Download AS90153 Geometric Reasoning

NCEA
LEVEL 1
MATHEMATICS
Part 5 - AS90153
Geometric Reasoning
QUESTIONS & ANSWERS
MAHOBE
Published by Mahobe Resources (NZ) Ltd
Distributed free at www.mathscentre.co.nz
2
NCEA Level 1 Mathematics, Questions & Answers
Part 5 - AS90151 Geometric Reasoning
Contributors: Robert Cole, Mohsen Davoudi, Kim Freeman, Dr Sophia Huang, Tara Kelly.
This edition is Part 5 of a 6 Part eBook series designed to help you study towards NCEA.
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3
NCEA Level 1 Mathematics - Questions & Answers
Contents
About this Book
5
Geometric Reasoning
8
More Geometric Reasoning
9
Geometric Reasoning Example
10
Circle Geometry
11
Circle Geometry Example
12
Geometric Reasoning Achievement Example
13
Exercises
14
Geometric Reasoning - Merit Example
17
Geometric Reasoning - Excellence
21
Sample Exam
25
Answers
31
YEAR 11 MATHEMATICS
MAHOBE
4
MAHOBE
YEAR 11 MATHEMATICS
5
About This Book
Q&A eResources are recognised as the leading study guides for NCEA. Each
freely available title has been compiled by a team of experienced educators to
meet the study and revision needs of NCEA students. They are proving to be
valuable resources in the hands of students who want to work ahead of their
regular class programme. They also serve as effective revision programmes in the
run up to the final examinations.
This book carefully explains the mathematical concepts that will be tested in
NCEA then illustrates them with Achievement, Merit and Excellence examplars. It
allows students to practise on NCEA-type questions and provides detailed
solutions. After working through this programme, all students will be well prepared
for their final assessments.
4cm
The student who wrote the above answer on a recent assessment paper did not
use a Q&A Level 1 Mathematics eResource.
YEAR 11 MATHEMATICS
MAHOBE
6
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7
MATHEMATICS 1.9 - AS90153
Geometric reasoning
Always understand what the examiner wants! A past examination answer is
shown below. The student who wrote this answer on a recent assessment paper
did not use a Q&A Level 1 Mathematics eResource.
Name the parts of a circle
Santa lives here
Intruding square
Inside Circle
Circle border
YEAR 11 MATHEMATICS
Running round
in circles
Outside Circle
Bottom of circle
MAHOBE
8
Geometric Reasoning
The Facts to Learn
Angles in a triangle add up to 180°.
a + b + c = 180°
Angles on a straight line add up to 180°.
a + b + c = 180°
Angles in a quadrilateral add up to 360°.
a + b + c + d = 360°
A cyclic quadrilateral has all its vertices (corners)
touching a circle.
Opposite angles of a cyclic quadrilateral add up to 180°
a + c = 180°, d + b = 180°
Angles round a point add up to 360°.
a + b + c + d = 360°
Exterior angles of a triangle
= sum of the opposite interior angles
a+b=d
60°
60°
Isosceles triangle
2 sides the same length
2 angles the same size
MAHOBE
60°
Equilateral triangle
3 sides the same length
3 angles the same size (60°)
YEAR 11 MATHEMATICS
9
More Geometric Reasoning
Vertically opposite angles are equal.
(Even when parallel lines are involved).
b°
b°
«
«
b°
b°
b°
b°
«
«
«
«
Alternate angles
(Z angles) are equal.
«
b°
Corresponding angles
(F angles) are equal.
«
Co-interior angles
(C angles) add to 180°.
Exterior angles of a polygon add up to 360°.
Interior angles
Exterior angle
The sum of the interior angles of a polygon = (n - 2) × 180º
(Where n = number of sides)
This formula comes from splitting a polygon
into triangles by using diagonals. There will
always be 2 less triangles then there are sides.
When given a geometry problem don’t just concentrate on the angle you have been
asked to find. Try and find all the obvious angles according to the rules that you
can remember. Sooner or later the answer you need will be found.
