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Gr 9 Maths: Content Area 3 & 4
Geometry & Measurement (2D)
QUESTIONS
• Geometry of Straight Lines
• Triangles: Basic facts
• Congruent Δs
• Similar Δs
• Quadrilaterals
• Polygons
Theorem of Pythagoras Area and Perimeter of 2D shapes
Mostly past ANA exam content
All questions have been
graded to facilitate
concept development.
GOOD LUCK!
Compiled by
Anne Eadie & Gretel Lampe
THE ANSWER SERIES
tel: (021) 671 0837
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Questions: Geometry of Straight lines
Copyright © The Answer Q1
GEOMETRY OF STRAIGHT LINES
( Solutions on page A1)
1. Calculate the sizes of the angles marked a to d.
Give reasons for your answers. 1.1
(3)
1.2
(2)
1.3
(3)
2. Calculate the size
of the largest angle.
Show all your steps
with reasons. (4)
3. Complete the following:
3.1 Angles which add up to 90º are called
. . . . . . . angles. (1)
3.2 Angles around a point add up to . . . . . . . (1)
4. Complete each of the following statements:
4.1 ˆD and ˆF are complementary angles if
____________________________________ . (1)
4.2 The sum of the interior angles of a triangle is
equal to _____________________________ . (1)
4.3 The sum of the exterior angles of any polygon
is equal to ___________________________ . (1)
4.4 A trapezium is a quadrilateral with one pair
of ___________________ sides. (1)
4.5 The diagonals of a rectangle are _________
in length. (1)
5. In the figure, ˆ3
B = 35º and BE || CF.
Determine the size of ˆ1
B and ˆBCF.
Statement Reason
ˆ
1B =
ˆBCF =
(3)
6.
In the figure above, AB || TC, ˆ1
C = 65º and ˆ2
C = 43º.
Calculate the size of ˆA , ˆ1
B and ˆ2
B .
Statement Reason
(4)
58°c
12°
d
T S R
112°
P Q
A
E
C B 3
F
21
A
D CB
3
T
2
1 12
A
D C
B 43°
a
b
x – 6°
x – 9° x + 15°
Refer to page Q13 for details
on parallel lines & angles.
Questions: Geometry of Straight lines
Q2 Copyright © The Answer
7. Give reasons for each of your statements in the
questions below.
In the figure PQ || RS, ˆ1
Q , ˆ2
Q and ˆ3
Q
are equal to 2x, 3x and 4x respectively.
ˆR = y and ˆS = z.
7.1 Calculate the value of x. (3) 7.2 Calculate the value of y. (3) 7.3 Calculate the value of z. (3)
8. Calculate, with reasons, the value of x.
(4)
9. State, giving reasons,
whether PQ || RS.
(4)
10. Find the size of angles a to g (in that order) ,
giving reasons.
(7)
11. In the sketch, AB is a straight line.
Determine the value of x + y.
(4)
12. Calculate, with reasons, the value of x.
(4)
Hint: Draw a third line, through B,
parallel to the given parallel lines.
T
P Q
R Sy
1
2 3
z
P R
T W
Q S
76°
V U 104°
g
b
c d
a
ef
35° 60°
A B
x + yyx
A
C
B
120°
110°
x
For further practice in this topic –
see The Answer Series
Gr 9 Mathematics 2 in 1 on p. 1.32
A
DC
B3x – 10°
x + 30°
STRAIGHT LINE GEOMETRY
Important Vocabulary
An acute angle is one that lies between 0º and 90º.
An obtuse angle is one that lies between 90º and 180º.
A reflex angle is one that lies between 180º and 360º.
A right angle = 90º
A straight angle = 180º
A revolution = 360º
When the sum of 2 angles = 90º, we say the angles are
complementary. When the sum of 2 angles = 180º, we say the angles are
supplementary.
When 2 lines intersect,
4 angles are formed:
ˆ ˆ ˆ ˆ1, 2, 3, 4
Adjacent angles have a common vertex and a common
arm, e.g. ˆˆ1 and 2, ˆ ˆ2 and 3, ˆ ˆ3 and 4 or ˆ ˆ1 and 4.
Vertically opposite angles lie opposite each other,
e.g. ˆˆ1 and 3 or ˆ ˆ2 and 4.
The FACTS
When 2 lines intersect:
� adjacent angles are supplementary
� vertically opposite angles are equal.
See the end of the questions
for more on straight lines.
1 2
3 4
Questions: Triangles
Copyright © The Answer Q3
TRIANGLES: BASIC FACTS
( Solutions on page A3)
Reasons must be provided for all Geometry statements.
1. In the figure below, ΔANT is an equilateral triangle.
Calculate the size of ˆ1T and ˆ
2T .
(4)
2. In the figure below, CS || HN, ˆEAW = 70º;
AE = AW and ˆCAE = x.
Determine the value of x.
(3)
3. In ΔPRT alongside,
M is the midpoint of PR
and MR = MT.
If ˆP = 25º, calculate
with reasons:
3.1 The size of ˆ1T (1)
3.2 The size of ˆ2
M (1)
4. In ΔEDF, DF is produced to C.
The size of ˆE is . . . ?
A 40º B 60º
C 140º D 20º (1) [10]
5.
In ΔABC, AB = AC and ˆC = x.
Determine the size of ˆA in terms of x. (3)
6.
In the figure above, ˆB = 50º and ˆACD = 110º.
The size of ˆA is . . . . . . A 50º B 60º
C 110º D 160º
7. Using the figure below, calculate the size of the
angles a, b and c (in this order). AD = BD = BC;
ˆADB = 72º
(6)
8. Determine the values of x, a, b and c in the figures
below.
8.1
(2)
8.2
(6)
A
P N T
2 1
D
E
F C
3x 4x 5x
1
1
2
2
P
M
R T
B
CA
B C
A
D50° 110°
b
c
a
28°44°
106°x
44°
A
D
B
a
72° b
c
C
CA
S
W E H
1 2
70°
x
2 1
N
Questions: Triangles
Q4 Copyright © The Answer
9. Calculate the values of x and y if
ˆ
2B = x, ˆ
2D = y, ˆ
1D = 44º, ˆ
1C = 75º and AD || BC.
(3)
10.
In the above figure AB || ED, ˆACD = 95º
and ˆD = 30º.
Determine the size of ˆE and ˆA . (3)
CLASSIFICATION OF TRIANGLES . . .
Triangles are classified according to their sides or
their angles (or both).
