Download Revision for Nov Exam Part 2

Gr 9 Maths: Content Area 3 & 4
Geometry & Measurement (2D)
QUESTIONS
Mostly past ANA exam content
• Geometry of Straight Lines
• Triangles: Basic facts
• Congruent Δ
• Similar Δ
s
s
All questions have been
graded to facilitate
concept development.
GOOD LUCK!
• Quadrilaterals
• Polygons
Theorem of Pythagoras
Area and Perimeter of 2D shapes
Compiled by
Anne Eadie & Gretel Lampe
THE ANSWER SERIES
tel: (021) 671 0837
fax: (021) 671 2546
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Questions: Geometry of Straight lines
2.
GEOMETRY OF STRAIGHT LINES
( Solutions on page A1)
Calculate the size
of the largest angle.
5.
1
x – 6°
Show all your steps
with reasons.
x – 9° x + 15°
In the figure, Bˆ 3 = 35º and BE || CF.
ˆ
Determine the size of Bˆ and BCF.
(4)
A
Refer to page Q13 for details
on parallel lines & angles.
3.
1.
Complete the following:
. . . . . . . angles.
C
43°
Statement
Reason
Bˆ 1 =
(1)
ˆ =
BCF
B
(3)
D
b
F
E
(1)
3.2 Angles around a point add up to . . . . . . .
1.1
a
1 2
3.1 Angles which add up to 90º are called
Calculate the sizes of the angles marked a to d.
Give reasons for your answers.
A
C
B 3
(3)
4.
6.
Complete each of the following statements:
A
T
4.1 D̂ and F̂ are complementary angles if
____________________________________ .
1.2
c
4.2 The sum of the interior angles of a triangle is
58°
equal to _____________________________ .
12°
(2)
P
Q
d
(1)
(1)
2
1
3
2
C
1
D
Calculate the size of  , Bˆ 1 and Bˆ 2 .
Statement
4.4 A trapezium is a quadrilateral with one pair
of ___________________ sides.
B
In the figure above, AB || TC, Cˆ 1 = 65º and Cˆ 2 = 43º.
4.3 The sum of the exterior angles of any polygon
is equal to ___________________________ .
1.3
(1)
Reason
(1)
4.5 The diagonals of a rectangle are _________
S
Copyright © The Answer
112°
R
T
(3)
in length.
(1)
(4)
Q1
Questions: Geometry of Straight lines
7.
10.
Give reasons for each of your statements in the
questions below.
Find the size of angles a to g (in that order),
giving reasons.
g
In the figure PQ || RS, Qˆ 1 , Qˆ 2 and Qˆ 3
are equal to 2 x, 3 x and 4 x respectively.
R̂ = y and Ŝ = z.
3
An acute angle is one that lies between 0º and 90º.
35°
a
b
1
2
Important Vocabulary
60°
T
P
STRAIGHT LINE GEOMETRY
c
Q
d
An obtuse angle is one that lies between 90º and 180º.
e
A reflex angle is one that lies between 180º and 360º.
f
(7)
A straight angle = 180º
11.
In the sketch, AB is a straight line.
A revolution = 360º
When the sum of 2 angles = 90º, we say the angles are
complementary.
Determine the value of x + y.
R
8.
y
z
When the sum of 2 angles = 180º, we say the angles are
supplementary.
S
7.1 Calculate the value of x.
(3)
7.2 Calculate the value of y.
(3)
7.3 Calculate the value of z.
(3)
x+y
x + 30°
B
120°
P
State, giving reasons,
whether PQ || RS.
D
U
Q
2
4
3
Adjacent angles have a common vertex and a common
ˆ 2ˆ and 3,
ˆ 3ˆ and 4ˆ or 1ˆ and 4.
ˆ
arm, e.g. 1̂ and 2,
A
Vertically opposite angles lie opposite each other,
x
C
R
76°
4 angles are formed :
1
ˆ 2,
ˆ 3,
ˆ 4
ˆ
1,
ˆ
e.g. 1̂ and 3ˆ or 2ˆ and 4.
(4)
The FACTS
When 2 lines intersect:
Hint: Draw a third line, through B,
parallel to the given parallel lines.
T
(4)
110° B
(4)
9.
When 2 lines intersect,
y
A
B
3x – 10°
C
x
12. Calculate, with reasons, the value of x.
Calculate, with reasons, the value of x.
A
A right angle = 90º
V 104°
S
W
(4)
For further practice in this topic –
see The Answer Series
Gr 9 Mathematics 2 in 1 on p. 1.32
Q2
adjacent angles are supplementary
vertically opposite angles are equal.
See the end of the questions
for more on straight lines.
Copyright © The Answer
Questions: Triangles
4.
TRIANGLES: BASIC FACTS
7.
In ΔEDF, DF is produced to C.
The size of Ê is . . . ?
( Solutions on page A3)
Using the figure below, calculate the size of the
angles a, b and c (in this order).
AD = BD = BC;
E
A
a
ˆ = 72º
ADB
Reasons must be provided for all Geometry statements.
1.
In the figure below, ΔANT is an equilateral triangle.
Calculate the size of Tˆ and Tˆ .
1
3x
2
D
A
2 1
P
2.
T
B 60º
C 140º
D 20º
x
H
W
3.1 The size of Tˆ
ˆ
3.2 The size of M
2
Copyright © The Answer
44°
(3)
A
106°
x
(2)
(3)
P
B
110°
C
50°
D
8.2
a
ˆ = 110º.
In the figure above, B̂ = 50º and ACD
M 1
The size of  is . . . . . .
2
R
1
Determine the values of x, a, b and c in the figures
below.
8.1
6.
N
In ΔPRT alongside,
M is the midpoint of PR
and MR = MT.
If P̂ = 25º, calculate
with reasons:
C
Determine the size of  in terms of x.
1 2
E
(6)
In ΔABC, AB = AC and Ĉ = x.
S
70°
1 2
c
C
8.
A
A
(1) [10]
D
B
ˆ = 70º;
In the figure below, CS || HN, EAW
ˆ = x.
AE = AW and CAE
C
C
72°
b
(4)
Determine the value of x.
3.
A 40º
5.
N
B
4x 5x
F
2
1
T
(1)
A 50º
B 60º
C 110º
D 160º
(1)
Q3
c
28°
b
44°
(6)
Questions: Triangles
9.
Calculate the values of x and y if
TRIANGLES: Study the following very carefully
Bˆ 2 = x, Dˆ 2 = y, Dˆ 1 = 44º, Cˆ 1 = 75º and AD || BC.
A
2
y
1
1
44°
2
D
CLASSIFICATION OF TRIANGLES . . .
B
Triangles are classified according to their sides or
their angles (or both).
x 2
1
75°
• Sides
C
E
(3)
equilateral Δ
isosceles Δ
INTERIOR AND EXTERIOR ANGLES . . .