YEAR 11 MATHEMATICS
MAHOBE
10
Geometric Reasoning Example
Find all the other angles in the diagram below.
B
C
›
30°
A
30º
15º
›
F
E
Triangle FAB is isosceles therefore
FAB =
D
FBA = 75°
AB and FC are parallel. ABFC is a Z-shape.
Therefore if
FBA = 75° then
BFC = 75° (alternate or “Z” angles)
EFC = 60° (straight line angles sum to 180°)
BC and FE are parallel, therefore
FCB = 60° (alternate or “Z” angles)
FBC = 45° (triangle angles sum to 180°)
FEC = 90° (triangle angles sum to 180°)
Angles on a straight line mean that
CED = 90° (straight line angles
sum to 180°)
If needed extend all the lines on the diagram. This can sometimes help
you see all the related angles - especially when parallel lines are
concerned.
45°
60°
75°
75°
60°
45°
90°
›
60°
30°
30°
60°
90°
75°
30º
15º
60°
MAHOBE
75°
60°
75° 30º
15º
›
90° 90°
90°
90°
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11
tangent
Circle Geometry
radius
Angle in a semicircle = 90º.
Tangent and radius meet at 90º.
bº
A triangle formed using two radii is an
isosceles triangle.
bº
The angle at the centre
is twice the angle
at the circumference.
2aº
Angles in the same
segment are equal.
i.e angles formed at the
circumference by chords
are equal.
A
B
C
ACB = 61°
When writing angles use three letters.
The middle letter is where the angle is.
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MAHOBE
12
Circle Geometry Examples
1.
In the diagram below, O is the centre of the circle.
Find the angles: U, V, W.
The diagram is not drawn to scale.
B
Find all the obvious angles.
A
Angles in the same segment
CAD =
V
O
70º
CBD = 70º
W
Angles in a semicircle = 90º
ABC = 90º,
U
C
D
DAB = 90º
Isosceles Triangles. Sides are radii.
AOD, BOC, BOA are all isosceles triangles.
BDA = 70º,
CAB = 20º and
V = 20º
(Angle in a semicircle, angles in an isosceles triangle.)
Angle U = 70º (identical isosceles triangle to AOD)
BOC and
AOD = 40º (angles in a triangle = 180º)
W = 140º (straight line = 180º or isosceles triangle AOB)
2.
Find, with geometric reasons, the size of angle dº.
B
AB and CD are parallel lines.
4x + 5x add to 180º.
(co-interior angles.)
9x = 180º therefore x = 20º.
A
5x
D
If x = 20º, 5x = 100º.
This means that d = 80º.
Opposite angles in a cyclic
4x
C
triangle are supplementary.
MAHOBE
YEAR 11 MATHEMATICS
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Geometric Reasoning - Achievement Examples
1.
The diagram shows the side view of a bridge over a stream.
B
125º
A
AB = BC, BCD = 125º.
Calculate the size of angle
BCA = 180° - 125°
C
D
ABC
(supplementary angles sum to 180°)
= 55°
BAC = 55° (base angles of an isosceles triangle are equal)
ABC = 70° (triangle angles sum to 180°)
The diagram below shows the side view of a water slide.
Find the angle CAB the angle between the ladder and the slide.
A
Slid
e
Support
2.
126°
C
E
160°
Level Ground
B
ACE = 54° as angles on a straight line sum to 180°
ABE = 20° as angles on a straight line sum to 180°
CAB = 180° - 20° - 54°
= 106° as angles in a triangle sum to 180°
YEAR 11 MATHEMATICS
MAHOBE
14
Exercises
1.
a.
A
Calculate the size of angle
CDE
. . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . ..
C
67º
42º
E
B
. . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . ..
. . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . ..
D
b.
H
I
Calculate the size of angle
HIJ.
. . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . ..
. . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . ..
52º
K
. . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . ..
J
c.
Calculate the size of angle c.
. . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . ..
84º
. . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . ..
. . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . ..
cº
d.