• Sides
• Angles
• Sides and Angles
INTERIOR AND EXTERIOR ANGLES . . .
An exterior angle is formed between one side of
a triangle and the produced (extension) of another.
4 BASIC FACTS
• FACT 1
The sum of the
interior angles
of a triangle = 180°
• FACT 2:
The exterior angle
of a triangle equals
the sum of the
interior opposite angles.
• FACT 3
In an isosceles triangle,
the base angles
are equal.
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°.
44° 75° y
x
A B
D C E
1
1
2
22
1
For further practice in this topic –
see The Answer Series
Gr 9 Mathematics 2 in 1 on p. 1.24
A B
D
C
E
95°
30°
1
60°
60° 60°
TRIANGLES: Study the following very carefully
This is an
isosceles,
right-angled
triangle
This is an
isosceles,
acute-angled
triangle
This is a
scalene,
obtuse-angled
triangle.
3 acute angles 1 obtuse angle1 right angle (90°)
equilateral Δ isosceles Δ scalene Δ
3 sides equal 2 sides equal no sides equal
acute-angled Δ right-angled Δ obtuse-angled Δ
ˆˆ ˆA + B + C = 180°
A
B C
If AB = AC,
then ˆ1 = ˆ2
Converse:
If ˆ1 = ˆ2,
then AB = AC
1 2
A
B C
2
3 1
ˆ ˆˆ1 = 2 + 3
ˆ1 , ˆ2 and ˆ3
are interior
angles of the
triangle x is an exterior ø
y is not an exterior ø
1
2 3 x y
Questions: Congruent Triangles
Copyright © The Answer Q5
CONGRUENT ΔS
( Solutions on page A4)
1.
Which triangle is congruent to ΔPQR?
Statement Reason
(2)
2. State which triangle is congruent to ΔABC.
(2)
3. Why is ΔABC ≡ ΔDCB?
A S, S, S B 90º , Hyp, S (RHS)
C S, ø , S D ø , ø , S
4. In the figure below ˆ1
D = ˆ2
B = 90º and AD = BC.
Prove that ΔABD ≡ ΔCDB.
5. In the figure below, AB = AC and BD = CD.
5.1 Prove that ΔABD ≡ ΔACD. (4)
5.2 Prove that DA bisects ˆBAC (2)
6. 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.
Prove with reasons that ΔKNQ ≡ ΔMPQ. (4)
7. ΔABC, D and E are points on BC such that BD = EC
and AD = AE.
7.1 Why is BE = CD? (1)
7.2 Which triangle is congruent to ΔABE? (1)
B
P A
C Q R
F E
D
A D
B C
A D
B C
1
1 2
2
D
1
1
2
2
B C
A
DB C
A
E
M N
K
Q
1
1 2
2
P
See the notes on Congruency
and Similarity on page A5
V
S
T
R
P
QB
C
A
Questions: Congruent Triangles
Q6 Copyright © The Answer
8. In the given figure, P and T
are points on a circle with
centre M. N is a point on a
chord PT such that
MN ⊥ PT.
Prove that PN = NT.
Statement Reason
(8)
9.
In the above diagram, AC = DF, AB = DE and BF = CE. 9.1 Prove that BC = EF.
Statement Reason
(2)
9.2 Prove that ΔABC ≡ ΔDEF.
Statement Reason
(5)
9.3 Why is ˆ ˆB = E ?
Statement Reason
ˆ ˆB = E
(1)
9.4 Use your answer in Question 9.3 to derive a
further relationship between AB and ED.
Note: It has (already) been given that AB = ED.
Statement Reason
(2)
10. In the figure
alongside AB = AC and BD = CD
10.1 Prove that ΔABD ≡ ΔACD. (4)
10.2 Prove that ΔABE ≡ ΔACE. (4)
10.3 Prove that ˆ1
E = ˆ2
E = 90º. (3)
10.4 Hence, state the relationship
between AE and BC. (1)
11. In the figure below, PS || QR. Which ONE of the
following statements is true for this figure?
A ΔPTS ≡ ΔPQT
B ΔPTS ≡ ΔRTQ
C ΔPTS ||| ΔSRT
D ΔPTS ||| ΔRTQ (1)
For further practice in this topic –
see The Answer Series
Gr 9 Mathematics 2 in 1 on p. 1.28
M
1
1
2
2
P TN
A
1
1
E
D
C
F
B
A
C
D
B
E
1
21
1
1
1 2
2
P
T
Q
S
R
Questions: Similar Triangles
Copyright © The Answer Q7
SIMILAR ΔS
( Solutions on page A6)
1. Examine ΔDEF and ΔKLM.
Complete the following calculations if ΔDEF ||| ΔKLM.
DE
KL =
EF
LM =
DF (proportional sides of similar triangles)
� 14
7 =
x
� x = ________ (3)
2. Calculate the length of AB if ΔABC ||| ΔEDF:
(4)
3. In ΔPQR and ΔSTR in the
figure alongside, PQ || ST,
PR = 10 cm, ST = 3 cm and
SR = 6 cm.
3.1 Prove that
ΔPQR ||| ΔSTR (4) 3.2 Calculate the length of PQ. (3)
4. 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.
4.1 Complete the following (give reasons for
the statements):
Prove with reasons that ΔQPN ||| ΔLMN.
In ΔQPN and ΔLMN
1. ˆN ……………………………
2. ˆ
1P = ……………. ……………………………
3. ˆ
1Q = ……………. ……………………………
∴ ΔQPN ||| Δ …. ………………………….. (4)
4.2 Hence, calculate the length of PN. (3)
5.
In the figure,
ˆB = ˆC , AD = 9 cm, AE = 7 cm and CE = 21 cm.
5.1 Prove that ΔABD ||| ΔACE.
Statement Reason
(6)
5.2 Calculate the length of BD.
Statement Reason
(5)
For further practice in this topic –
see The Answer Series
Gr 9 Mathematics 2 in 1 on p. 1.28
7 cm
10 cm
12 cm
K
L M E
D
F20 cm
14 cm x cm
E
D F
6 cm
10 cm
4 cm
A
B C15 cm
A
1
B
E
C D
1
2
2
F
L
1Q
M
P
N
1
2 2
P
QR
S
T
10
36
See the notes on Congruency
and Similarity on page A5
Questions: Quadrilaterals
Q8 Copyright © The Answer
QUADRILATERALS
( Solutions on page A7)
1. ABCD is a parallelogram. Calculate the size of ˆB .
(4)
2. In the figure below, DEFG is a rhombus and ˆE = 156°.
Calculate the size of :
2.1 ˆEFG
2.2 ˆ
2F
2.3 ˆG
Statement Reason
2.1 ˆEFG = (2)
2.2 ˆ
2F = (2)
2.3 ˆG = (2)
3. In the figure below, ABCD is a square and ATB is an
equilateral triangle.