1
2
3
1̂ , 2̂ and 3̂
are interior
angles of the
triangle
y
x
x̂ is an exterior ø
ŷ is not an exterior ø
scalene Δ
An exterior angle is formed between one side of
a triangle and the produced (extension) of another.
A
10.
C
1
3 sides equal
B
95°
2 sides equal
no sides equal
• Angles
acute-angled Δ
right-angled Δ
obtuse-angled Δ
30°
E
D
3 acute angles
ˆ = 95º
In the above figure AB || ED, ACD
and D̂ = 30º.
Determine the size of Ê and  .
For further practice in this topic –
see The Answer Series
Gr 9 Mathematics 2 in 1 on p. 1.24
1 right angle (90°)
1 obtuse angle
• Sides and Angles
(3)
This is an
isosceles,
right-angled
triangle
This is an
isosceles,
acute-angled
triangle
This is a
scalene,
obtuse-angled
triangle.
4 BASIC FACTS
• FACT 1
The sum of the
interior angles
of a triangle = 180°
A
Aˆ + Bˆ + Cˆ = 180°
B
C
• FACT 2 :
2
The exterior angle
of a triangle equals
3
the sum of the
interior opposite angles.
• FACT 3
In an isosceles triangle,
the base angles
are equal.
B
1̂ = 2ˆ + 3ˆ
1
A
1
If AB = AC,
then 1̂ = 2̂
2
If 1̂ = 2̂,
then
AB = AC
The converse states:
If 2 angles of a triangle are equal,
then the sides opposite them are equal.
• FACT 4
The angles of an equilateral triangle
all equal 60°.
Q4
Converse:
C
60°
60°
60°
Copyright © The Answer
Questions: Congruent Triangles
3.
CONGRUENT ΔS
6.
Why is ΔABC ≡ ΔDCB?
A
( Solutions on page A4)
D
In the figure below ΔKNQ and ΔMPQ have a common
vertex Q.
P is a point on KQ and N is a point on MQ.
KQ = MQ and PQ = QN.
B
See the notes on Congruency
and Similarity on page A5
P
1.
K
A S, S, S
B 90º, Hyp, S (RHS)
C S, ø , S
D ø, ø, S
E
F
A
C
1
4.
In the figure below Dˆ 1 = Bˆ 2 = 90º and AD = BC.
2
P
1 2
B
Q
R
B
2
D
C
C
1
Q
2
A
Reason
(2)
State which triangle is congruent to ΔABC.
C
(4)
1
D
Prove that ΔABD ≡ ΔCDB.
2.
N
Prove with reasons that ΔKNQ ≡ ΔMPQ.
Which triangle is congruent to ΔPQR?
Statement
M
7.
ΔABC, D and E are points on BC such that BD = EC
and AD = AE.
A
5.
In the figure below, AB = AC and BD = CD.
A
P
1 2
1 2
A
B
Q
B
S
T
Copyright © The Answer
D
R
B
V
(2)
C
5.1 Prove that ΔABD ≡ ΔACD.
(4)
ˆ
5.2 Prove that DA bisects BAC
(2)
Q5
D
E
C
7.1 Why is BE = CD?
(1)
7.2 Which triangle is congruent to ΔABE?
(1)
Questions: Congruent Triangles
8.
In the given figure, P and T
are points on a circle with
centre M.
P
N
T
10.
9.2 Prove that ΔABC ≡ ΔDEF.
Statement
2 1
Reason
AB = AC
2 1
N is a point on a
chord PT such that
MN ⊥ PT.
1
A
and
M
B
In the figure
alongside
1
2
2
1
BD = CD
D
1
C
Prove that PN = NT.
Statement
Reason
(5)
9.3 Why is Bˆ = Eˆ ?
Statement
Reason
Bˆ = Eˆ
11.
(1)
(8)
9.
A
1
C
Prove that ΔABD ≡ ΔACD.
(4)
10.2
Prove that ΔABE ≡ ΔACE.
(4)
10.3
Prove that Eˆ 1 = Eˆ 2 = 90º.
(3)
10.4
Hence, state the relationship
between AE and BC.
(1)
In the figure below, PS || QR. Which ONE of the
following statements is true for this figure?
P
S
T
Note: It has (already) been given that AB = ED.
Statement
1
F
10.1
9.4 Use your answer in Question 9.3 to derive a
further relationship between AB and ED.
E
D
Q
Reason
R
A ΔPTS ≡ ΔPQT
B
B ΔPTS ≡ ΔRTQ
In the above diagram, AC = DF, AB = DE and BF = CE.
(2)
9.1 Prove that BC = EF.
Statement
2
1
E
C ΔPTS ||| ΔSRT
D ΔPTS ||| ΔRTQ
Reason
(1)
For further practice in this topic –
see The Answer Series
Gr 9 Mathematics 2 in 1 on p. 1.28
(2)
Q6
Copyright © The Answer
Questions: Similar Triangles
3.
SIMILAR Δ
S
( Solutions on page A6)
In ΔPQR and ΔSTR in the
figure alongside, PQ || ST,
PR = 10 cm, ST = 3 cm and
SR = 6 cm.
3.1
See the notes on Congruency
and Similarity on page A5
3.2
B
5.
P
10
E
R
Q
T
A
3
6
Prove that
ΔPQR ||| ΔSTR
1
S
Calculate the length of PQ.
F
1 2
D
(4)
(3)
2
C
In the figure,
B̂ = Ĉ , AD = 9 cm, AE = 7 cm and CE = 21 cm.
1.
Examine ΔDEF and ΔKLM.
4.
D
In ΔNML below, P and Q are points on the sides
MN and LN respectively such that QP || LM.
MN = 16 cm, QP = 3 cm and LM = 8 cm.
K
14 cm
7 cm
E
20 cm
L
F
10 cm
M
Complete the following calculations if ΔDEF ||| ΔKLM.
x = ________
Copyright © The Answer
D
2. Pˆ = …………….
3.
F
Qˆ = …………….
1
∴ ΔQPN ||| Δ ….
4 cm
10 cm
Calculate the length of BD.
Statement
Reason
1. N̂ ……………………………
E
C
5.2
Complete the following (give reasons for
the statements):
1
6 cm
P
In ΔQPN and ΔLMN
A
15 cm
1
Prove with reasons that ΔQPN ||| ΔLMN.
(3)
Calculate the length of AB if ΔABC ||| ΔEDF:
B
2
1
N
4.1
14
x
=
7
2
(6)
DE
EF
DF
=
=
(proportional sides of similar triangles)
KL
LM
2.
Reason
12 cm
Q
Prove that ΔABD ||| ΔACE.
Statement
M
L
x cm
5.1
(4)
4.2
……………………………
…………………………..
Hence, calculate the length of PN.
Q7
(5)
……………………………
(4)
(3)
For further practice in this topic –
see The Answer Series
Gr 9 Mathematics 2 in 1 on p. 1.28
Questions: Quadrilaterals
3.
QUADRILATERALS
( Solutions on page A7)
D
2
1
1
1.
