2. pythagoras' theorem

2

Contents:

Pythagoras’ TheoremPythagoras’ Theorem

A

B

C

D

E

Pythagoras’ Theorem

The converse of Pythagoras’ Theorem

Problem solving using Pythagoras’ Theorem

Three-dimensional problems

More difficult problems (Extension)

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Y:\HAESE\SA_10-6ed\SA10-6_02\033SA10-6_02.CDR Monday, 4 September 2006 2:17:55 PM PETERDELL

34 PYTHAGORAS’ THEOREM (Chapter 2)

OPENING PROBLEM

�

take hold of knots at arrows make rope taut

corner

line of one sideof building

Tee Caddy

Ball

150 m marker

Hole

70 m250 m

Right angles (90o angles) are used in the con-

struction of buildings and in the division of areas

of land into rectangular regions.

The ancient Egyptians used a rope with 12equally spaced knots to form a triangle with sides

in the ratio 3 : 4 : 5: This triangle has a right

angle between the sides of length 3 and 4 units,

and is, in fact, the simplest right angled triangle

with sides of integer length.

Karrie is playing golf in the US Open but hits a wayward tee shot on the

opening hole. Her caddy paces out some distances and finds that Karrie has

hit the ball 250 m, but is 70 m from the line of sight from the tee to the hole.

A marker which is 150 m from the pin is further up the fairway as shown:

Consider the following questions:

1 How far is the caddy away from the tee?

2 From where the caddy stands on the fairway, what distance is left to the 150 m marker

if he knows the hole is 430 m long?

3 How far does Karrie need to hit her ball with her second shot to reach the hole?


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Y:\HAESE\SA_10-6ed\SA10-6_02\034SA10-6_02.CDR Wednesday, 26 July 2006 11:46:26 AM PETERDELL

PYTHAGORAS’ THEOREM (Chapter 2) 35

A right angled triangle is a triangle which

has a right angle as one of its angles.

The side opposite the right angle is called

the hypotenuse and is the longest side of the

triangle.

The other two sides are called the legs of the

triangle.

Around 500 BC, the Greek mathematician

Pythagoras formulated a rule which connects

the lengths of the sides of all right angled tri-

angles. It is thought that he discovered the

rule while studying tessellations of tiles on

bathroom floors. Such patterns, like the one

illustrated, were common on the walls and

floors of bathrooms in ancient Greece.

In geometric form, the Theorem of Pythagoras is:

In any right angled triangle, the area of

the square on the hypotenuse is equal to

the sum of the areas of the squares on the

other two sides.

INTRODUCTION

hypotenuse

legs

Can you see how

Pythagoras may have

discovered the rule by

looking at the tile

pattern above?

PYTHAGORAS’ THEOREMA

In a right angled triangle, with

hypotenuse c and legs a and b,

c2 = a2 + b2.

c

b

a

cc

c2

a

a a2

b2b

b

GEOMETRY

PACKAGE


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1 Find the length of the hypotenuse in the following triangles, leaving your answer in surd

(square root) form if appropriate:

a b c

b

a b

a

a

ac

c

c

c

b

b

If , then

, but

we reject

as lengths must

be positive!

x k

x k

k

2 == §

p

¡p`

`

Find the length of the hypotenuse in:

The hypotenuse is opposite the right angle and has length x cm.

) x2 = 32 + 22

) x2 = 9 + 4

) x2 = 13

i.e., x =p

13 fas x > 0g

) the hypotenuse isp

13 cm.

3 cm

2 cmx cm

Example 1

EXERCISE 2A

4 cm

7 cm x cm5 cm

x cm8 km

13 km

x km

Over 400 different proofs of the Pythagorean Theorem exist. Here is one of them:

Proof:

On a square we draw 4 identical (congruent) right

angled triangles, as illustrated. A smaller square in

the centre is formed.

Since total area of large square

= 4 £ area of one triangle + area of smaller square,

(a+ b)2 = 4(12ab) + c2

) a2 + 2ab+ b2 = 2ab+ c2

) a2 + b2 = c2

Note:

Suppose the legs are of length and and the

hypotenuse has length

a b

c:

p7:

When using Pythagoras’ Theorem we often see , which are square root

numbers like

surds

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Y:\HAESE\SA_10-6ed\SA10-6_02\036SA10-6_02.CDR Thursday, 3 August 2006 10:07:08 AM PETERDELL


Find the length of the third

side of:

The hypotenuse has length 6 cm.