Calculate the size of
VUW
. . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . ..
U
V
. . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . ..
X
55º
Z
MAHOBE
W
Y
. . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . ..
Note: Diagrams are not drawn to scale.
YEAR 11 MATHEMATICS
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2.
Shape A is a regular pentagon. Shape H is a regular heptagon. Both shapes
share the side AB.
a.
Calculate the size of the angle CBD.
. . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . ..
A
Shape P
C
. . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . ..
Shape H
B
. . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . ..
D
The diagram below gives the angles of a playground slide.
b.
Calculate the size of the shaded angle XYZ.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
X
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
35º
U
Y
15º
Z
W
Two cyclic quadrilaterals are shown below. The size of angle
c.
Find the size of the angles OMN, and OPN.
d.
How do we know that JK is parallel to OP?
JKN is 95º.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
M
J
O
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
95º
K
N
YEAR 11 MATHEMATICS
P
MAHOBE
16
3.
a.
PQRS is a cyclic quadrilateral. O is the centre of the circle.
Find the size of angles PRQ and PSQ.
.. . . . . . . . . . . . . . . . . . . . . . . . . . . .
R
.. . . . . . . . . . . . . . . . . . . . . . . . . . . .
.. . . . . . . . . . . . . . . . . . . . . . . . . . . .
.. . . . . . . . . . . . . . . . . . . . . . . . . . . .
120º
P
Q
.. . . . . . . . . . . . . . . . . . . . . . . . . . . .
S
b.
In the diagram below (which is not drawn to scale), C is the centre of
the circle, and XY is a tangent to the circle. Angle ABY = 70º.
Complete the sentences below to find, in 4 logical steps the angle
which equals 50º.
A
C
70º
X
B
Y
Angle
Size
Reason
XBC
90º
..........................................
CBA
......
Adjacent angles on a line add up to 180º.
CAB
20º
..........................................
XCB
40º
..........................................
......
50º
..........................................
MAHOBE
YEAR 11 MATHEMATICS
17
Geometric Reasoning - Merit Examples
1.
The diagram below shows a pentagram - a five sided star.
The shaded shape formed by ABCDE is a regular pentagon.
Z
Calculate the angle VCW.
Give geometric reasons with each of your calculations.
The total of all the interior angles
B
A
Y
V
of a regular pentagon can be
calculated by the formula:
C
E
(n-2) × 180 where n is the
D
number of sides (5).
Therefore 3 × 180° = 540°
X
W
Each angle = 108º (as 540° ÷ 5 = 108°).
VCW = 108º as it is vertically opposite.
N
Axis of symmetry
This means that
P
O
42º
2.
M
R
Q
The figure above shows a semicircle with centre R.
RO is an axis of symmetry. The angle NMP is 42º.
Find the angle NQM. Support your answer with geometric reasons.
QNM =
MPQ = 90º (angle in a semicircle)
This means that
NOP and
NOM = 48º (angles in a triangle sum to 180º)
MOQ = 132º (angles around a point sum to 360º)
(Vertically opposite angles are equal)
OQM =
OMQ = 24º (base angles of an isosceles triangle)
YEAR 11 MATHEMATICS
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18
4.
In the diagram below the points A, B, C and D lie on a circle with centre O.
Calculate the size of angle BCD. Give geometric reasons for your answer.
B
....... . . . . . . . . . . . . . . . .. . . . . . .
47º
....... . . . . . . . . . . . . . . . .. . . . . . .
A
....... . . . . . . . . . . . . . . . .. . . . . . .
58º
O
....... . . . . . . . . . . . . . . . .. . . . . . .
....... . . . . . . . . . . . . . . . .. . . . . . .
D
....... . . . . . . . . . . . . . . . .. . . . . . .
....... . . . . . . . . . . . . . . . .. . . . . . .
5.
The diagram below shows a circle with two tangents at W and X.
O is the centre of the circle.
Calculate the angle WYX. Give geometric reasons for your answer.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
O
W
150º
X
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Y
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
MAHOBE
YEAR 11 MATHEMATICS
C
19
6.