ABCD is a parallelogram. Calculate the size of B̂ .
A
x + 50°
5.
In the figure below, ABCD is a square and ATB is an
equilateral triangle.
1
4
A
2
D
C
3
T
1
B
1
2
C
2
B
Look at parallelogram ABCD above and complete
the table.
1
A
2x – 20°
C
2.
D
(4)
In the figure below, DEFG is a rhombus and Ê = 156°.
D
1
2
156°
E
4.
2
B
Statement
3.1 Name two isosceles triangles.
(2)
3.2 Calculate the size of Dˆ 2 .
(3)
3.3 Calculate the size of Tˆ 4 .
(2)
In ΔADB and ΔCBD
PRTW is a square. ΔPQR and ΔRTS are equilateral.
ˆ
Calculate x (RQS)
Q
G
2
Reason
Dˆ 1 = ______
Alternate ø's and AD || BC
Bˆ 1 = ______
Alternate ø's and AB || DC
BD = BD
Common side
1
F
â ΔADB ≡ Δ______ ____________
x
Calculate the size of:
P
ˆ
2.1 EFG
â AD = ______ and Corresponding sides of
s
AB = ______
congruent Δ
R
2.2 Fˆ2
S
(4)
2.3 Ĝ
W
Statement
Reason
2.1
ˆ =
EFG
(2)
2.2
Fˆ2 =
(2)
Ĝ =
(2)
2.3
T
NB: Study 'Quadrilaterals'
on page Q12 very carefully.
Q8
(7)
6.
A parallelogram with at least one angle equal to 90°
is called a __________
A
kite.
B
rhombus.
C
trapezium.
D
rectangle.
(1)
Copyright © The Answer
Questions: Quadrilaterals
7.
A
B
2
T
3
1
9.
2
P
90°
1
B
120°
C
C
100°
D
108°
The bisectors of B̂ and Ĉ of parallelogram ABCD
intersect at T. Points B, T and D do not lie on
a straight line. P is a point on DC such that
ˆ = 90º.
TPD
10.
7.1
Prove that Tˆ 2 = 90º.
(5)
7.2
Which triangle is similar to ΔBCT?
(2)
7.3
If BC = 2TC and TP = 4 cm, calculate the length
of BT.
8.
(1)
What is the size of each angle in a regular hexagon?
A
90°
B
120°
C
100°
D
108°
(1)
(3)
For further practice in this topic –
see The Answer Series
Gr 9 Mathematics 2 in 1 on p. 1.26
In the given quadrilateral AE = ED and BE = EC,
therefore:
A
NOTES
What is the size of each angle in a regular
pentagon?
A
1
2
D
POLYGONS
D
E
B
A
ΔAEB ||| ΔCED
B
ΔAED ||| ΔBEC
C
ΔAEB ≡ ΔDEC
D
ΔAED ≡ ΔBEC
Copyright © The Answer
C
(2)
Q9
Questions: Theorem of Pythagoras
5.
THEOREM OF PYTHAGORAS
Calculate the length of AD.
( Solutions on page A9)
1.
8 cm
Triangle
B
12 cm
C
D
B
(2)
C
Show that B̂ = 90°.
base
b
ℓ
A
A right-angled triangle has
6 cm
C
y
D
2.1 Calculate x.
(3)
2.2 Calculate y.
(3)
Give reasons.
3.
The area of
2
ΔTUW = 30 cm
and UW = 12 cm.
ℓ: length
b: breadth
T
2
The perimeter of a rectangle
= ℓ + b + ℓ + b
= 2ℓ + 2b
= 2(ℓ + b)
The area of a rectangle
= ℓ % b = ℓb
Square
B
The perimeter of a square
= 4 % s = 4s
C
s
The area of a square
= s % s = s2
This theorem states:
In a right-angled triangle . . .
the square of the hypotenuse equals
the sum of the squares on the other two sides.
See the Quadrilaterals on page Q12
for the areas of all other quadrilaterals.
i.e. In ΔABC, B̂ = 90°
U
3.2 the perimeter of ΔTUW
12 cm
W
(2)
(3)
2
2
Circle
2
So: AC = AB + BC
The converse theorem states the reverse:
A
If in any ΔABC,
4.
hypotenuse
The Theorem of Pythagoras
Calculate:
3.1 TU
one angle of 90º. Here, B̂ = 90°.
The side opposite the right angle (90°)
is called the hypotenuse.
Here, AC is the hypotenuse.
base × height
base
Rectangle
A Right-angled triangle
B
The area of
height a triangle =
base
17 cm
x
C
a
height
height
In ΔABC: AB = 9 cm, BC = 12 cm and AC = 15 cm.
For further practice in this topic –
see The Answer Series
Gr 9 Mathematics 2 in 1 on p. 1.33
A
b
10 cm
5 cm
6.
The perimeter
of this triangle
= (a + b + c) units.
A
c
C 12,8 cm
B
8 cm
PERIMETER AND AREA FORMULAE
B 6 cm
D 14 cm
Determine the length of
AC if AB = 5 cm and
BC = 12 cm.
(4)
2.1
A
A 2 cm
A
In ΔABC, AB ⊥ BC.
In rectangle ABCD, AB = 8 cm and diagonal AC = 10 cm.
A ladder is standing against the wall. If the ladder
reaches a height of 12 m up the wall and has its foot
5 m away from it, calculate the length of the ladder. (3)
2
2
B
Q10
centre
radius (r)
2
AC = AB + BC ,
then B̂ = 90°.
circumference
C
diameter
The circumference
of a circle:
= πd = π(2r) = 2 π r
The area of a circle:
= π r2
Copyright © The Answer
Questions: Measurement: 2D
4.
MEASUREMENT: 2D
In the figure below, AP = 5 m,
AS = SB = 2 m and PS ⊥ AB.
( Solutions on page A10)
1.
2.
AB, the diameter of the
given circle, is 12 cm.
Use π = 3,14 to answer
the following questions,
correct to two decimal places.
3.
A
5m
C
A
(3)
If the length of the side of a square is
0,12 cm then the area =
8.
4.1 Calculate the length of PS correct to
2 decimal places.
(3)
4.2 Calculate the length of PT if PT = 3 % AB.
(1)
4.3 What kind of quadrilateral is APBT ?
(2)
(2)
A
5.
(1)
The circumference of a circle is 52 cm. Calculate
the area of the circle correct to 2 decimal places.
(4)
12 cm
Two circles have the
same centre.
The smaller circle has a
radius of 20 cm.
The larger circle has a
radius of 30 cm.
(2)
B
3 cm
(2)
8.2 The area of the shaded section.
(3)
D
R
T
C
r
In parallelogram ABCD, AB = 5 cm, AD = 12 cm,
BT = 3 cm and AT ⊥ BC.
60 m
3.1 How many times must he run around the field
in order to run a distance of at least 4 km?
30
8.1 the circumference of the smaller circle.
