Gr 9 Maths: Content Area 3 & 4 Geometry & Measurement (2D) · PDF fileGr 9 Maths: Content Area...

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

fax: (021) 671 2546

faxtoemail: 088 021 671 2546

www.theanswer.co.za

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


TRIANGLES: BASIC FACTS


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, ÊAW = 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 Ê is . . . ?

A 40º B 60º

C 140º D 20º (1) [10]

5.

In ΔABC, AB = AC and ˆC = x.

Determine the size of Â in terms of x. (3)

6.

In the figure above, ˆB = 50º and ÂCD = 110º.

The size of Â 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;

ÂDB = 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


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, ÂCD = 95º

and ˆD = 30º.

Determine the size of Ê and Â . (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




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 Â + 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


CONGRUENT ΔS


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


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)




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


SIMILAR ΔS


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)




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



Questions: Quadrilaterals


QUADRILATERALS


1. ABCD is a parallelogram. Calculate the size of ˆB .

(4)

2. In the figure below, DEFG is a rhombus and Ê = 156°.

Calculate the size of :

2.1 ÊFG

2.2 ˆ

2F

2.3 ˆG

Statement Reason

2.1 ÊFG = (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


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




A

C

B

D

1

1

2

2

1

2

3

T

P

A D

B C

E

Questions: Theorem of Pythagoras


THEOREM OF PYTHAGORAS


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




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


30

20

MEASUREMENT: 2D


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




60 m

100 m

A

B

S

2 m5 m

2 m

P T


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


OR

A rhombus with one

angle of 90º

Properties of a rectangle

The Sides


sides parallel


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


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


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


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

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Solutions: Geometry of Straight lines

Copyright © The Answer A1

GEOMETRY OF STRAIGHT LINES

1.1 â = 43º � . . . vertically opposite øs

ˆb = â . . . 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.

Â = 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. â = 60º � . . . vertically opposite angles

ˆb = 35º � . . . alternate øs

; || lines

ˆc = 35º � . . . base øs

of isosceles Δ

ˆd = 180º – ( â + ˆc ) . . . sum of the øs

of a Δ

= 180º – (60º + 35º)

= 85º �

ê = â – 35º . . .

= 25º �

ˆf = ( ˆb + ˆc ) . . . corresponding øs

; || lines

= 70º �

or ˆf = ˆc + 35º . . . ext ø of Δ = sum of int. opp. øs

= 70º �

ˆg = ê . . . 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


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

ÊFC = Ê + ˆD . . .

â 5x = Ê + 3x

â Ê = 2x

= 40º �

â Answer: A �

5. ˆB (= ˆC ) = x . . .

â Â = 180º – 2x � . . . sum of the øs of Δ = 180º

6. Â = 110º – 50º . . .

= 60º �

â Answer: B �

7. In ΔABD: â = ÂBD . . .

â â = 1

2(180º – 72º) . . .

= 1

2(108º)

= 54º �

ˆb = 72º + â . . .

= 126º �

ˆc = ˆBDC . . .

ˆc = 1

2(180º – ˆb )

= 1

2(54º)

= 27º �

8.1 x = 106º – 44º . . .

= 62º

8.2 â + 44º = 90º . . . sum of the øs of a Δ = 180º

â â = 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. Ê = 95º – 30º . . .

= 65º �

Â = 180º – Ê . . .

= 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: Ê = 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 ΔBDC = sum


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



Often, in geometry riders,

there are several possible methods.

Solutions: Congruent Triangles


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) ÂEB = ÂDC . . .

(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 = Ê because they are corresponding angles of the

congruent triangles in Question 9.2 �

9.4 ˆB and Ê are alternate angles

& ˆB = Ê 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



Solutions: Congruent Triangles


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


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) Â is common

(2) Â DB = Â 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 , Ê = ˆL and ˆF = ˆM

and this determines the proportional sides

proportional sides of

similar triangles


similar triangles


similar triangles


similar triangles


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


QUADRILATERALS

1. (x + 50º) + (2x – 20º) = 180º . . .

â 3x + 30º = 180º

Subtract 30º: â 3x = 150º

Divide by 3: â x = 50º

â Â = 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 ÊFG = 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 ÂDC = 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


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!


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


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


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


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

Gr 9 Maths: Content Area 3 & 4 Geometry & Measurement (2D) · PDF fileGr 9 Maths: Content Area...

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