3.1 Name two isosceles triangles. (2)
3.2 Calculate the size of ˆ2
D . (3)
3.3 Calculate the size of ˆ4
T . (2)
4. PRTW is a square. ΔPQR and ΔRTS are equilateral.
Calculate x ˆ(RQS)
(7)
5.
Look at parallelogram ABCD above and complete
the table.
Statement Reason
In ΔADB and ΔCBD
ˆ
1D = ______ Alternate ø's and AD || BC
ˆ
1B = ______ Alternate ø's and AB || DC
BD = BD Common side
â ΔADB ≡ Δ______ ____________
â AD = ______ and
AB = ______
Corresponding sides of
congruent Δs
(4)
6. A parallelogram with at least one angle equal to 90°
is called a __________ A kite.
B rhombus.
C trapezium.
D rectangle. (1)
A
C
B
D
x + 50°
2x – 20°
E
G F
D 156° 1
1
2
2
C
A B
D
T
1
3
2
2
1
1
1
2
4
Q
W T
S
RP
x
A
C B
D
1
1 2
2
NB: Study 'Quadrilaterals'
on page Q12 very carefully.
Questions: Quadrilaterals
Copyright © The Answer Q9
7.
The bisectors of ˆB and ˆC 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
ˆTPD = 90º.
7.1 Prove that ˆ2T = 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. (3)
8. In the given quadrilateral AE = ED and BE = EC,
therefore:
A ΔAEB ||| ΔCED
B ΔAED ||| ΔBEC
C ΔAEB ≡ ΔDEC
D ΔAED ≡ ΔBEC (2)
POLYGONS
9. What is the size of each angle in a regular
pentagon? A 90°
B 120°
C 100°
D 108° (1)
10. What is the size of each angle in a regular hexagon? A 90°
B 120°
C 100°
D 108° (1)
NOTES
For further practice in this topic –
see The Answer Series
Gr 9 Mathematics 2 in 1 on p. 1.26
A
C
B
D
1
1
2
2
1
2
3
T
P
A D
B C
E
Questions: Theorem of Pythagoras
Q10 Copyright © The Answer
THEOREM OF PYTHAGORAS
( Solutions on page A9)
1. In ΔABC, AB ⊥ BC. Determine the length of
AC if AB = 5 cm and
BC = 12 cm. (4)
2.1
2.1 Calculate x. (3)
2.2 Calculate y. (3) Give reasons.
3. The area of
ΔTUW = 30 cm2
and UW = 12 cm.
Calculate:
3.1 TU (2)
3.2 the perimeter of ΔTUW (3)
4. 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)
5. In rectangle ABCD, AB = 8 cm and diagonal AC = 10 cm. Calculate the length of AD.
A 2 cm
B 6 cm
C 12,8 cm
D 14 cm (2)
6. In ΔABC: AB = 9 cm, BC = 12 cm and AC = 15 cm.
Show that ˆB = 90°.
A
B C12 cm
5 cm
10 cm
8 cm A B
CD
12 cm
T
W U
For further practice in this topic –
see The Answer Series
Gr 9 Mathematics 2 in 1 on p. 1.33
A
CB D
17 cm 8 cm
6 cm y
x
A Right-angled triangle
A right-angled triangle has
one angle of 90º. Here, ˆB = 90°.
The side opposite the right angle (90°)
is called the hypotenuse.
Here, AC is the hypotenuse.
The Theorem of Pythagoras
This theorem states:
In a right-angled triangle . . .
the square of the hypotenuse equals
the sum of the squares on the other two sides.
i.e. In ΔABC, ˆB = 90°
So: AC2 = AB
2 + BC
2
The converse theorem states the reverse:
If in any ΔABC,
AC2 = AB
2 + BC
2,
then ˆB = 90°.
A
CB
hypotenuse
A
CB
PERIMETER AND AREA FORMULAE
Triangle The perimeter
of this triangle
= (a + b + c) units.
The area of
a triangle =
base × height
2
Rectangle
The perimeter of a rectangle
= ℓ + b + ℓ + b
= 2ℓ + 2b
= 2(ℓ + b)
The area of a rectangle
= ℓ % b = ℓb
Square
The perimeter of a square
= 4 % s = 4s
The area of a square
= s % s = s2
Circle
The circumference
of a circle:
= πd = π(2r) = 2πr
The area of a circle:
= πr2
A
CB a
bc
height
base
height
base
height
base
ℓ: length
b: breadth
b
ℓ
s
centre
radius (r) diameter
circumference
See the Quadrilaterals on page Q12
for the areas of all other quadrilaterals.
Questions: Measurement: 2D
Copyright © The Answer Q11
30
20
MEASUREMENT: 2D
( Solutions on page A10)
1. AB, the diameter of the
given circle, is 12 cm.
Use π = 3,14 to answer
the following questions,
correct to two decimal places.
1.1 Calculate the area of the circle. (4) 1.2 Calculate the perimeter of the
semi–circle ACB. (3)
2. If the length of the side of a square is
0,12 cm then the area =
A 0,24 cm2
B 0,144 cm2
C 1,44 cm2
D 0,0144 cm2 (2)
3. Peter runs around the field with the following
dimensions:
3.1 How many times must he run around the field
in order to run a distance of at least 4 km?
Use π = 3,14. (4)
3.2 Calculate the area of this field, correct to
two decimal places. (4)
4. In the figure below, AP = 5 m,
AS = SB = 2 m and PS ⊥ AB.
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)
4.4 Calculate the area of the figure correct to
2 decimal places. (2)
5.
In parallelogram ABCD, AB = 5 cm, AD = 12 cm,
BT = 3 cm and AT ⊥ BC.
5.1 Calculate the length of AT. (3)
5.2 Determine the area of the parallelogram. (3)
5.3 Calculate
5.3.1 the perimeter of trapezium ADCT. (1)
5.3.2 the area of trapezium ADCT. (3)
6. The length of a rectangle is doubled.
Write down the value of k if the area of the
enlarged rectangle = k % the area of the
original rectangle. (1)
7. The circumference of a circle is 52 cm. Calculate
the area of the circle correct to 2 decimal places. (4)
8. Two circles have the
same centre.