5 cm
100 m
20
Calculate:
9.
Peter runs around the field with the following
dimensions:
Copyright © The Answer
7.
4.4 Calculate the area of the figure correct to
2 decimal places.
2
3.2 Calculate the area of this field, correct to
two decimal places.
Write down the value of k if the area of the
enlarged rectangle = k % the area of the
original rectangle.
B
1.2 Calculate the perimeter of the
semi–circle ACB.
Use π = 3,14.
T
S
2m
B
(4)
0,24 cm
2
0,144 cm
2
1,44 cm
2
0,0144 cm
The length of a rectangle is doubled.
2m
P
1.1 Calculate the area of the circle.
A
B
C
D
6.
5.1 Calculate the length of AT.
(3)
5.2 Determine the area of the parallelogram.
(3)
9.1 Show that the
area of the
shaded ring
is equal to
2
2
π(R – r ).
(2)
9.2 Determine the
area of the
shaded ring in
terms of π if
R = 14 cm and
r = 8 cm.
(2)
5.3 Calculate
(4)
(4)
5.3.1 the perimeter of trapezium ADCT.
(1)
5.3.2 the area of trapezium ADCT.
(3)
Q11
For further practice in this topic –
see The Answer Series
Gr 9 Mathematics 2 in 1 on p. 1.26
QUADRILATERALS
The arrows indicate
various ‘ROUTES’
from ‘any’
quadrilateral to the
square, the ‘ultimate
quadrilateral’.
Pathways of definitions and properties
A Rectangle
A Parallelogram
A Trapezium
Definition of a rectangle
A parallelogram
with one right angle
Definition of a parallelogram
A quadrilateral with 2 pairs
of opposite sides parallel
Definition of a trapezium
a
f
c
e
parallel
• 2 pairs of opposite sides
The Angles
equal
The Sides
• 1 pair of opposite sides parallel
• all 4 angles equal 90º
The Angles
Definition of a square
The Diagonals . . .
• 2 pairs of opposite
angles equal
d
The Square
sides parallel
• 2 pairs of opposite
sides equal
• 2 pairs of opposite sides
Properties of a trapezium
b
• 2 pairs of opposite
The Sides
A quadrilateral with 1 pair
of opposite sides parallel
Properties of a rectangle
The Sides
Properties of a parallelogram
'Any'
Quadrilateral
Quadrilaterals play a prominent role
right through to Grade 12!
• bisect each other equally
(the diagonals are equal to each other!)
The Diagonals . . .
s
Sum of the ø of
any quadrilateral = 360°
A rectangle with one pair
of adjacent sides equal
OR
• bisect each other
A rhombus with one
angle of 90º
Properties of a square
A square contains ALL the accumulated
A Kite
properties of sides, angles and diagonals!!!
Properties of a kite
A Rhombus
The Sides
• 2 pairs of adjacent
sides equal
The Sides
• all 4 sides
equal
The Angles
• the following pair of angles
will be equal because of
isosceles triangles as a result
of adjacent sides equal
Definition of a kite
A quadrilateral with 2 pairs
of adjacent sides equal
Properties of a rhombus
Definition of a rhombus
A parallelogram with one pair
of adjacent sides equal
The Diagonals . . .
• cut perpendicularly
• the LONG DIAGONAL bisects the
short diagonal and the opposite angles
OR
A kite with 2 pairs of
opposite sides parallel
Q12
QUADRILATERALS
See how the properties
accumulate as you
move from left to right.
i.e. the first quad has
no special properties
and each successive
quadrilateral has all
preceding properties.
The Angles
• 2 pairs of opposite angles
equal
The Diagonals . . .
• cut perpendicularly
• bisect each other
• bisect the opposite angles
Copyright © The Answer
Geometry of Straight lines
MORE STRAIGHT LINE GEOMETRY
Angles that 'alternate'
are on opposite sides of the transversal.
Important Vocabulary
When 2 lines are cut
by another line (a transversal),
two families of angles are formed:
1 2
4 3
1
4
8
5
6
interior 'alternate' angles.
These are
These are
interior
exterior
angles.
angles.
3
5
They are NOT necessarily equal.
the transversal
5 6
These are
pairs of
5 6
8 7
ˆ 3,
ˆ 4ˆ and 5,
ˆ 7,
ˆ 2,
ˆ 6,
ˆ 8ˆ
1,
4 3
4
1
1 2
7
8 7
exterior 'alternate' angles.
3
Not usually
used.
8
These pairs of angles correspond.
1
Each of
these groups
are 'co-' angles
2
These are
pairs of
2
2
5
6
6
4
3
7
8
Note:
They are NOT
necessarily
equal.
7
The FACTS
i.e. they are on the same side of the transversal
4
5
These are
pairs of
co-interior angles.
When 2 PARALLEL lines are cut by a transversal, then
the corresponding angles are equal,
the (interior) alternate angles are equal, and
the co-interior angles are supplementary.
3
6
& conversely:
They are NOT necessarily supplementary.
1
8
Copyright © The Answer
7
5 6
8 7
If the corresponding angles are equal, or if
the (interior) alternate angles are equal, or if
the co-interior angles are supplementary, then the lines are parallel.
2
These are
pairs of
co-exterior angles.
1 2
4 3
Recognise these
angles in
unfamiliar situations.
Not usually
used.
Q13
Gr 9 Maths: Content Area 3 & 4
Geometry & Measurement (2D)
ANSWERS
• Geometry of Straight Lines
• Triangles: Basic facts
• Congruent Δ
• Similar Δ
s
s
• Quadrilaterals
• Polygons
Theorem of Pythagoras
Area and Perimeter of 2D shapes
Compiled by
Anne Eadie & Gretel Lampe
THE ANSWER SERIES
tel: (021) 671 0837
fax: (021) 671 2546
faxtoemail: 088 021 671 2546
www.theanswer.co.za
Solutions: Geometry of Straight lines
GEOMETRY OF STRAIGHT LINES
5.
s
Bˆ 1 (= Bˆ 3 ) = 35º . . . vertically opposite ø
s
ˆ = Bˆ
BCF
1
. . . corresponding ø ; BE || CF
. . . vertically opposite ø
b̂ = â
. . . corresponding ø ; AB || CD
ĉ = 180º – (12º + 58º)
= 110º 1.3
2.
s
B 3
A
35°
. . . alternate ø ; PQ || SRT
s
â d̂ = 180º – 112º . . . co-interior ø supplementary;
PS || QR
= 68º 6.
7.1
F
E
s
A
7.2
T
3.2
360º D̂ + F̂ = 90º 4.2
180º 4.2
360º 4.4
parallel 4.5
equal Copyright © The Answer
S
s
2x + 3x + 4x = 180º
. . . ø on a straight line
s
y (= Qˆ 2 ) = 3x
. . . alternate ø ; PQ || RS
. . . x = 20º in Question 7.1
s
7.3
z (= Qˆ 1 ) = 2x = 40º 8.