) x2 + 52 = 62 fPythagorasg

) x2 + 25 = 36

) x2 = 11

) x =p

11 fas x > 0g

) third side isp

11 cm long.

2 Find the length of the third side of the following right angled triangles.

Where appropriate leave your answer in surd (square root) form.

a b c

3 Find x in the following:

a b c

Example 2

6 cm

5 cm

x cm

6 cm11 cm

x cm

1.9 km2.8 km

x km

9.5 cm

x cm

Find x in the following: The hypotenuse has length x cm.

) x2 = 22 + (p

10)2 fPythagorasg

) x2 = 4 + 10

) x2 = 14

) x = §p

14

But x > 0, ) x =p

14.

Example 3

~`1`0 cm

x cm2 cm

~`1`0 cm

x cm

3 cm

~`2 cmx cm

~`7 cm

~`5 cm

x cm

Solve for x: x2 +¡1

2

¢2= 12 fPythagorasg

) x2 + 1

4= 1

) x2 = 3

4

) x = §q

3

4

) x =q

3

4fas x > 0g

Example 4

1 cm

x cm

Qw_ cm

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4 Solve for x:

a b c

5 Find the value of x:

a b c

1 cm

x cm

Qw_ cm

x cm

Qw_ cm

Ew_ cm

Find the value of x: (2x)2 = x2 + 62 fPythagorasg

) 4x2 = x2 + 36

) 3x2 = 36

) x2 = 12

) x = §p12

But x > 0, ) x =p12.

Example 5

2 mx

x m

6 m

2 cmx

x cm

9 cm

26 cm

3 cmx

2 cmx

2 mx

3 mx

~`2`0 m

Find the value of any unknowns:

) x2 = 52 + 12 fPythagorasg

) x2 = 26

) x = §p26

) x =p26 fx > 0g

In ¢ACD, the hypotenuse is 6 cm.

) y2 + (

p26)2 = 62 fPythagorasg

) y2 + 26 = 36

) y2 = 10

) y = §p10

) y =p10 fy > 0g

Example 6

5 cm

1 cm

6 cm

x cmy cm

A B

CD

In triangle ABC, the hypotenuse is x cm.

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x m

1 m

m2

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Y:\HAESE\SA_10-6ed\SA10-6_02\038SA10-6_02.CDR Wednesday, 30 August 2006 11:13:17 AM DAVID3


6 Find the value of any unknowns:

a b c

7 Find x:

a b

8 Find the length of AC in:

9 Find the distance AB, in the following figures.

(Hint: It is necessary to draw an additional line or two on the figure in each case.)

a b c

If a triangle has sides of length a, b and c units and a2 + b2 = c2,

then the triangle is right angled.

x cm

3 cm

2 cm

y cm

4 cm

7 cm

2 cm

x cm

y cm

( )x�� cm

13 cm5 cm

4 cm3 cm

x cm

5 m9 m

A

B DC

1 cm

3 cm

2 cm

y cmx cm

B

NM

A

1 m

5 m

3 m

A B

4 m

6 m 7 m

D C

A

B

3 cm4 cm

THE CONVERSE OF PYTHAGORAS’ THEOREMB

THE CONVERSE OF PYTHAGORAS’ THEOREMGEOMETRY

PACKAGE

If we have a triangle whose three sides have known lengths, we can use the converse of

Pythagoras’ Theorem to test whether (or not) it is right angled.


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1 The following figures are not drawn accurately. Which of the triangles are right angled?

a b c

d e f

2 If any of the following triangles (not drawn accurately) is right angled, find the right

angle:

a b c

The simplest right angled triangle with sides of integer

length is the 3-4-5 triangle.

The numbers 3, 4, and 5 satisfy the rule 32 + 42 = 52.

Other examples are: f5, 12, 13g, f7, 24, 25g, f8, 15, 17g.

Is the triangle with sides 6 cm, 8 cm and 5 cm right angled?