In the figure below ABDC is an isosceles trapezium.
Find the length of BD. Give geometric reasons for your answer.
12
cm
E
A
B
10 cm
C
D
15 cm
....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . ....
....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . ....
....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . ....
....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . ....
7.
In the diagram below ABC is an isosceles triangle.
O is the centre of the circle. The angle OBC = 31º
a.
Calculate the size of BAC.
b.
Calculate the size of ODC.
Give reasons for your answers.
A
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .....
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .....
O
B
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .....
31º
C
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .....
D
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .....
YEAR 11 MATHEMATICS
MAHOBE
20
8.
The diagram below is NOT drawn to scale.
O is the centre of the circle.
P, K, Q and N are points on the circumference.
QT is the tangent to the circle at Q.
Angle POQ = 108°
Angle NPQ = 35°
Calculate the angles x and y.
P
35º
N
108º
O
K
T
Q
....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . ....
....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . ....
....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . ....
....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . ....
....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . ....
....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . ....
....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . ....
....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . ....
MAHOBE
YEAR 11 MATHEMATICS
21
Geometric Reasoning - Excellence Examples
1.
The diagram below shows a circle with points A, B and C on the circumference.
Centre of the circle is O.
B
Calculate the size of CAB.
Give geometric reasons for your answer.
C
32º
A
Triangle AOC is isosceles (OA = OC)
OAC = 32º
O
(base angles of an isosceles triangle)
AOC = 116º
(angle sum of a triangle = 180º)
The reflex angle about the centre is 360º - 116º = 244º
ABC = 122º (angle at the circumference = ½ angle at centre)
BCA = 32º. (alternate angles are equal i.e.
BCA and
OAC)
BAC = 26º (angle sum of a triangle = 180º)
2.
In the figure below, O is the centre of the circle. XZ is the diameter.
Prove that c = 2b. Give geometric reasons for your answer.
XYZ = 90º (angle in a semicircle)
Y
OX, OZ and OY are radii therefore there
Z
are two isosceles triangles formed, OYZ
and OYX. This means that
O
and
OYZ = aº
OXY = bº.
b = 90 - a (angle in a semicircle)
X
c = 180 - 2a (angles in a triangle sum)
Therefore c = 2b
YEAR 11 MATHEMATICS
MAHOBE
22
9.
In the figure below, line KM forms an axis of symmetry.
Length QN = Length QK = Length QL.
Angle NQM = 120°.
Angle NMQ = 30°.
Prove that the quadrilateral KLMN is cyclic.
K
L
N
Q
120º
30º
M
....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . ...
....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . ...
....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . ...
....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . ...
....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . ...
....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . ...
....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . ...
....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . ...
MAHOBE
YEAR 11 MATHEMATICS
23
10.
The circle below has a centre O and a tangent XY at point B.
The points A and D lie on the circle.
Prove that
YBC =
BAC
X
B
A
Y
O
C
....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . ...
....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . ...
....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . ...
....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . ...
....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . ...
....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . ...
....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . ...
....... . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . ...
YEAR 11 MATHEMATICS
MAHOBE
24
MAHOBE
YEAR 11 MATHEMATICS
25
MAHOBE
Level 1 Mathematics - Sample Exam
AS90153 Geometric Reasoning
Published by Mahobe Resources (NZ) Ltd
Distributed free at www.mathscentre.co.nz
YEAR 11 MATHEMATICS
MAHOBE
26
MAHOBE
YEAR 11 MATHEMATICS
27
You are advised to spend 25 minutes answering the questions in this section.
QUESTION ONE
Pauline’s garden table is illustrated in the diagram.
A
ABEF is a straight and horizontal table top.
XY is an axis of symmetry.
Angle BED = 110°
Diagram is not to scale.
X
B
E
F
110°
Y
C
D
Calculate the size of angle ABC giving reasons for your answers.
................................................................
................................................................
................................................................
................................................................
.....................................