The smaller circle has a
radius of 20 cm.
The larger circle has a
radius of 30 cm.
Calculate:
8.1 the circumference of the smaller circle. (2)
8.2 The area of the shaded section. (3)
9. 9.1 Show that the
area of the
shaded ring
is equal to
π(R2 – r
2). (2)
9.2 Determine the
area of the
shaded ring in
terms of π if
R = 14 cm and
r = 8 cm. (2)
R
r
A B
C
12 cm
5 cm
3 cm
A
B T C
D
For further practice in this topic –
see The Answer Series
Gr 9 Mathematics 2 in 1 on p. 1.26
60 m
100 m
A
B
S
2 m5 m
2 m
P T
Q12 Copyright © The Answer
QUADRIL
ATERALS
QUADRILATERALS
The arrows indicatevarious ‘ROUTES’
from ‘any’ quadrilateral to thesquare, the ‘ultimate
quadrilateral’.
See how the propertiesaccumulate as you
move from left to right.
i.e. the first quad has no special properties and each successive quadrilateral has all preceding properties.
Properties of a rhombus
The Sides
• all 4 sides
equal
The Angles
• 2 pairs of opposite angles
equal
The Diagonals . . .
• cut perpendicularly
• bisect each other
• bisect the opposite angles
Properties of a trapezium
The Sides
• 1 pair of opposite sides parallel
Definition of a kite
A quadrilateral with 2 pairs
of adjacent sides equal
Definition of a parallelogram
A quadrilateral with 2 pairs
of opposite sides parallel
Definition of a rectangle
A parallelogram
with one right angle
Definition of a trapezium
A quadrilateral with 1 pair
of opposite sides parallel
Properties of a parallelogram
The Sides
• 2 pairs of opposite sides
parallel
• 2 pairs of opposite sides
equal
The Angles
• 2 pairs of opposite
angles equal
The Diagonals . . .
• bisect each other
Definition of a square
A rectangle with one pair
of adjacent sides equal
OR
A rhombus with one
angle of 90º
Properties of a rectangle
The Sides
• 2 pairs of opposite
sides parallel
• 2 pairs of opposite
sides equal
The Angles
• all 4 angles equal 90º
The Diagonals . . .
• bisect each other equally
(the diagonals are equal to each other!)
Definition of a rhombus
A parallelogram with one pair
of adjacent sides equal
OR
A kite with 2 pairs of
opposite sides parallel
Properties of a kite
The Sides
• 2 pairs of adjacent
sides equal
The Angles
• the following pair of angles
will be equal because of
isosceles triangles as a result
of adjacent sides equal
The Diagonals . . .
• cut perpendicularly
• the LONG DIAGONAL bisects the
short diagonal and the opposite angles
Properties of a square
A square contains ALL the accumulated
properties of sides, angles and diagonals!!!
Pathways of definitions and properties
Quadrilaterals play a prominent role right through to Grade 12!
The Square
A Rectangle
'Any'
Quadrilateral
Sum of the øs
of
any quadrilateral = 360°
A Parallelogram
A Rhombus
A Trapezium
a
b
c
de
f
A Kite
Geometry of Straight lines
Copyright © The Answer Q13
MORE STRAIGHT LINE GEOMETRY
Angles that 'alternate'
are on opposite sides of the transversal.
The FACTS
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.
& conversely:
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.
12
34
5 6
78
the transversal
When 2 lines are cut
by another line (a transversal),
two families of angles are formed:
ˆ ˆˆ ˆ1, 2, 3, 4 and ˆ ˆˆ ˆ5, 6, 7, 8
These are
exterior
angles. 78
12These are
interior
angles.5 6
3 4
i.e. they are on the same side of the transversal
Each of
these groups
are 'co-' angles5
4
8
1
6
3
7
2
These pairs of angles correspond.
5 6
7 8
1 2
3 4
These are
pairs of
co-exterior angles.
7
2
Not usually
used.8
1
Note:
They are NOT
necessarily
equal.
These are
pairs of
exterior 'alternate' angles. Not usually
used.
2
87
1
5
1
6
23
7
4
8
They are NOT necessarily equal.
These are
pairs of
interior 'alternate' angles.6
4 3
5
They are NOT necessarily supplementary.
These are
pairs of
co-interior angles.5
4
6
3
Recognise these
angles in
unfamiliar situations.
Important Vocabulary
Gr 9 Maths: Content Area 3 & 4
Geometry & Measurement (2D)
ANSWERS
• Geometry of Straight Lines
• Triangles: Basic facts
• Congruent Δs
• Similar Δs
• 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
Copyright © The Answer A1
GEOMETRY OF STRAIGHT LINES
1.1 ˆa = 43º � . . . vertically opposite øs
ˆb = ˆa . . . corresponding øs
; AB || CD
= 43º �
1.2 ˆc = 180º – (12º + 58º) . . .
= 110º �
1.3 ˆPQR = 112º . . . alternate øs
; PQ || SRT
â ˆd = 180º – 112º . . .
= 68º �
2. x – 9º + x – 6º + x + 15º = 360º . . .
â 3x = 360º
â x = 120º �
â The largest angle = x + 15° = 135° �
3. complementary 3.2 360º �
4.1 ˆD + ˆF = 90º � 4.2 180º �
4.2 360º � 4.4 parallel �
4.5 equal �
5. ˆ
1B (= ˆ
3B ) = 35º � . . . vertically opposite ø
s
ˆBCF = ˆ
1B . . . corresponding ø
s
; BE || CF
= 35º �
6.
ˆA = ˆ2
C . . . alternate øs
; AB || TC
= 43º �
ˆ
1B = ˆ
1C . . . corresponding ø
s
; AB || TC
= 65º �
ˆ
2B = 180º – ˆ
1B . . . ø
s
on a straight line
= 115º �
7.
7.1 2x + 3x + 4x = 180º . . . øs
on a straight line
â 9x = 180º
â x = 20º �
7.2 y (= ˆ2
Q ) = 3x . . . alternate øs
; PQ || RS
= 60º � . . . x = 20º in Question 7.1
7.3 z (= ˆ1
Q ) = 2x � . . . corresponding øs
; PQ || RS
= 40º �
8. (3x – 10º) + (x + 30º) = 180º . . .
