chapter 2chapter 2 pythagoraspythagoras - …€¦ · chapter 2 this is page 19 pythagoras 11 cm 7...

Chapter 2 this is page 17 Pythagoras

Pythagoras PythagorasChapter 2Chapter 2

Pythagoras

Pythagoras was a famous Greek Mathematician who discovered an amazing connection between the three sides of any right angled triangle.

This relationship, which connects the 3 sides, means it is possible to CALCULATE the length of one side of a right angled triangle as long as you know the lengths of the other two.

Look at this right angled triangle with sides 3 cm, 4 cm and 5 cm.

If you add the two smaller sides (3 cm and 4 cm) together, do you get the longer side (5 cm) ? – NO.

Can you see that • 32 = 9,

• 42 = 16,

• 52 = 25 ?

Can you also see that:- 32 + 42 = 9 + 16 = 25 = 52 ?

Pythagoras found that this connection between the three sides of a right angled triangle was true for every right angled triangle.

3 cm

4 cm

5 cm

Introductory Exercise 2.0 (confirmation - possibly orally)

1. The three sides of this right angled triangle are 6 cm, 8 cm and 10 cm.

(a) Write down the values of 62 , 82 and 102 .

(b) Find the value of 62 + 82 .

(c) Check that 62 + 82 = 102 .




(c) Check that 92 + 122 = 152 .




(c) Check that 52 + 122 = 132 .

8 cm

6 cm

10 cm

12 cm

13 cm5 cm

9 cm15 cm

12 cm


Exercise 2·1

1. Use Pythagoras’ Rule to calculate the length of the hypotenuse in this triangle :-

c2 = a2 + b2

=> c2 = 152 + ...

=> c2 = 225 + ...

=> c = ..... = ....

Copy and complete the working.

2. Use Pythagoras’ Rule to calculate the length of the hypotenuse in the right angled

triangle shown below.

16 cm

12 cm(c)

(b)

(a)

a cm

c cm b cm

hypotenuse

15 cm

8 cmc

3. Use Pythagoras’ Rule (referred to as PYTHAGORAS’ THEOREM) to calculate the length of the hypotenuse in each of these 3 triangles :-

(a)

(b)

(c)

36 cm

15 cmc

20 cm

15 cmc

6 cm

11·25 cm

cc

c 24 cm

7 cm

use your “√” button on the calculator

Pythagoras Theorem

Pythagoras came up with a simple rule which shows the connection between the three sides of any right angled triangle.

The longest side of a right angled triangle is called the HYPOTENUSE.

If the three sides are a cm, b cm and c cm (the hypotenuse), then Pythagoras’ Rule says :-

=> c2 = a2 + b2

We can use this rule to calculate the length of the hypotenuse of a right angled triangle if we know the lengths of the two smaller sides.

Example 1 :- The two smaller sides of this right angledtriangle are 12 centimetres and 16 centimetres.

To calculate the length of the hypotenuse, use Pythagoras’ Rule.

=> c2 = a2 + b2

=> c2 = 162 + 122

=> c2 = 256 + 144 = 400

=> c = 400 = 20 cm. This is how you set down the working.


11 cm

7 cmc

use your “√” button on the calculator

4. Use Pythagoras’ Theorem to calculate the length of the hypotenuse in this triangle .

5. Use Pythagoras’ Theorem to calculate the length of the hypotenuse in the right angled triangle shown .

6. Calculate the length of the hypotenuse marked p cm.

7. Calculate the length of the line marked q cm.

10 cm

c cm6 cm

14 cm

c cm8 cm

8. Calculate the length of the hypotenuse in this right angled triangle.

9. Sketch the following right angled triangles :-

Use Pythagoras’ Theorem to calculate the length of the hypotenuse in each case.

(a) (b)

(c) (d)

(e) (f)

15 cm

p cm

10 cm

17 cm22 cm

q cm

13·9 cm

6·1 cmx cm

9 cm

5 cmh cm

11 cm

18 cmh cm

5·2 m

12·3 m

h m

33 mm25 mm

h mm

16 cm

32 cm

h cm

4·5 cm

h cm8·5 cm

In most cases, the 3 sides are not exact values.

Example 2 :- => c2 = a2 + b2

=> c2 = 112 + 72

=> c2 = 121 + 49 = 170

=> c = 170 = 13·0384048...

= 13·04 cm

(to 2 decimal places),

(For the remainder of this exercise, give your answers correct to 2 decimal places).


Pythagorasc2 = a2

+ b2 Whenever you come across a problem involving finding a missing side in a right angled triangle, you should always consider using Pythagoras’ Theorem to calculate its length.

Problems involving Pythagoras’ Theorem

Exercise 2·2

(The triangles in these questions are right-angled)

1. A strong wire is used to supporta pole while the cement, holdingit at its base, dries.

Calculate the length of the wire.

2. A ramp is used to help push wheelchairs into the back of an ambulance.

Calculate the length of the ramp.