(3x – 10º) + (x + 30º) = 180º
. . . corresponding ø ; PQ || RS
43°
B
 = Cˆ 2
2
1
2
3
C
1
65°
D
s
. . . alternate ø ; AB || TC
Bˆ 1 = Cˆ 1
â 4x + 20º = 180º
s
. . . corresponding ø ; AB || TC
Subtract 20º :
â 4x = 160º
Divide by 4:
â x = 40º s
. . . co-interior ø ;
AB || CD
= 65º Bˆ 2 = 180º – Bˆ 1
4.1
4x
z
= 60º = 43º complementary
Q
â 9x = 180º
â x = 20º s
x – 9º + x – 6º + x + 15º = 360º . . . ø about a point
add up to 360º
â 3x = 360º
â x = 120º 3
y
R
. . . adjacent ø on a straight
line add up to 180º
ˆ = 112º
PQR
2
C
1 2
s
â The largest angle = x + 15° = 135° 3.
3x
s
â = 43º = 43º 1.2
2x 1
P
= 35º 1.1
T
7.
s
. . . ø on a straight line
9.
= 115º ˆ = 180º – 76º
PUV
s
. . . ø on a straight line
= 104º
Be sure to study
'Straight Line Geometry' (page Q2)
- vocabulary and facts and 'More Straight Line Geometry' (page Q13)
- vocabulary and facts A1
ˆ = PUV
ˆ
â RVW
â PQ || RS s
. . . corresponding ø equal
Solutions: Geometry of Straight lines
10.
â = 60º . . . vertically opposite angles
b̂ = 35º . . . alternate ø ; || lines
ĉ = 35º . . . base ø of isosceles Δ
NOTES
s
s
d̂ = 180º – ( â + ĉ )
s
. . . sum of the ø of a Δ
= 180º – (60º + 35º)
= 85º exterior ø of a Δ = the sum of
s
ê = â – 35º
...
the interior opposite ø
= 25º f̂ = ( b̂ + ĉ )
s
. . . corresponding ø ; || lines
= 70º or f̂ = ĉ + 35º
= 70º . . . ext ø of Δ = sum of int. opp. ø
s
s
ĝ = ê
. . . alternate ø ; || lines
= 25º s
. . . ø on a
straight line
x + ( x + y) + y = 180º
11.
â 2x + 2y = 180º
i.e.
â 2(x + y) = 180º
Divide by 2:
12.
â x + y = 90º 120º + 110º + x = 2 % 180º
s
. . . 2 pairs of co-interior ø ;
parallel lines
â 230º + x = 360º
Subtract 230º: â x = 130º
120°
A
110° B
x
C
A2
Copyright © The Answer
Solutions: Triangles
5.
TRIANGLES: BASIC FACTS
s
. . . ø opposite equal sides
in an isosceles Δ
B̂ (= Ĉ ) = x
2.
s
6.
. . . ø on a straight line
are supplementary
 = 110º – 50º
...
= 60º 8.2
exterior ø of Δ = sum
s
of interior opposite ø
NB :
. . . AE = AW; ø opposite equal
sides in an isosceles Δ
ˆ ) = 55º
â Eˆ 2 (= W
1
â x (= Eˆ 2 ) = 55º 7.
s
. . . alternate ø ; CS || HN
ĉ = b̂ + 90º
s
. . . the base ø of an
isosceles Δ are equal
ˆ
In ΔABD: â = ABD
1
(180º – 72º)
2
1
= (108º)
2
â â =
Often, in geometry riders,
there are several possible methods.
= 16º + 90º
= 106º s
. . . sum of the ø
of a Δ
9.
= 54º 3.1
Tˆ1 = 25º 3.2
ˆ = P̂ + Tˆ
M
2
1
s
. . . ø opposite equal sides MT
and MP in an isosceles triangle
b̂ = 72º + â
= 126º . . . exterior ø of Δ MPT
ˆ
ĉ = BDC
= 2(25º)
= 50º 4.
4x + 5x = 180º â 9x = 180º
â x = 20º
ˆ = Ê + D̂ OR: Ê = 5x – 3x
EFC
= 2x
â 5x = Ê + 3x
â Ê = 2x
s
the interior ø
= 40º â Answer: A Copyright © The Answer
10.
Ê = 95º – 30º
= 65º = 115º exterior ø of Δ
. . . equals the sum of
A3
exterior ø of Δ = sum
s
of interior opposite ø
. . . exterior ø of ΔBDC =s sum
of interior opposite ø
s
= 31º Â = 180º – Ê
= 27º ...
. . . alternate ø ; AD || BC
s
. . . the base ø of an
isosceles Δ are equal
1
(180º – b̂ )
2
1
= (54º)
2
s
x̂ = 75º – 44º
= 31º ŷ = x̂
ext ø of ΔABD = the sum of
...
s
the interior opposite ø
ĉ =
. . . ø on a straight line
s
. . . ø opposite equal sides
in an isosceles Δ
â b̂ = 44º – 28º
= 16º An interior angle = the exterior ø – the other interior ø
s
s
. . . sum of the ø of a Δ = 180º
b̂ + 28º = 44º
. . . sum of the ø of a Δ
= 110º
â + 44º = 90º
â â = 90º – 44º
= 46º â Answer: B s
ˆ = 180º – 70º
Eˆ 2 + W
1
ˆ
But Eˆ 2 = W
1
NB : See comment in Question 6.
. . . ø of an equilateral Δ all = 60º
â Tˆ2 = 120º . . . exterior ø of Δ = sums
of interior opposite ø
. . . sum of the ø of Δ = 180º
s
Tˆ1 = 60º x = 106º – 44º
= 62º
s
â Â = 180º – 2x 1.1
8.1
. . . exterior ø of ΔDECs = sum of
interior opposite ø
s
. . . co-interior ø are supplementary
because AB || ED
Solutions: Congruent Triangles
CONGRUENT TRIANGLES
(Symbol: ≡)
1.1
ΔDEF 6.
In ΔKNQ and ΔMPQ
(1) NQ = PQ
. . . given
(2) KQ = MQ
. . . given
(3) Nˆ 2 = Nˆ 1 = 90º
ΔSTV . . . SøS
3.1
SøS â Answer: C â ΔMPN ≡ ΔMTN 1
M
7.1
But: BD = EC
. . . given
(3) BD is common
. . . 90º, Hyp, S (RHS)
Note:
Observe the layout of a congruency proof.
5.1
. . . given
(2) BD = CD
. . . given
N
7.2
5.2
â Aˆ 1 = Aˆ 2
. . . given
. . . proved in Question 7.1
s
ø opposite equal sides
ˆ = ADC
ˆ
(2) AEB
...
in isosceles ΔADE
(3) AE = AD
. . . given
. . . SøS
Study the proof
carefully!