â 4x + 20º = 180º
Subtract 20º : â 4x = 160º
Divide by 4: â x = 40º �
9. ˆPUV = 180º – 76º . . . øs
on a straight line
= 104º
â ˆRVW = ˆPUV
â PQ || RS � . . . corresponding øs
equal
co-interior øs
supplementary;
PS || QR
adjacent øs
on a straight
line add up to 180º
øs
about a point
add up to 360º
co-interior øs
;
AB || CD
A
DCB
3
T
2
1 12
43°
65°
35°
A
E
CB 3
F
21
T
P Q
R Sy
1
2 3
z
2x
3x4x
Be sure to study
'Straight Line Geometry' (page Q2)
- vocabulary and facts -
and 'More Straight Line Geometry' (page Q13)
- vocabulary and facts -
Solutions: Geometry of Straight lines
A2 Copyright © The Answer
10. ˆa = 60º � . . . vertically opposite angles
ˆb = 35º � . . . alternate øs
; || lines
ˆc = 35º � . . . base øs
of isosceles Δ
ˆd = 180º – ( ˆa + ˆc ) . . . sum of the øs
of a Δ
= 180º – (60º + 35º)
= 85º �
ˆe = ˆa – 35º . . .
= 25º �
ˆf = ( ˆb + ˆc ) . . . corresponding øs
; || lines
= 70º �
or ˆf = ˆc + 35º . . . ext ø of Δ = sum of int. opp. øs
= 70º �
ˆg = ˆe . . . alternate øs
; || lines
= 25º �
11. x + (x + y) + y = 180º . . .
â 2x + 2y = 180º
i.e. â 2(x + y) = 180º
Divide by 2: â x + y = 90º �
12. 120º + 110º + x = 2 % 180º . . .
â 230º + x = 360º
Subtract 230º: â x = 130º
exterior ø of a Δ = the sum of
the interior opposite øs
2 pairs of co-interior øs
;
parallel lines
øs
on a
straight line
A
C
B
120°
110°
x
NOTES
Solutions: Triangles
Copyright © The Answer A3
TRIANGLES: BASIC FACTS
1.1 ˆ
1T = 60º � . . . ø
s of an equilateral Δ all = 60º
â ˆ
2T = 120º � . . .
2. ˆ
2E + ˆ
1W = 180º – 70º . . . sum of the ø
s of a Δ
= 110º
But ˆ
2E = ˆ
1W . . .
â ˆ2E (= ˆ
1W ) = 55º
â x (= ˆ2E ) = 55º � . . . alternate ø
s; CS || HN
3.1 ˆ
1T = 25º � . . .
3.2 ˆ
2M = ˆP + ˆ
1T . . . exterior ø of Δ MPT
= 2(25º)
= 50º �
4. 4x + 5x = 180º � . . . øs on a straight line
â 9x = 180º
â x = 20º
ˆEFC = ˆE + ˆD . . .
â 5x = ˆE + 3x
â ˆE = 2x
= 40º �
â Answer: A �
5. ˆB (= ˆC ) = x . . .
â ˆA = 180º – 2x � . . . sum of the øs of Δ = 180º
6. ˆA = 110º – 50º . . .
= 60º �
â Answer: B �
7. In ΔABD: ˆa = ˆABD . . .
â ˆa = 1
2(180º – 72º) . . .
= 1
2(108º)
= 54º �
ˆb = 72º + ˆa . . .
= 126º �
ˆc = ˆBDC . . .
ˆc = 1
2(180º – ˆb )
= 1
2(54º)
= 27º �
8.1 x = 106º – 44º . . .
= 62º
8.2 ˆa + 44º = 90º . . . sum of the øs of a Δ = 180º
â ˆa = 90º – 44º
= 46º �
ˆb + 28º = 44º . . .
â ˆb = 44º – 28º
= 16º �
ˆc = ˆb + 90º . . .
= 16º + 90º
= 106º �
9. x = 75º – 44º . . .
= 31º �
y = x . . . alternate øs; AD || BC
= 31º �
10. ˆE = 95º – 30º . . .
= 65º �
ˆA = 180º – ˆE . . .
= 115º �
NB:
An interior angle = the exterior ø – the other interior ø
NB: See comment in Question 6.
øs
on a straight line
are supplementary
AE = AW; øs opposite equal
sides in an isosceles Δ
øs opposite equal sides MT
and MP in an isosceles triangle
exterior ø of Δ
equals the sum of
the interior øs
OR: ˆE = 5x – 3x
= 2x
øs opposite equal sides
in an isosceles Δ
exterior ø of Δ = sum
of interior opposite øs
øs opposite equal sides
in an isosceles Δ
exterior ø of Δ = sum
of interior opposite øs
exterior ø of ΔBDC = sum
of interior opposite øs
co-interior øs are supplementary
because AB || ED
exterior ø of ΔDEC = sum of
interior opposite øs
the base øs of an
isosceles Δ are equal
ext ø of ΔABD = the sum of
the interior opposite øs
the base øs of an
isosceles Δ are equal
sum of the øs
of a Δ
exterior ø of Δ = sum
of interior opposite øs
Often, in geometry riders,
there are several possible methods.
Solutions: Congruent Triangles
A4 Copyright © The Answer
CONGRUENT TRIANGLES
(Symbol: ≡)
1.1 ΔDEF � . . . SøS
2. ΔSTV � . . . SøS
3.1 SøS � â Answer: C �
4. In ΔABD and ΔCDB
(1) ˆ
2B = ˆ
1D = 90º . . . given
(2) AD = CB . . . given
(3) BD is common
â ΔABD h ΔCDB � . . . 90º, Hyp, S (RHS)
5.1 In ΔABD and ΔACD
(1) AB = AC . . . given
(2) BD = CD . . . given
(3) AD is common
â ΔABD h ΔACD � . . . SSS
5.2 â ˆ1
A = ˆ2
A . . .
i.e. DA bisects ˆBAC �
6. In ΔKNQ and ΔMPQ
(1) NQ = PQ . . . given
(2) KQ = MQ . . . given
(3) ˆQ is common
â ΔKNQ ≡ ΔMPQ � . . . SøS
7.1 BE = BD + DE
& CD = EC + DE
But: BD = EC . . . given
â BE = CD �
7.2 In ΔABE and ΔACD
(1) BE = CD . . . proved in Question 7.1
(2) ˆAEB = ˆADC . . .
