Figure 10
a
O
17 Solve question 16 in the general case (see Figure 11) and
FS
Figure 9
y
PR
O
show that y = 2a. This result is important for space
navigation (imagine the circle to be the Earth) in that an
object at y can be seen by people at x and z at the same
time.
O
PA
G
19 The triangle that is similar to triangle BAD is:
C BCD
DBDC
E AOB
TE
D
A COD
BCAB
y
x
Figure 11
E
Questions 18, 19 and 20 refer to Figure 12 at right. The line
BA is a tangent to the circle at point B. Chord CD is
extended to meet the tangent at A.
18 Find the values of the angles x and y.
z
C
x
y D
O
z 50° 45° A
B
Figure 12
20 The value of the angle z is:
C 95°
D100°
E 130°
21 Find the values of the angles x, y and z in Figure 13 on the below. The line AB is
tangent to the circle at B. The line CD is a diameter.
22 Solve question 24 in the general case; that is, express angles x, y and z in terms
of a (see Figure 14 below).
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B85°
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A 50°
C
x
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B
y D
z
25°
Figure 13
24 cm
A
14°
18 cm
x
Figure 14
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REVIEW QUESTIONS
Units 1 & 2
Geometry in the plane
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Topic 4
Concept #
Sit Topic test
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178
MATHS QUEST 11 SPECIALIST MATHEMATICS VCE Units 1 and 2
c04GeomentryInThePane.indd 178
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4 Answers
EXERCISE 4.2
EXERCISE 4.3
1a56°
1aSimilar
b Congruent
b 67.5°
c x = 38°, y = 32°
2aSimilar
2 x = 157°
b Not enough information
3 Check with your teacher.
4 Check with your teacher
4 Check with your teacher.
5 D
5a x = 7.4 cm, y = 48 cm
6 C
b x = 3.1 cm, y = 3.1 cm
7 Answers will vary.
c x = 12.5 cm, y = 19.5 cm
8 Answers will vary.
d x = 7.5 cm, y = 10 cm
9a x = 60°, y = 114°
e x = 4
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3 Answers will vary.
6a x = 75°, y = 75°, z = 4 cm
b x = 47°, y = 55°, z = 78°
c x = 59°, y = 121°, z = 59°
b x = 13.8, y = 67°23′, z = 39
b x = 45°, y = 42°
d x = 24, y = 11.3
c x = 15, y = 12
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10a x = 75°, y = 30°, z = 150°
7a A
c 60°
11Answers will vary.
12122°
14x = 25°
15AB = BC, therefore ABC is isosceles.
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13x = 130°, y = 70°
E
D
C
C
A
b
A
EC
∠BAC = ∠BCA = 45° (property of isosceles
triangles).
B
In triangle OAB, ∠OAB + ∠ABO + ∠BOA = 180°
(sum of angles in a triangle).
Number of sides
360°
C
180°
4
5
540°
6
720°
7
900°
8
1080°
9
1260°
10
1440°
20
3240°
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Sum of interior angles
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16a
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∠OAB = ∠ABO = 45°, so ∠BOA = 90°.
b General formula: 180°(n − 2)
D
B
E
c
d
E
A
D
A
C
B
Topic 4 Geometry in the plane c04GeomentryInThePane.indd 179
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e
EXERCISE 4.5
A
A
1 Perpendicular bisectors meet outside the triangle.
2 Construct the perpendicular bisector of the base; vary the
radius to get different triangles.
3
C
B
B
D
8 A
9 B
4 Answers will vary.
10D
6 Check with teacher
12D
7 Find the angle bisectors of the angles formed at the centre
13a a = b = c = d = 78°
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of the circle. Join them to the circumference to create 12
vertices. Join the 12 vertices to create the dodecagon.
8 Check with teacher
9 Constructions
10Create perpendicular bisectors, then angle bisectors
to create 45° angles from the centre of circle, and
a diameter.
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b a = c = 112°, b = 68°
c a = b = d = e = 74°, c = 32°
d a = b = e = 102°, c = d = 78°
e x = 63°, y = 36°
f a = e = 122°, b = 32°, c = d = 26°
1517 cm
16a 2.4 metres. Need to find an alternative route.
b 4.5 metres
c 18 metres. Need a longer ladder.
12Answers will vary.
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D
17a x = 5.5, y = 3.75, z = 2.5
13a F
b 34.38, 23.44, 15.63
182.9 metres
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EXERCISE 4.4
1 Constructions. Check with your teacher.
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2–12 Constructions. Check with your teacher.
13Draw an arc with A as the centre, crossing the line
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at B and C.
With B as the centre, draw an arc above the line.
Repeat with the same radius at C. Join the points where
the arcs meet to point A.
14–16 Constructions. Check with your teacher.
17 a Constructions. Check with your teacher.
b The bisectors meet at the same point.
18 a Constructions. Check with your teacher.
b The bisectors still meet at a point.
19
2r
r
c T
d F
e F
1440°, 40°, 140°
1575°
1630°
17Check with teacher
18Check with teacher
19Check with teacher
20Check with teacher
21Answers will vary.
22Constructions
EXERCISE 4.6
1 x = 44°, y = 44°, z = 44°
2 C
4 AED = BEC (opposite angles)
4r
180 b F
3 x = 46°, y = 34°, z = 46°
A
r
B r
11Perpendicular bisectors meet on the hypotenuse.
d 55 metres
r
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1421.5 metres
20
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5 Check with teacher
11B
6r
r C
ADE = BCE (circle theorem 2)
Thus, similar triangles, so x = z.
MATHS QUEST 11 SPECIALIST MATHEMATICS VCE Units 1 and 2
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5 x = 126°, y = 96°
EXERCISE 4.7
x
6 y = 180° −
2
7 74°
8 B
9 B
1090
11x = 30°, y = 60°
12a Theorem 3, 90°
b Theorem 1, 25.5°
1 Construction
2 Check with your teacher
3 x = 85°, y = 20°, z = 85°
4 x = 42°, y = 62°
5a CBD, CEA
b 85°, 70°
c 7.36 cm
6 4.17, 5.83
c Theorem 4, 99°
7
ΔPQR and ΔPQS
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13x = y = z = 20°
14Answers will vary.
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e Theorem 1, 58°
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d Theorem 2, 51°
9 18.92°
16Answers will vary.
1048.87 cm
17a + b + c = 180° — sum of angles in a triangle
11x = 42°, y = 132°
c + d = 180°— straight angle
a + b + c = c + d — equating the two equations
a + b = d QED
13B
14D
E
12MAC, NAC, FDA, FBA, EDG, EBG
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18y + ECO = EBC — isosceles triangle
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8 ABE, DCE and ADE, BCE
15x = y = z = 42°
15x = 180° − a − b
16x = 80°, y = 20°, z = 80°
17Answers will vary.
18x = 50°, y = 95
19B
20C
21x = 25°, y = 65°, z = 40°
22x = a, y = 90° − a, z = 90° − 2a
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BEC + EBC + y = 180° — sum of angles in triangle EBC
BEC + y + ECO + y = 180° — substitution
2y + ECO = 180°− BEC — rearranging
x + OAE + AEO = 180° — sum of angles in
triangle OAE
AEO = BEC — opposite angles
x + OAE = 180° − BEC — rearranging
2y + ECO = x + OAE — equating two equations
involving 180° − BEC
ECO = OAE — isosceles triangle
2y = x QED
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