ˆ i.e. DA bisects BAC
. . . given
â BC = EF In ΔABE and ΔACD
(1) BE = CD
BC = BF + FC
But: BF = CE
9.2
â Answer: ΔACD In ΔABC and ΔDEF
(1) AB = DE
. . . given
(2) AC = DF
. . . given
(3) BC = EF
. . . proved in Question 9.1
â ΔABC ≡ ΔDEF 9.3
9.4
. . . SSS
B̂ = Ê because they are corresponding angles of the
congruent triangles in Question 9.2 B̂ and Ê are alternate angles
& B̂ = Ê in Question 9.3
Order is important in congruency layout:
. . . SSS
...
9.1
NB: Always
give reasons!
(3) AD is common
â ΔABD h ΔACD Q
& EF = CE + FC
â ΔABE h ΔACD
In ΔABD and ΔACD
(1) AB = AC
. . . RHS
. . . corresponding sides of
congruent triangles
â PN = NT â BE = CD . . . given
â ΔABD h ΔCDB 2
P
BE = BD + DE
& CD = EC + DE
In ΔABD and ΔCDB
(2) AD = CB
. . . given that MN ⊥ PT
1 2
2.
(1) Bˆ 2 = Dˆ 1 = 90º
. . . radii of the circle
(2) MN is common
â ΔKNQ ≡ ΔMPQ . . . SøS
. . . SøS
We need to prove congruent triangles!
In ΔMPN and ΔMTN
(1) MP = MT
K
(3) Q̂ is common
NB: The letters must be in the correct order so
that equal sides and angles correspond.
4.
8.1
NB: Always
give reasons!
s
*corresponding ø of
s
congruent Δ in Question 5.1
* Nothing to do with
s
corresponding ø on || lines
• The letters must be in the same order in both triangles,
corresponding to the equal sides and angles of the
triangles;
• In the facts (1), (2) and (3), the sides and angles of the
first triangle must come first.
A4
â AB || ED . . . converse fact
See the notes on Congruency
and Similarity on page A5
Copyright © The Answer
Solutions: Congruent Triangles
10.1 In ΔABD and ΔACD
(1) AB = AC
. . . given
(2) BD = CD
. . . given
≡ means 'is congruent to'
11.
. . . given
P
1
P
B
C
Q
R
 = P̂ , B̂ = Q̂ and Ĉ = R̂
and AB = PQ, AC = PR and BC = QR
S
Note the order of the lettering
Two triangles are congruent if they have
1
1
R
• 3 sides the same length
. . . SSS
• 2 sides & an included angle equal
. . . SøS
• a right angle, hypotenuse & a side equal . . . RHS
s
. . . corresponding ø of congruent
triangles in Question 10.2
But Eˆ1 + Eˆ 2 = 180º
i.e. ΔABC ≡ ΔPQR means that
T
. . . SøS
Q
10.3 Eˆ1 = Eˆ 2
1
A
have the same shape and size.
All 3 angles and all 3 sides are equal.
The sketch with all equal angles filled in.
(There are no equal sides)
(2) Aˆ 1 = Aˆ 2
. . . corresponding angles in congruent
triangles in Question 10.1
(3) AE is common
â ΔABE ≡ ΔACE Congruent Triangles . . .
||| means 'is similar to'
(i.e. same SHAPE but not necessarily same SIZE
. . . SSS
10.2 In ΔABE and ΔACE
(1) AB = AC
of triangles
whereas:
(3) AD is common
â ΔABD ≡ ΔACD Congruency (≡) and Similarity (|||)
(i.e. same SHAPE and SIZE)
• 2 angles and a side equal
Answer: D . . . øøS
. . . angles on a straight line
â Eˆ1 = Eˆ 2 = 90º Study the (easy) logic very carefully!
10.4 AE ⊥ BC [i.e. AE is perpendicular to BC ]
Proof:
In ΔPTS and ΔRTQ
s
Similar Triangles . . .
s
have the same shape, but not necessarily the same size.
(1) Pˆ1 = Rˆ 1
. . . alternate ø ; PS æ QR
(2) Sˆ 1 = Qˆ 1
. . . alternate ø ; PS æ QR
ˆ = RTQ
ˆ
& (3) PTS
. . . vertically opposite ø
â ΔPTS ||| ΔRTQ s
. . . øøø
A
All 3 angles are equal.
i.e. ΔABC ||| ΔPQR means that
 = P̂ , B̂ = Q̂ and Ĉ = R̂
B
P
C
Q
R
The sides are not necessarily equal, but are proportional:
AB
AC
BC
=
=
PQ
PR
QR
Note the order of the lettering
Copyright © The Answer
A5
Solutions: Similar Triangles
3.1
SIMILAR Δ
(Symbol: |||)
'S
In ΔPQR and ΔSTR
(1) P̂ = Ŝ
. . . alternate ø ; PQ || ST
(2) Q̂ = T̂
. . . alternate ø ; PQ || ST
s
. . . vertically opposite ø
â ΔPQR ||| ΔSTR 3.2
Multiply by 16 :
. . . øøø
â PN =
( )
PQ
PR
QR
=
=
. . . proportional sides of
similar triangles
ST
SR
TR
PQ
10
â
=
Choose the sides for which
3
6
â
( )
PN
QP
QN
=
=
. . . proportional sides of
MN
LM
LN
similar triangles
PN
3
=
â
Choose the sides for which
16
8
â
the lengths have been given.
s
ˆ = SRT
ˆ
& (3) PRQ
Be sure to read the
notes on Congruency
and Similarity on pg. A5
4.2
s
2
3 × 16
8
â PN = 6 cm the lengths have been given.
1.1
Multiply by 3:
DF
DE
EF
then
=
=
KL
LM
KM
10 × 3
â PQ =
6
(1) Â is common
â PQ = 5 cm (2) ABˆ D = ACˆ E
â
14
7
â x
=
=
x
12
24 . . . proportional sides of
similar triangles
( )
. . . 14 = 2 = ?
7
1
12
5.1
(3) 3 L:
Note the ORDER of the letters in similar Δ :
ΔDEF ||| ΔKLM ¥ D̂ = K̂ , Ê = L̂ and F̂ = M̂
and this determines the proportional sides
4.1
(1) N̂ is common
s
(2) Pˆ1 = M̂
. . . corresponding ø ; QP || LM
& (3) Qˆ 1 = L̂
. . . corresponding ø ; QP || LM
s
If ΔABC ||| ΔEDF,
â
ED
DF
AB
15
=
6
10
( EF )
Multiply by 6:
â AB =
. . . øøø
. . . proportional sides of
similar triangles
Choose the sides
for which you have
the lengths!
8 cm
Q
1
A
Dˆ 1 = Eˆ1
( )
â
BD
AD
AB
=
=
CE
AE
AC
â
BD
9
=
21
7
2
F
21 cm
1 2
D
C
s
. . . sum of the ø of a Δ
. . . øøø
. . . proportional sides of
similar triangles
Choose the sides
with known lengths.