(3) AE = AD . . . given
â ΔABE h ΔACD . . . SøS
â Answer: ΔACD �
8.1
In ΔMPN and ΔMTN
(1) MP = MT . . . radii of the circle
(2) MN is common
(3) ˆ
2N = ˆ
1N = 90º . . . given that MN ⊥ PT
â ΔMPN ≡ ΔMTN � . . . RHS
â PN = NT � . . .
9.1 BC = BF + FC
& EF = CE + FC
But: BF = CE . . . given
â BC = EF �
9.2 In ΔABC and ΔDEF
(1) AB = DE . . . given
(2) AC = DF . . . given
(3) BC = EF . . . proved in Question 9.1
â ΔABC ≡ ΔDEF � . . . SSS
9.3 ˆB = ˆE because they are corresponding angles of the
congruent triangles in Question 9.2 �
9.4 ˆB and ˆE are alternate angles
& ˆB = ˆE in Question 9.3
â AB || ED . . . converse fact
Order is important in congruency layout: • 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.
We need to prove congruent triangles!
*corresponding øs
of
congruent Δs
in Question 5.1
* Nothing to do with
corresponding øs
on || lines
øs
opposite equal sides
in isosceles ΔADE
Study the proof
carefully!
corresponding sides of
congruent triangles
NB: The letters must be in the correct order so
that equal sides and angles correspond.
Note:
Observe the layout of a congruency proof.
NB: Always
give reasons!
NB: Always
give reasons!
M N
K
Q
1
1 2
2
P
See the notes on Congruency
and Similarity on page A5
Solutions: Congruent Triangles
Copyright © The Answer A5
10.1 In ΔABD and ΔACD
(1) AB = AC . . . given
(2) BD = CD . . . given
(3) AD is common
â ΔABD ≡ ΔACD � . . . SSS
10.2 In ΔABE and ΔACE
(1) AB = AC . . . given
(2) ˆ
1A = ˆ
2A . . .
(3) AE is common
â ΔABE ≡ ΔACE � . . . SøS
10.3 ˆ
1E = ˆ
2E . . .
But ˆ1E + ˆ
2E = 180º . . . angles on a straight line
â ˆ1E = ˆ
2E = 90º �
10.4 AE ⊥ BC � [ i.e. AE is perpendicular to BC � ]
11. The sketch with all equal angles filled in.
(There are no equal sides)
Answer: D � Proof: In ΔPTS and ΔRTQ
(1) ˆ
1P = ˆ
1R . . . alternate ø
s
; PS æ QR
(2) ˆ
1S = ˆ
1Q . . . alternate ø
s
; PS æ QR
& (3) ˆPTS = ˆRTQ . . . vertically opposite øs
â ΔPTS ||| ΔRTQ � . . . øøø
Study the (easy) logic very carefully!
corresponding øs
of congruent
triangles in Question 10.2
corresponding angles in congruent
triangles in Question 10.1
≡ means 'is congruent to'
(i.e. same SHAPE and SIZE) whereas:
||| means 'is similar to'
(i.e. same SHAPE but not necessarily same SIZE
P
T
Q
S
R
1 1
1 1
Congruency (≡) and Similarity (|||)
of triangles
Congruent Triangles . . .
have the same shape and size.
All 3 angles and all 3 sides are equal.
i.e. ΔABC ≡ ΔPQR means that
ˆA = ˆP , ˆB = ˆQ and ˆC = ˆR and AB = PQ, AC = PR and BC = QR
Note the order of the lettering
Similar Triangles . . .
have the same shape, but not necessarily the same size.
All 3 angles are equal.
i.e. ΔABC ||| ΔPQR means that
ˆA = ˆP , ˆB = ˆQ and ˆC = ˆR
The sides are not necessarily equal, but are proportional:
= =
AB AC BC
PQ PR QR
Note the order of the lettering
B C
A
Q R
P
B C
A
Q R
P
Two triangles are congruent if they have
• 3 sides the same length . . . SSS
• 2 sides & an included angle equal . . . SøS
• a right angle, hypotenuse & a side equal . . . RHS
• 2 angles and a side equal . . . øøS
Solutions: Similar Triangles
A6 Copyright © The Answer
SIMILAR Δ'S
(Symbol: |||)
1.1 If ΔDEF ||| ΔKLM, i.e. ΔDEF is similar to ΔKLM,
then DE
KL =
EF
LM =
KM
DF � . . .
â 14
7 =
x
12 . . . ( )14 2
= =7 1 12
?
â x = 24 �
2. If ΔABC ||| ΔEDF,
then ( )==
AB BC
ED DF
AC
EF . . .
â =
AB 15
6 10
Multiply by 6:
â AB = ×15 6
10
â AB = 9 cm �
3.1 In ΔPQR and ΔSTR
(1) ˆP = ˆS . . . alternate øs
; PQ || ST
(2) ˆQ = ˆT . . . alternate øs
; PQ || ST
& (3) ˆPRQ = ˆSRT . . . vertically opposite øs
â ΔPQR ||| ΔSTR � . . . øøø
3.2 â ( )==
PQ PR
ST SR
QR
TR . . .
â =
PQ 10
3 6
Multiply by 3:
â PQ = ×10 3
6
â PQ = 5 cm �
4.1 In ΔQPN and ΔLMN
(1) ˆN is common
(2) ˆ
1P = ˆM . . . corresponding ø
s
; QP || LM
& (3) ˆ
1Q = ˆL . . . corresponding ø
s
; QP || LM
â ΔQPN ||| ΔLMN . . . øøø
4.2 â ( )==
PN QP
MN LM
QN
LN . . .
â =
PN 3
16 8
Multiply by 16:
â PN = 3 16×
2
8
â PN = 6 cm �
5.1 In ΔABD and ΔACE
(1) ˆA is common
(2) ˆA DB = ˆA EC
(3) 3rd
L : ˆ1
D = ˆ1E . . . sum of the ø
s
of a Δ
â ΔABD ||| ΔACE � . . . øøø
5.2 â ( )==
BD AD
CE AE
AB
AC . . .
â =
BD 9
21 7
Multiply by 21:
â BD = 9 21×
3
7
â BD = 27 cm �
Note the ORDER of the letters in similar Δs
:
ΔDEF | | | ΔKLM � ˆD = ˆK , ˆE = ˆL and ˆF = ˆM
and this determines the proportional sides
proportional sides of
similar triangles
proportional sides of
similar triangles
proportional sides of
similar triangles
proportional sides of
similar triangles
proportional sides of
similar triangles
Choose the sides for which you have
the lengths!