The two shorter sides have lengths 5 cm and 6 cm, and 52 + 62

= 25 + 36

= 61But 82 = 64

) 52 + 62 6= 82 and hence the triangle is not right angled.

Example 7

EXERCISE 2B

7 cm

4 cm5 cm

9 cm

15 cm

12 cm5 cm

8 cm

9 cm

15 m

8 m

17 m

3 cm

~`1`2 cm

~`7 cm

~`7`5 m

~`4`8 m~` `7� m

A

B C

8 m ~`2`0`8 m

12 m

A

B C

1 cm2 cm

~`5 cmA

B

C

5 km

7 km

~`2`4 km

PYTHAGOREAN TRIPLES

53

4

The set of integers fa, b, cg is a Pythagorean triple if it obeys the rule

a2 + b2 = c2:

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INVESTIGATION PYTHAGOREAN TRIPLES SPREADSHEET

3 Determine if the following are Pythagorean triples:

a f8, 15, 17g

d f14, 48, 50g

b f6, 8, 10g

e f1, 2, 3g

c f5, 6, 7g

f f20, 48, 52g

4 Find k if the following are Pythagorean triples:

a f8, 15, kg

d f15, 20, kg

b fk, 24, 26g

e fk, 45, 51g

c f14, k, 50g

f f11, k, 61g

5 For what values of n does fn, n+ 1, n+ 2g form a Pythagorean triple?

6 Show that fn, n+ 1, n+ 3g cannot form a Pythagorean triple.

Well known Pythagorean triples are f3, 4, 5g, f5, 12, 13g,f7, 24, 25g and f8, 15, 17g.

Formulae can be used to generate Pythagorean triples.

An example is 2n+ 1, 2n2 + 2n, 2n2 + 2n+ 1 where n is a positive integer.

A spreadsheet will quickly generate sets of Pythagorean triples using such formulae.

Show that

f5, 12, 13g

is a Pythagorean triple.

We find the square of the largest number first.

132 = 169

and 52 + 122 = 25 + 144 = 169

) 52 + 122 = 132

i.e., is a Pythagorean triple.

Example 8

Find k if f9, k, 15g is a Pythagorean triple.

Let 92 + k2 = 152 fPythagorasg

) 81 + k2 = 225

) k2 = 144

) k = §p

144

) k = 12 fk > 0g

Example 9

SPREADSHEET

f5, 12, 13g

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Y:\HAESE\SA_10-6ed\SA10-6_02\041SA10-6_02.CDR Friday, 28 July 2006 3:29:17 PM PETERDELL


1 Open a new spreadsheet

and enter the following:

2 Highlight the formulae in B2, C2 and D2 and fill

down to Row 3. You should now have generated

two sets of triples:

3 Highlight the formulae in Row 3 and fill down

to Row 11 to generate 10 sets of triples.

4 Check that each set of numbers is actually a triple by adding two more columns to

your spreadsheet.

In E1 enter the heading

‘aˆ2+bˆ2’ and in F1 enter the

heading ‘cˆ2’.

In E2 enter the formula

=B2ˆ2+C2ˆ2 and in F2 enter

the formula =D2ˆ2.

5 Highlight the formulae in E2 and F2 and fill down to Row 11. Is each set of numbers

a Pythagorean triple? [Hint: Does a2 + b2 = c2?]

6 Your task is to prove that the formulae f2n+ 1, 2n2 + 2n, 2n2 + 2n+ 1g will

produce sets of Pythagorean triples for positive integer values of n.

We let a = 2n+ 1, b = 2n2 + 2n and c = 2n2 + 2n+ 1.

Simplify c2¡b2 = (2n2+2n+1)2¡(2n2+2n)2 using the difference of two squares

factorisation, and hence show that it equals (2n+ 1)2 = a2.

Right angled triangles occur frequently in problem solving and often the presence of right

angled triangles indicates that Pythagoras’ Theorem is likely to be used.

All of these figures contain right angled triangles where Pythagoras’ Theorem applies:

In a rectangle, a right angle exists between

adjacent sides.

Construct a diagonal to form a right angled

triangle.

What to do:

fill down

PROBLEM SOLVING USING

PYTHAGORAS’ THEOREMC

SPECIAL GEOMETRICAL FIGURES

rectangle

diagonal


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Things to remember

² Draw a neat, clear diagram of the situation.