QUESTION TWO
Stan’s step-ladder is shown in the diagram.
L
Angle LMN = 122°
KL = LN
Calculate the size of the inside angle of the top of the
step-ladder KLM giving reasons.
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122°
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YEAR 11 MATHEMATICS
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28
QUESTION THREE
The diagram shows a regular pentagon inside a circle.
A
B
E
36°
C
D
Calculate the size of BEA and DCB giving reasons for your answers.
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QUESTION FOUR
In the diagram, PQRO is a rhombus QR is parallel to ST.
Angle TSR = 63°. Calculate the angle of ROT.
Give a geometric reason for each step leading to your answer.
Q
S
P
R
W
O
63°
T
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YEAR 11 MATHEMATICS
29
QUESTION FIVE
In the diagram V, W, X and Y lie on a circle, centre O.
ZW is a tangent to the circle at W.
Angle YVW = 65°
Angle XWZ = x°
V
65°
Find the size of angle OYX.
Y
O
X
x°
W
Z
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YEAR 11 MATHEMATICS
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QUESTION SIX
KLM and MNO are two triangles.
Angle KLM = 32°
Angle MNO = 24°
L
K
J
32°
M
O
Find the size of angle OMN.
45°
N
Give geometric reasons for each step leading to your answer.
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YEAR 11 MATHEMATICS
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MAHOBE
The Answers
YEAR 11 MATHEMATICS
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32
The Answers
Page 14
1.
A
a.
d.
67º
E
B
55º
70º
C
42º
67º
U
71º
V
55º
D
X
55º
CED = 67° - vertically opposite
W
Y
Z
Angles in a triangle sum to 180°
180° - 67° - 42° = 71°
See diagram above for the
corresponding angles. The 70° is
b.
then calculated from the isosceles
triangle WUV.
52º H
I
52º
64º
Page 15
2.
a.
Sum interior angles = (n-2) × 180º
Sum interior angles of pentagon
52º
K
(5 - 2) × 180º = 540º
64º
Each angle = 108º
J
Sum interior angles heptagon
(7-2) × 180º = 900º
Each angle = 128.57º
Angle IHJ = 52° alternate angles,
Angle CBD = 360 - (108 + 128.57)
parallel lines.
= 123.43º
HIJ = 64° - Angles in an isosceles
b.
triangle.
X
c.
84º
75º
35º
96º 84º
84º
U
90º Y
15º
Z
W
Angle UYW = 90º right angled
rectangle.
See the diagram above for all the
corresponding angles.
c = 96° angles on a straight line
MAHOBE
The angle ZYW = 75º angle sum
of a triangle.
XYZ = 360º - (75º + 90º + 35º)
= 160º
YEAR 11 MATHEMATICS
33
2.
C.
M
J
XBC = Angle between a tangent
b.
Page 15 (cont)
85º
and a radius
O
CBA = 20°
95º
CAB = Base angles of an isosceles
triangles are equal
K
85º
95º
N
XCB = 40° angles on straight line
P
CXB = Angles in a triangle sum to
180°
NMJ = 85º opposite angles of a
cyclic triangle sum to 180º
Page 18
This means angle NMO = 95º
4.
B
47º
angles on a straight line and
NPO = 85º as it is the opposite
47º
61º
A
angle in the cyclic triangle.
d.
JK is parallel to OP as all the
O
72º
C
61º
angles indicate that they are
D
corresponding.
R
Page 16
86º
58º
OB, OA and OD are all radii.
This means that there are 2 isosceles
60º
triangles AOB (47°, 47°, 86°)
and AOD (61°, 61°, 58°)
120º
P
BAO + DAO: 61° + 47° = 108°
Q
120º
BCD
S
3.
a.
= 72° opposite angles cyclic quad.
or BOA + AOD: 86° + 58° = 144°
BCD = 72° angle at circumference
PRQ = 60º angle at the centre
PSQ = 120º opposite angles in a
5.
cyclic quadrilateral.
B.