Choose the sides for which the lengths have been given.
Choose the sides for which the lengths have been given.
Choose the sides with known lengths.
L
1Q
M
P
N
1
2 2
8 cm
3 cm 16 cm
A
1
B
E
C D
1
2
2
F
7 cm
21 cm
9 cm
Be sure to read the
notes on Congruency
and Similarity on pg. A5
Solutions: Quadrilaterals
Copyright © The Answer A7
QUADRILATERALS
1. (x + 50º) + (2x – 20º) = 180º . . .
â 3x + 30º = 180º
Subtract 30º: â 3x = 150º
Divide by 3: â x = 50º
â ˆA = 50º + 50º = 100º
â ˆB = 180º – 100º . . . co-interior øs
; AC || BD
= 80º �
OR: ˆC = 2(50º) – 20º = 80º
â ˆB = 80º � . . . opposite øs
of a ||m
are equal
2.1 ˆEFG = 180º – 156º . . .
= 24º �
2.2 ˆ
2F = 1
2(24º) . . .
= 12º �
2.3 ˆG = 156º � . . .
3.1 ΔATD and ΔBTC �
3.2 ˆ
2A = 60º . . . ø of equilateral Δ
â ˆ1
A = 30º . . . ˆDAB = 90º in square
â ˆ2
D + ˆ1T = 180º – 30º . . . sum of the ø
s
of a Δ
= 150º
But ˆ2
D = ˆ1T . . . ø
s
opposite equal sides in Δ ATD
â ˆ2
D = 75º � . . . half of 150º
3.3 Similarly : ˆ3T = 75º
and ˆ2T = 60º . . . ø of equilateral Δ ATB
â ˆ4T = 360º – (75º + 60º + 75º) . . .
= 360º – 210º
= 150º �
OR: ˆ1
D = 90º – 75º . . . ˆ
2D = 75º and ˆADC = 90º
= 15º
& Similarly : ˆ1
C = 15º
â ˆ4T = 180º – 2(15º) . . . sum of the ø
s
of Δ DTC
= 180º – 30º
= 150º �
4.
ˆQRP = ˆTRS = 60º . . . øs
of equilateral Δs
& ˆPRT = 90º . . . ø of square
â ˆQRS = 360º – (60º + 90º + 60º) . . .
= 150º
QR = PR . . . sides of equilateral ΔPQR
= RT . . . sides of square
= RS . . . sides of equilateral ΔRTS
â x = ˆRSQ . . . angles opposite equal sides in ΔQRS
= 12
(180º – 150º) . . .
= 12
(30º)
= 15º �
5. In ΔADB and ΔCBD:
ˆ
1D = ˆ
2B . . . alternate ø
s
; AD || BC in parallelogram
ˆ
1B = ˆ
2D . . . alternate ø
s
; AB || DC in parallelogram
BD = BD . . . common side
â ΔADB ≡ ΔCBD . . . øøS
â AD = BC � . . .
& AB = DC �
6. D rectangle �
co-interior øs
;
AB || CD
co-interior øs
;
DE || GF in rhombus
the diagonals of a rhombus
bisect the øs
of the rhombus
opposite øs
of a rhombus
(or ||m
) are equal
. . . AT = AB . . . sides of equilateral Δ
= AD . . . sides of square & Similarly: BT = AB = BC
sum of øs
about
a point = 360º
sum of øs
about
a point = 360º
ø of isosceles Δ; sum
of the øs
of a Δ = 180º
Note: We have just proved, using congruency,
that both pairs of opposite sides of a
parallelogram are equal in length.
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.'
corresponding sides
of congruent Δ
s
C
A B
D
T
1
3
2
2
1
1
1
2
4
30°
60°
60°
60°
Q
60°
60° 60°
60°
60°
60°
W T
S
RP
x
Solutions: Quadrilaterals
A8 Copyright © The Answer
7.1 Something new to experience!
( ˆ1
B + ˆ2
B ) + ( ˆ1
C + ˆ2
C ) = 180º
. . .
Let ˆ1
B = ˆ2
B = x and ˆ1
C = ˆ2
C = y . . .
. . . given that the angles were bisected
â 2x + 2y = 180º
i.e. 2(x + y) = 180º
Divide by 2: â x + y = 90º
â In ΔBTC: ˆ1
B + ˆ1
C = 90º
â ˆ2T = 90º . . . sum of the ø
s of a Δ
7.2 ΔTCP �
ˆTPC = 180º – 90º = 90º . . . angles on a straight line
In ΔBCT and ΔTCB:
1) ˆ
1C = ˆ
2C (= y)
2) ˆ
2T = ˆTPC (= 90º)
3) â ˆ1
B = ˆ1T . . . 3
rd ø of Δ
â ΔBCT ||| ΔTCP � . . . øøø
7.3 â ⎞= ⎟⎠
= ⎛⎜⎝
CBT BC
TP TC
T
CP . . .
â =
BT 2TC
4 TC
Multiply by 4:
â BT = 2 % 4
= 8 cm �
8. Answer: C ΔAEB ≡ ΔDEC �
In ΔAEB and ΔDEC:
1) AE = BE . . . given
2) EB = EC . . . given
3) ˆAEB = ˆDEC . . . vertically opposite øs
â ΔAEB ≡ ΔDEC . . SøS
9.
(a + i + h) + (b + g + f) + (c + d + e)
= 3 % 180º
= 540º
â Each ø = °540
5 = 108º
â Answer: D �
10.
The sum of the øs
of the hexagon
= 4 % 180º
= 720º
â Each ø = °720
6 = 120º
â Answer: B �
A very useful
technique in
geometry!
proportional sides of
similar triangles
co-interior øs
;
AB || DC in
parallelogram
Choose the sides whose lengths you have been given.
bc d
e
f
gh
a
i
A
C
B
D
1
1
2
2
1
2 3
T
P
x
x
y
y
NOTES
Solutions: Theorem of Pythagoras
Copyright © The Answer A9
THEOREM OF PYTHAGORAS
1. AC = 13 cm . . . 5 : 12 : 13 Pythagoras 'trip'
OR: In ΔABC: AC2 = AB
2 + BC
2 . . .
= 52 + 12
2
= 25 + 144
= 169
â AC = 13 cm
2.1 In ΔABC: x = 10 cm � . . .
2.2 In ΔABD: BD = 15 cm . . . Pythag 'trip': 8 : 15 : 17
â y = 15 cm – 6 cm
= 9 cm �
OR: 2.1 x2 = 8
2 + 6
2, etc.