² Mark on known lengths and right angles.

² Use a symbol, such as x, to represent the unknown length.

² Write down Pythagoras’ Theorem for the given information.

² Solve the equation.

² Write your answer in sentence form (where necessary).

A rectangular gate is 3 m wide and has a 3:5 m diagonal. How high is the gate?

Let x m be the height of the gate.

Now (3:5)2 = x2 + 32 fPythagorasg

) 12:25 = x2 + 9

) 3:25 = x2

) x =p

3:25 fas x > 0g

) x + 1:80

Thus the gate is approximately

1 A rectangle has sides of length 8 cm and 3 cm. Find the length of its diagonals.

2 The longer side of a rectangle is three times the length of the shorter side. If the length

of the diagonal is 10 cm, find the dimensions of the rectangle.

3 A rectangle with diagonals of length 20 cm has sides in the ratio 2 : 1. Find the

a perimeter b area of the rectangle.

Example 10

3.5 m

3 m

x m

EXERCISE 2C.1

In a and a , the

.

square rhombus

diagonals bisect each other at

right angles

In an and an

, the

.

isosceles triangle

equilateral triangle altitude

bisects the base at right angles

equilateral triangle

square rhombus

isosceles triangle

1 80: m high.

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A rhombus has diagonals of length 6 cm and 8 cm.

Find the length of its sides.

The diagonals of a rhombus bisect at right angles.

Let a side be x cm.

) x2 = 32 + 42 fPythagorasg

) x2 = 9 + 16

) x2 = 25

) x = §p

25

) x = 5 fx > 0g

i.e., the sides are 5 cm in length.

4 A rhombus has sides of length 6 cm. One of its diagonals is 10 cm long. Find the length

of the other diagonal.

5 A square has diagonals of length 10 cm. Find the length of its sides.

6

A man travels due east by bicycle at 16 kmph. His son travels due south on his

bicycle at 20 kmph. How far apart are they after 4 hours, if they both leave point

A at the same time?

After 4 hours the man has travelled 4£ 16 = 64 km

and his son has travelled 4£ 20 = 80 km.

Thus x2 = 642 + 802 fPythagorasg

i.e., x2 = 4096 + 6400

) x2 = 10496

) x =p

10 496 fas x > 0g

) x + 102

)

7 A yacht sails 5 km due west and then 8 km due south.

How far is it from its starting point?

8 Town A is 50 km south of town B and town C is 120km east of town B. Is it quicker to travel directly from

A to C by car at 90 kmph or from A to C via B in a

train travelling at 120 kmph?

Example 11

3 cm 4 cm

x cm

Example 12

N

S

W E

manA 64 km

80 kmx km

son

they are km apart after hours.102 4

A rhombus has diagonals of length cm and cm. Find its perimeter.8 10

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9 Two runners set off from town A at the same time. If

one runs due east to town B and the other runs due

south to town C at twice the speed, they arrive at B and

C respectively two hours later. If B and C are 50 km

apart, find the speed at which each runner travelled.

10 Find any unknowns in the following:

a b c

11 An equilateral triangle has sides of length 12 cm. Find the length of one of its altitudes.

12 An isosceles triangle has equal sides of length 8 cm and a base of length 6 cm. Find the

area of the triangle.

13 When an extension ladder rests against a wall it

reaches 4 m up the wall. The ladder is extended

a further 0:8 m without moving the foot of the lad-

der and it now rests against the wall 1 m further up.

How long is the extended ladder?

An equilateral triangle has sides of length 6 cm. Find its area.

The altitude bisects the base at right angles.

) a2 + 32 = 62 fPythagorasg

) a2 + 9 = 36

) a2 = 27

) a = §p

27

) a =p

27 fa > 0g

Now, area = 1

2base £ height

= 1

2£ 6£

p27

= 3p

27 cm2

+ 15:6 cm2 f1 d.p.g

So, area is 15:6 cm2:

Example 13

45°

y°

2 cmx cm

12 cm

h cm

x cm

7 cm

1 cm

30°y cm

x cm60°

1 m

4 m

a cm

6 cm

3 cm

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14 An equilateral triangle has area 16p

3 cm2. Find the length of its sides.