W
O
150º
X
90º
90º
Isosceles triangle (2 radii)
A
C
20º
140º
30º
20º
70º
X
Y
Y
B
90º angle between tangent and radius
OWY AND OXY are 90° as the angle
between the tangent and radius are
perpendicular.
WYX = 30° interior angles of a
quadrilateral sum to 360°
YEAR 11 MATHEMATICS
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34
Page 19
Page 20
E
6.
P
12
cm
35º
N
A
126º
B
10 cm
C
T
108º
O
54º
K
19º
35º
D
15 cm
36º
36º
Q
EAB and ECD are similar triangles.
EA
12
This means
=
10
15
8.
Lines OP and OQ are radii.
This means that EA = 8 cm
This means that OPQ is an isosceles
BD = CE - EA
triangle and the base angles OPQ and
(CE = DE, AE = BE)
= 12 cm - 8 cm
OQP = 36°
= 4 cm
A
7.
Angle PKQ = 54° as it is angle at the
circumference (½ POQ)
Angle x (PNQ) = 126° as opposite angles
59°
118º
O
B
31º
31º
C
of a cyclic quadrilateral sum to 180°.
OQT is the angle between a radius and a
tangent = 90°.
NQP = 19° angle sum of a triangle
D
NQT (y° )
= 90° - 36° - 19°
= 35°
BCO = 31° base angles isosceles triangle.
COB = 118° angle sum of a triangle
BAC = 59° angle at circumference is half
that at the centre.
BAC is isosceles triangle
BDC is an isosceles triangle
Angle BDC = 121° opposite angles cyclic
quadrilateral. ODC is half this 60.5°.
MAHOBE
YEAR 11 MATHEMATICS
35
K
Page 22
L
Page 27
60º 60º
60º
60º
60º
60º Q
30º
N
Question One
FED = 70° angles on a straight line
30º
ABC = 70° symmetry through XY
120º
120º
Question Two
LMK = 58° angles on a straight line
LKM = 58° as triangle is isosceles
30º30º
KLM = 64° angle sum of a triangle
Page 28
M
Question Three
9.
KQN = 60° angles on a straight line
BEA = 36° angles in the same segment
QKN = 60° isosceles triangle QN = QK
DCB = 108° interior angles of a regular
LKN = 120° line KM is a line of
pentagon.
symmetry
Question Four
LKN = 2 × QKN
OTS is an isosceles triangle
LMN = 60° due to line of symmetry KM
STO = 63° base angles isosceles triangle
LKN + LMN = 180°
ROT
= 180° - 2 × 63°
= 54°
KLMN = a cyclic triangle (opposite angles
of a cyclic quadrilateral sum to 180°)
Page 29
Page 23
Question Five
X
OW is at 90° to WZ
10.
It is a radius and tangent
B
This means that 0WX = 90° - x
ß
A
O
Y
YXW = 115° opposite angles of a
cyclic quadrilateral
WOY = 130° angle at centre is twice the
C
angle at the circumference
OB is a radius and the tangent XY is 90°
Angle sum of a quadrilateral is 360°
to the tangent.
OYX = 360 - 130 - 115 - (90 - x)
Let YBC = ß
= (25 + x)°
OBC = 90 - ß tangent 90º to radius
OCB = 90 - ß base angles of isos triangle
OB = OC
BOC
Page 30
Question Six
= 180 - (90 - ß) - (90 - ß)
Let KML = x°, then OMN = x° as they are
= 2ß
vertically opposite.
BAC = ß angle at circumference is half
JOM = x + 45°, JKM = x + 32° as the
that at the centre
exterior angles of a triangle = the sum of
BAC = ß, YBC = ß
the two opposite interior angles
Therefore YBC = BAC
JOM + JKM = 180° (cyclic quadrilateral)
x° + 45° + x° + 32° = 180°
x = 51.5°
YEAR 11 MATHEMATICS
MAHOBE
36
MAHOBE
YEAR 11 MATHEMATICS
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YEAR 11 MATHEMATICS
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