â BD =
9 × 21
7
3
â BD = 27 cm 3 cm
2
1
Multiply by 21 :
M
L
5.2
7 cm
E
9 cm
â ΔABD ||| ΔACE In ΔQPN and ΔLMN
â ΔQPN ||| ΔLMN
then AB = BC = AC
In ΔABD and ΔACE
rd
s
2.
B
If ΔDEF ||| ΔKLM, i.e. ΔDEF is similar to ΔKLM,
2
1
P
16 cm
15 × 6
10
â AB = 9 cm N
A6
Copyright © The Answer
Solutions: Quadrilaterals
3.2
QUADRILATERALS
1.
s
â 3x + 30º = 180º
Subtract 30º:
Divide by 3:
But Dˆ 2 = Tˆ1
â 3x = 150º
â x = 50º
â Dˆ 2 = 75º 3.3
â Â = 50º + 50º = 100º
â B̂ = 180º – 100º
s
. . . co-interior ø ; AC || BD
Ĉ = 2(50º) – 20º = 80º
s
â B̂ = 80º . . . opposite ø of a ||
2.1
= 12º 2.3
3.1
Ĝ = 156º the diagonals of a rhombus
...
s
bisect the ø of the rhombus
s
= AD
Cˆ 1 = 15º
s
. . . sum of the ø of Δ DTC
1
2
1
4
T
A 2
Copyright © The Answer
60°
s
Dˆ 1 = Bˆ 2
. . . alternate ø ; AD || BC in parallelogram
Bˆ 1 = Dˆ 2
. . . alternate ø ; AB || DC in parallelogram
BD = BD
. . . common side
s
1
C
P
60°
Note: We have just proved, using congruency,
that both pairs of opposite sides of a
parallelogram are equal in length.
60° R
60°
3
60° S
60°
W
60°
. . . øøS
corresponding sides
s
. . . of congruent Δ
& AB = DC 2
30°
In ΔADB and ΔCBD:
â ΔADB ≡ ΔCBD
x
& Similarly: BT = AB = BC
1
5.
60°
. . . sides of square
ø of isosceles Δ; sum
s
of the ø of a Δ = 180º
...
= 15º Q
. . . sides of equilateral Δ
s
. . . angles opposite equal sides in ΔQRS
â AD = BC 4.
D
. . . ø of square
= 1 (180º – 150º)
2
= 1 (30º)
2
s
= 180º – 30º
= 150º opposite ø of a rhombus
m
(or || ) are equal
s
. . . sides of equilateral ΔPQR
. . . sides of square
. . . sides of equilateral ΔRTS
ˆ
â x = RSQ
ˆ = 90º
. . . Dˆ 2 = 75º and ADC
â Tˆ4 = 180º – 2(15º)
ΔATD and ΔBTC . . . AT = AB
QR = PR
= RT
= RS
. . . half of 150º
= 15º
& Similarly:
...
s
Dˆ 1 = 90º – 75º
s
. . . ø of equilateral Δ
of ø about
ˆ = 360º – (60º + 90º + 60º) . . sum
.
â QRS
a point = 360º
= 150º
. . . ø opposite equal sides in Δ ATD
Similarly: Tˆ3 = 75º
and Tˆ2 = 60º . . . ø of equilateral Δ ATB
OR:
s
= 24º 2.2
are equal
. . . co-interior ø ;
DE || GF in rhombus
ˆ = 180º – 156º
EFG
Fˆ2 = 1 (24º)
2
m
ˆ = 90º
& PRT
sum of ø about
â Tˆ4 = 360º – (75º + 60º + 75º) . . .
a point = 360º
= 360º – 210º
= 150º = 80º OR:
ˆ = TRS
ˆ = 60º
QRP
. . . ø of equilateral Δ
ˆ = 90º in square
â Aˆ 1 = 30º . . . DAB
s
â Dˆ 2 + Tˆ1 = 180º – 30º . . . sum of the ø of a Δ
= 150º
. . . co-interior ø ;
AB || CD
(x + 50º) + (2x – 20º) = 180º
Aˆ 2 = 60º
60°
T
B
A7
6.
D rectangle If one angle equals 90º, then, because of co-interior
angles and parallel lines, so do the others equal 90º.
â We only need 'at least one angle equal to 90º.'
Solutions: Quadrilaterals
7.1
Something new to experience!
( Bˆ 1 + Bˆ 2 ) + ( Cˆ 1 + Cˆ 2 ) = 180º
A
T 2
s
co-interior ø ;
. . . AB || DC in
parallelogram
3
P
D
Let Bˆ 1 = Bˆ 2 = x and Cˆ 1 = Cˆ 2 = y . . .
...
1
given that the angles were bisected
x 2
1
x
B
8.
Answer: C
C
A very useful
technique in
geometry!
1)
AE = BE
. . . given
2)
EB = EC
. . . given
3)
ˆ = DEC
ˆ
AEB
9.
Divide by 2: â x + y = 90º
ab
7.2
g
= 540º
. . . angles on a straight line
â Each ø =
In ΔBCT and ΔTCB:
540°
= 108º
5
â Answer: D 1)
Cˆ 1 = Cˆ 2 (= y)
2)
ˆ
(= 90º)
Tˆ2 = TPC
3)
â Bˆ 1 = Tˆ1
rd
. . . 3 ø of Δ
â ΔBCT ||| ΔTCP . . . øøø
BT
BC ⎛ CT ⎞
=
⎜=
⎟
TP
TC ⎝ CP ⎠
BT
2TC
=
â
4
TC
...
Multiply by 4:
â BT = 2 % 4
= 8 cm d
(a + i + h) + (b + g + f) + (c + d + e)
= 3 % 180º
ΔTCP â
. . SøS
e
f
h
s
. . . sum of the ø of a Δ
ˆ = 180º – 90º = 90º
TPC
7.3
c
i
Bˆ 1 + Cˆ 1 = 90º
â Tˆ2 = 90º
s
. . . vertically opposite ø
â ΔAEB ≡ ΔDEC
i.e. 2(x + y) = 180º
â In ΔBTC:
NOTES
In ΔAEB and ΔDEC:
y
y2 1
â 2x + 2y = 180º
ΔAEB ≡ ΔDEC proportional sides of
similar triangles
Choose the sides whose
lengths you have been given.
10.
s
The sum of the ø of the hexagon
= 4 % 180º
= 720º
â Each ø =
720°
= 120º
6
â Answer: B A8
Copyright © The Answer
Solutions: Theorem of Pythagoras
THEOREM OF PYTHAGORAS
1.
AC = 13 cm
. . . 5 : 12 : 13 Pythagoras 'trip'
2
OR: In ΔABC:
2
2
AC = AB + BC
2
2
= 5 + 12
= 25 + 144
= 169
...