15

When using true bearings we measure the direction of travel by comparing it with the true

north direction. Measurements are always taken in the clockwise direction.

Imagine you are standing at point A, facing north. You turn

clockwise through an angle until you face B. The bearing

of B from A is the angle through which you have turned.

That is, the bearing of B from A is the measure of the angle

between AB and the ‘north’ line through A.

In the diagram at right, the bearing of B from A is 72o

from true north. We write this as 72oT or 072o.

If we want to find the true bearing of A from B, we place

ourselves at point B and face north and then measure the

clockwise angle through which we have to turn so that we

face A. The true bearing of A from B is 252o.

From the diagram alongside, in triangle SAB

]SAB = 90o.

x2 = 1122 + 1342 fPythagorasg

) x2 = 12544 + 17 956

) x2 = 30500

) x =p

30 500 fas x > 0g

) x + 175

i.e.,

1 Two bushwalkers set off from base camp at the same time. If one walks on a true bearing

of 049o at an average speed of 5 kmph and the other walks on a true bearing of 319o at

an average speed of 4 kmph, find their distance apart after 3 hours.

TRUE BEARINGS

A

B

north

72°

A

B

north

252°

In bearings

problems, notice the

use of the properties

of parallel lines for

finding angles.

Example 14

x km

134 km

112 km

74°

16°

74°

164°A

S

B

N N

EXERCISE 2C.2

outpost B is km from base station S.175

Let SB be km.x

Revisit the on page and answer the questions posed.Opening Problem 34

Self Tutor

A helicopter travels from base station S on a true bearing of

for km to outpost A. It then travels km on a true

bearing of to outpost B. How far is outpost B from base

station S?

074 112 134164

o

o


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Y:\HAESE\SA_10-6ed\SA10-6_02\046SA10-6_02.CDR Thursday, 3 August 2006 10:08:10 AM PETERDELL


2 James is about to tackle an orienteering course. He has been given these instructions:

² the course is triangular and starts and finishes at S

² the first checkpoint A is in a direction 056o from S

² the second checkpoint B is in a direction 146o from A

² the distance from A to B is twice the distance from S to A

² the distance from B to S is 2:6 km.

Find the length of the orienteering course.

3 A fighter plane and a helicopter set off from airbase A at the same time. If the helicopter

travels on a bearing of 152o and the fighter plane travels on a bearing of 242o at three

times the speed, they arrive at bases B and C respectively 2 hours later. If B and C are

1200 km apart, find the average speed of the helicopter.

Pythagoras’ Theorem is often used when finding lengths in three-dimensional solids.

A 50 m rope is attached inside an empty cylindrical

wheat silo of diameter 12 m as shown. How high

is the wheat silo?

Let the height be h m. ) h2 + 122 = 502 fPythagorasg

) h2 + 144 = 2500

) h2 = 2356

) h =p

2356 fas h > 0g

) h + 48:5 fto 1 dec. placeg

i.e., the wheat silo is 48:5 m high.

1 A cone has a slant height of 17 cm and a base radius of 8 cm. How high is the cone?

2 Find the length of the longest nail that could be put entirely within a cylindrical can of

radius 3 cm and height 8 cm.

3 A 20 cm nail just fits inside a cylindrical can. Three identical spherical balls need to fit

entirely within the can. What is the maximum radius of each ball?

THREE-DIMENSIONAL PROBLEMSD

Example 15

12 m

h m50 m

50 m

12 m

EXERCISE 2D

In three-dimensional problem solving questions we often need the theorem of Pythagoras

. We look for right angled triangles which have two sides of known length.twice

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A room is 6 m by 4 m at floor level and the floor to ceiling height is 3 m. Find the

distance from a floor corner point to the opposite corner point on the ceiling.

The required distance is AD. We join BD.

In ¢BCD, x2 = 42 + 62 f

f

Pythagoras

Pythagoras

g

gIn ¢ABD, y2 = x2 + 32

) y2 = 42 + 62 + 32

) y2 = 61

) y = §p

61

But y > 0 ) the required distance isp

61 + 7:81 m.

Example 16

A

B

C D

4 m

3 m

y m

x m

6 m

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4 A cube has sides of length 3 cm. Find

the length of a diagonal of the cube.