2.2 BD2 = 17
2 – 8
2, etc.
3.1 ×TU 12
2 = 30 . . .
h × b
2= area of a Δ
Multiply by 2: . . .
â TU % 12 = 60
Divide by 12:
â TU = 5 cm �
3.2 TW = 13 cm . . . Pythag 'trip' 5 : 12 : 13
â The perimeter of ΔTUW = 5 cm + 12 cm + 13 cm
= 30 cm �
4.1 The length of the ladder = 13 m � . . .
OR: (length of the ladder)2
= 52 + 12
2, etc.
5. Answer: B 6 cm �
In ΔADC: DC = 8 cm . . . opposite sides of rectangle
â AD = 6 cm . . .
6.
AC2 = 15
2 = 225
& AB2 + BC
2 = 9
2 + 12
2 = 81 + 144 = 225
â AC2 = AB
2 + BC
2
â ˆB = 90° . . . the converse of the Theorem of Pythagoras
Thm of Pythag.; ˆB = 90º
Pythag 'trip':
3 : 4 : 5 = 6 : 8 : 10
Pythag 'trip':
3 : 4 : 5 = 6 : 8 : 10
*
Note: When applying the Theorem of Pythagoras, there
are some well-known 'trips' which are useful to
know and use instead of long calculations. e.g. 3
2 + 4
2 = 9 + 16 = 25 = 5
2
52 + 12
2 = 25 + 144 = 169 = 13
2
82 + 15
2 = 64 + 225 = 289 = 17
2
So: the TRIP(LET)S : 3 : 4 : 5 ; 5 : 12 : 13 ; 8 : 15 : 17 and even multiples: 6 : 8 : 10
*
. . . Theorem of Pythag
Pythag 'trip':
5 : 12 : 13
OR: × TU 12
6
2
= 30
â 6 % TU = 30
Divide by 6:
â TU = 5 cm �
NOTES
12 m
5 m
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.
Solutions: Measurement: 2D
A10 Copyright © The Answer
MEASUREMENT: 2D
1.1 Radius, r = 1
2 % diameter = 6 cm
â Area = �r2 = 3,14 % 6
2
= 113,04 cm2 �
1.2 The circumference of the (full) circle = 2�r
â The 'circumference' of the semi-circle = �r = 3,14 % 6
= 18,84 cm â The perimeter of the shape ACB
= 18,84 cm + 12 cm = 30,84 cm � . . . diameter, AB = 12 cm
2. The area of a square = s2 = (0,12)
2 = 0,0144 cm
2
â Answer: D �
3.1 The perimeter of the field
= 2 % 100 m + 2 % semi-circles
= 200 m + 2 % 3,14 % 30 m . . .
= 388,4 m
â Number of laps = 4 000 m
388,4 m = 10,298 . . .
â 11 laps � . . . for at least 4 km!
3.2 The area of the field
= area of rectangle + area of 2 semi-circles
= (100 % 60)m2 + �.30
2 m
2 . . . Area of circle = �r
2
= 6 000 m2 + �.900 m
2
l 8 827,43 m2 �
4.1 In ΔAPS: PS2 = 5
2 – 2
2 . . . Theorem of Pythagoras
= 25 – 4
= 21
â PS = 21
l 4,58 m �
4.2 PT = 3 % AB = 3 % 4 m = 12 m �
4.3 A kite � . . . 2 pairs of adjacent sides equal
4.4 Method 1: Using the formula
The area = 1
2 the product of the diagonals
= 1
2 (PT % AB)
= 1
2 (12 % 4)
= 24 m2 �
Method 2: Without the formula
The area = ΔPAT + ΔPBT
= 2 % ΔPAT . . . because the 2 Δs
are congruent
= 21
2×
PT AS⎛ ⎞⎜ ⎟⎝ ⎠
.
= 12 m % 2 m
= 24 m2 �
5.1 In ΔABT: AT = 4 cm � . . . Pythag 'trip' : 3 : 4 : 5
OR: AT2 = 5
2 – 3
2 . . . Theorem of Pythagoras
= 25 – 9
= 16
â AT = 4 cm �
5.2 The area of parallelogram ABCT
= base % height
= BC % AT
= AD % 4 cm
= 12 cm % 4 cm
= 48 cm2
5.3.1 DC = AB = 5 cm . . . opposite øs
of ||m
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 �
Observe carefully:
(0,12)2 =
212
100
⎛ ⎞⎜ ⎟⎝ ⎠
= 12 12 144
100 100 10 000× = = 0,0144
Circumference of
circle = 2�r
See the shifting of ΔABT
as shown:
Parallelogram ABCD = rectangle ATSD
& Area of rectangle ATSD
= length % breadth
= 12 cm % 4 cm
= 48 cm2
12 cm
5 cm
3 cm
A
B T C
D
S
Solutions: Measurement: 2D
Copyright © The Answer A11
5.3.2 Method 1: Using the formula
The area of trapezium ADCT
= 1
2 (sum of the || sides) % the distance between them
= 1
2(AD + TC) % AT
= 1
2(12 + 9) % 4
= 42 cm2 �
Method 2: Without the formula
The area of trapezium ADCT
= Area of ΔATC + Area of ΔADC
= 1
2(9 % 4) +
1
2(12 % 4)
= 18 + 24
= 42 cm2 �
6. 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 �
7. The circumference of a circle, 2�r = 52 cm
â r = π
52
2 =
π
26
â The area of the circle
= �r2 = � %
⎛ ⎞⎜ ⎟π⎝ ⎠
2
26
= � % π
2
2
26
= π
226
l 215,18 cm2 � . . . correct to 2 decimal places
8.1 The circumference of a circle = 2�r
â The circumference of the smaller circle
= 2 % � % 20
l 125,66 cm �
8.2 The area of the shaded region
= the area of the full circle – the area of the inner circle
= �302 – �20
2
l 1 570,80 cm2 �
9.1 The area of the shaded ring
= the area of the full circle – the area of the inner ring
= �R2 – �r
2
= �(R2 – r
2) �
9.2 Area of the shaded ring = �(142 – 8
2)
= 132� cm2 �
Note: Give the answer 'in terms of �'
1
2. ( 9 + 12) .4, like the formula above!
*
*
b
ℓ
4 cm
9 cm
A
T C
D12 cm
NOTES