3.1
*
Thm of Pythag.;
B̂ = 90º
So:
. . . OR: TU × 12 = 30
2
â 6 % TU = 30
6
Divide by 6:
3.2
TW = 13 cm
â TU = 5 cm . . . Pythag 'trip' 5 : 12 : 13
â The perimeter of ΔTUW = 5 cm + 12 cm + 13 cm
= 30 cm 2
The length of the ladder = 13 m OR: (length of the ladder)
2
2
= 5 + 12 , etc.
2
; 5 : 12 : 13 ; 8 : 15 : 17
12 m
Answer: B 6 cm In ΔADC: DC = 8 cm
â AD = 6 cm
and even multiples:
. . . opposite sides of rectangle
. . . Pythag 'trip':
3 : 4 : 5 = 6 : 8 : 10
6 : 8 : 10
2.1 In ΔABC: x = 10 cm ...
Pythag 'trip':
3 : 4 : 5 = 6 : 8 : 10
2.2 In ΔABD: BD = 15 cm
. . . Pythag 'trip': 8 : 15 : 17
â y = 15 cm – 6 cm
= 9 cm 2
2
2
OR: 2.1 x = 8 + 6 , etc.
2
2
5 : 12 : 13
5m
5.
the TRIP(LET)S :
. . . Pythag 'trip':
2
3 +4
= 9 + 16 = 25 = 5
2
2
2
5 + 1 2 = 25 + 144 = 169 = 1 3
2
2
2
8 + 1 5 = 64 + 225 = 289 = 1 7
3:4:5
h×b
= area of a Δ
2
â TU = 5 cm When applying the Theorem of Pythagoras, there
are some well-known 'trips' which are useful to
know and use instead of long calculations.
2
Multiply by 2:
Divide by 12:
4.1
e.g.
...
â TU % 12 = 60
â AC = 13 cm
* Note:
TU × 12
= 30
2
2
2.2 BD = 17 – 8 , etc.
. . . Theorem of Pythag
6.
This sum requires us to apply the converse of the
Theorem of Pythagoras,
i.e. If the square on one side of a triangle equals the
sum of the squares on the other two sides, then
the angle opposite the first side is a right angle.
2
2
AC = 15 = 225
2
2
2
2
& AB + BC = 9 + 12 = 81 + 144 = 225
2
2
2
â AC = AB + BC
â B̂ = 90° . . . the converse of the Theorem of Pythagoras
Copyright © The Answer
A9
NOTES
Solutions: Measurement: 2D
3.2
MEASUREMENT: 2D
1.1
= 6 000 m
2
5.1
area of 2 semi-circles
2
+
+
.900 m
= (100 % 60)m
1
% diameter = 6 cm
Radius, r =
2
2
The area of the field
= area of rectangle +
2
2
.30 m . . . Area of circle = r
2
2
2
. . . Theorem of Pythagoras
= 16
â AT = 4 cm 2
2
2
2
In ΔAPS: PS = 5 – 2
. . . Theorem of Pythagoras
= 25 – 4
= 21
The circumference of the (full) circle = 2r
â The 'circumference' of the semi-circle = r = 3,14 % 6
= 18,84 cm
â PS =
21
l 4,58 m â The perimeter of the shape ACB
= 18,84 cm + 12 cm = 30,84 cm . . . diameter, AB = 12 cm
2
2
. . . Pythag 'trip': 3 : 4 : 5
= 25 – 9
2
l 8 827,43 m 4.1
2.
AT = 4 cm OR: AT = 5 – 3
2
â Area = r = 3,14 % 6
2
= 113,04 cm 1.2
In ΔABT:
2
The area of a square = s = (0,12) = 0,0144 cm
2
â Answer: D Observe carefully:
2
12 ⎞
12
12
144
2
(0,12) = ⎛⎜
= 0,0144
×
=
⎟ =
100 100 10 000
⎝ 100 ⎠
5.2
The area of parallelogram ABCT
= base % height
= BC % AT
= AD % 4 cm
See the shifting of ΔABT
= 12 cm % 4 cm
as shown:
2
= 48 cm
4.2
PT = 3 % AB = 3 % 4 m = 12 m 4.3
A kite 4.4
Method 1: Using the formula
1
The area =
the product of the diagonals
2
1
=
(PT % AB)
2
1
=
(12 % 4)
2
2
= 24 m A
12 cm
D
. . . 2 pairs of adjacent sides equal
5 cm
B
3 cm
T
C
S
Parallelogram ABCD = rectangle ATSD
& Area of rectangle ATSD
= length % breadth
= 12 cm % 4 cm
2
= 48 cm
Method 2: Without the formula
The area = ΔPAT + ΔPBT
3.1
s
The perimeter of the field
= 2 % 100 m + 2 % semi-circles
= 200 m + 2 % 3,14 % 30 m
= 388,4 m
. . . Circumference of
circle = 2r
= 2 % ΔPAT . . . because the 2 Δ are congruent
⎛1
⎞
= 2 × ⎜ PT . AS⎟
2
⎝
⎠
= 12 m % 2 m
2
= 24 m 4 000 m
= 10,298 . . .
388,4 m
. . . for at least 4 km!
â Number of laps =
â 11 laps A10
5.3.1 DC = AB = 5 cm
s
m
. . . opposite ø of ||
TC = BC – BT = 12 cm – 3 cm = 9 cm
â The perimeter of trapezium ADCT
= AD + DC + TC + AT
= 12 cm + 5 cm + 9 cm + 4 cm
= 30 cm Copyright © The Answer
Solutions: Measurement: 2D
7.
5.3.2 Method 1: Using the formula
The circumference of a circle, 2r = 52 cm
52
26
=
2π
π
â r =
The area of trapezium ADCT
1
(sum of the || sides) % the distance between them
=
2
1
= (AD + TC) % AT
2
1
= (12 + 9) % 4
2
2
= 42 cm â The area of the circle
2
26
2
= r = % ⎛⎜ ⎞⎟
⎝ π⎠
= %
=
262
π
262
π2
2
l 215,18 cm Method 2: Without the formula
. . . correct to 2 decimal places
The area of trapezium ADCT
= Area of ΔATC + Area of ΔADC
1
1
= (9 % 4) + (12 % 4) *
2
2
= 18 + 24
* 1 . ( 9 + 12) . 4, like the formula above!
2
2
= 42 cm A
8.1
The circumference of a circle = 2r
â The circumference of the smaller circle
= 2 % % 20
l 125,66 cm 8.2
The area of the shaded region
D
12 cm
= the area of the full circle – the area of the inner circle
2
2
= 30 – 20
2
l 1 570,80 cm 4 cm
T
9 cm
C
9.1
6.
= the area of the full circle – the area of the inner ring
2
2
= R – r
2
2
= (R – r ) Let the area of the original rectangle = ℓ % b
If the length is doubled, then the area of the enlarged
rectangle = 2ℓ % b
= 2(ℓb)
â k = 2 b
ℓ
Copyright © The Answer
The area of the shaded ring
9.2
2
2
Area of the shaded ring = (14 – 8 )
2
= 132 cm Note: Give the answer 'in terms of '
A11
NOTES