5 A room is 7 m by 5 m and has a height of 3 m. Find the distance from a corner point

on the floor to the opposite corner of the ceiling.

6 A rectangular box is 2 cm by 3 cm by 2 cm (internally). Find the length of the longest

toothpick that can be placed within the box.

7 Determine the length of the longest piece of timber which could be stored in a rectangular

shed 6 m by 5 m by 2 m high.

A pyramid of height 40 m has a square base with edges 50 m.

Determine the length of the slant edges.

Let a slant edge have length s m.

Let half a diagonal have length x m.

Using x2 + x2 = 502 fPythagorasg

) 2x2 = 2500

) x2 = 1250

Using s2 = x2 + 402 fPythagorasg

) s2 = 1250 + 1600

) s2 = 2850

) s =p

2850 fas s > 0g

) s + 53:4 fto 1 dec. placeg

i.e., each slant edge is 53:4 m long.

diagonal

Example 17

50 m

x m x m

40 m

x m

s m

50 m

x m

s m40 m

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Y:\HAESE\SA_10-6ed\SA10-6_02\048SA10-6_02.CDR 17 August 2006 09:02:48 DAVID2


Click on the icon to obtain a printable exercise on more difficult

problems requiring a solution using Pythagoras’ Theorem.

8

9

10 A cube has sides of length 2 m. B is at the centre of one face,

and A is an opposite vertex. Find the direct distance from A

to B.

E

D

MA B

C

B

A

A symmetrical square-based pyramid has height cm and

slant edges of cm. Find the dimensions of its square base.

1015

ABCDE is a square-based pyramid. E, the apex of the pyramid

is vertically above M, the point of intersection of AC and BD.

If an Egyptian Pharoah wished to build a square-based pyramid

with all edges 100 m, how high (to the nearest metre) would

the pyramid reach above the desert sands?

MORE DIFFICULT PROBLEMS (EXTENSION)E

PRINTABLE

EXERCISE

REVIEW SET 2A

1 Find the lengths of the unknown sides in the following triangles:

a b c

2 Is the following triangle right angled?

Give evidence.

3 Show that f5, 11, 13g is not a Pythagorean triple.

4

4 cm

7 cmx cm 9 cm

x cm

2 cmx

A

B

C

1

4

��17

2 cm

x cm5 cm

5 An isosceles triangle has equal sides of length 12 cm and a base of length 8 cm. Find

the area of the triangle.

6 A boat leaves X and travels due east for 10 km. It then sails 10 km south to Y. Find

the distance and bearing of X from Y.

7 What is the length of the longest toothpick which can be placed inside a rectangular

box that is 3 cm £ 5 cm £ 8 cm?

A rectangle has diagonal cm and one side cm. Find the perimeter of the rectangle.15 8


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REVIEW SET 2B


8

1 Find the value of x in the following:

a b c

2 Show that the following triangle is right angled

and state which vertex is the right angle:

3 A rectangular gate is twice as wide as it is high. If a diagonal strut is 3:2 m long, find

the height of the gate to the nearest millimetre.

4 If a softball diamond has sides of length 30 m, determine the distance a fielder must

throw the ball from second base to reach home base.

5 Town B is 27 km in a direction 134oT from town A, and town C is 21 km in a direction

224oT from town B. Find the distance between A and C.

6 If a 15 m ladder reaches twice as far up a vertical wall as the base is out from the wall,

determine the distance up the wall to the top of the ladder.

7 Can an 11 m long piece of timber be placed in a rectangular shed of dimensions 8 m by

7 m by 3 m? Give evidence.

8 Straight roads from towns A and B inter-

sect at right angles at X. A and B are 52km apart. Mika leaves A at the same time

as Toshi leaves B. Mika cycles at 24 km/h

and Toshi jogs at 10 km/h, towards X, and

they arrive at X at the same time. For how

long were they travelling?

x cm

5 cm��7 cm

x m5 m

6 m

��

2x

5x

A

B

C

25

~`2`9

A

B

X

52 km

Two rally car drivers set off from town C

at the same time. A travels in a direction

T at kmph and B travels in a

direction T at kmph. How far

apart are they after one hour?

63 120333 135

o

o


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2. pythagoras' theorem

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