3. A plane left from Erin Isle airport. The pilot flew 175 kilometres West. He then flew 115 kilometres due North.

Calculate how far away the plane then was from Erin Isle.

4. A cable is used to help ferry supplies onto a yacht from the top of a nearby cliff.

Calculate the length of the cable used.

5. This wooden door wedge is 12·5 cm long.and 3·1 cm high.

Calculate the length of the sloping face.

6. This trapezium shape has a line of symmetryshown dotted on the figure.

Calculate the length of one of the sloping edges and hence calculate the perimeter of the trapezium.

ramp

0·9 m 5·2 m

North

175 km

115 km

Erin Isle

37 m

54 m

12·5 cm

3·1 cm

7·5 mwire

4 m

3·6 cm

4·2 cm

3·9 cm


7. A triangular corner unit (shown in yellow), is built to house a TV set.

Calculate the length of the long edge of the unit (x).

8. A lawn in Edinburgh’s Princes Street is in theshape of a rectangle 26 metres long by 14·5 metres wide.

A path runs diagonally through the lawn.

Calculate the length of the path.

9. The picture shows the side view of a conservatory.

Calculate the length of the sloping roof.

10. The roof of a garage is in the shape of anisosceles triangle.

Calculate the length of one side of the sloping roof.

11. Rhombus PQRS has its 2 diagonals, PR and QS,crossing at its centre C.

Calculate the PERIMETER of the rhombus.

12. Two wires are used to support a tree in dangerof falling down after a recent storm.

Calculate the total length of the support wires.

13. Calculate the PERIMETER of these 2 triangles.

(a)

(b)

40 cm

9 cm

26 m

14·5 mpath

P

Q

S

C20 cm R

25 cm

122 cm

x cm133 cm

3·2 m

1·9 m2·7 m

slopingroof

12·6 m

10·9 m 13·3 m

10·5 cm

28 cm

4·6 m

? m1·2 m


a cm

15 cm25 cm

Exercise 2·3

1. Calculate the length of the side of this right angled triangle marked with a t.

2. Calculate the size of each of the smaller sides in the following right angled triangles.

(a) (b)

3. A wheelchair ramphas a sloping side 8·2 m long and a horizontal base 7·1 m long.

Calculate the height of the ramp.

4. A helium balloon is tethered by a rope to the ground as shown opposite.

Calculate the height of the balloon.

5. This isosceles triangle has a base of 96 cmand a sloping edge of 52 cm.

Calculate the area of the triangle.

6. Shown is the side view of a wooden bread tin.

Calculate the length (x) of the base of the bin.

7. Calculate the perimeter of this right angledtriangle.

8. Shown is a right angled isosceles triangle ABC.

Calculate the value of t.

45 cm

27 cm

t cm

a2 = c2 – b2

=> t2 = 452 – 272

=> t2 = 2025 – 729

=> t2 = ........=> t = .... cm

e cm

8 cm15 cm

17 cm

f cm26 cm

52 cm

96 cm

H

C

BA t cm

30 cmt cm

40 cm

41 cm

x cm

36 cm

38 cm32 cm

8·2 m

7·1 m

?

can you see why the “–” sign ?

Calculating the Length of one of the Smaller Sides

You can use Pythagoras’ Theorem to calculate one of the smaller sides of a right angled triangle.

This time, you are asked to find the length of the smaller side (a) :-

=> a2 = c2 – b2

=> a2 = 252 – 152

=> a2 = 625 – 225 = 400

=> a = 400 = 20 cm

85 mh m

47 m


Mixed Examples

In the following exercise, if you are asked to find :-

the hypotenuse —> use c2 = a2 + b2 .

a shorter side —> use a2 = c2 – b2 .

You must decide which formula you have to use.

c

a

b

12 cm

x cm17 cm

(here, you are looking for a short side)

Example 1 :- Example 2 :- 15 cm

7 cm y cm(here, you are looking for the

long side)

note note

Exercise 2·4

1. Use the appropriate formula to find the value of x each time :–

(a) (b) (c)

(d) (e) (f)

2. Andy was answering this question in a class test.

His working was set down as shown :–

Why should Andy have known that his answer had to be wrong, by just comparing it to the lengthof the hypotenuse of the triangle ?

41 mm

33 mm

x mmx m

7·6 m

5·1 m

x cm

123 cm

142 cm

8 cm

4 cmx cm

x cm

9 cm20 cm

11·5 m

x m7 m

9 cm

14 cmx cm

x2 = 142 + 92

x2 = 196 + 81

x2 = 277

x = 277 = 16·6 cm

x2 = 172 – 122

x2 = 289 – 144

x2 = 145

x = 145 = 12·04 cm

y2 = 152 + 72

y2 = 225 + 49

y2 = 274

y = 274 = 16·55 cm


3. ONE of the following two answers is known to be the correct value for y in this question.

Without actually doing the calculation, saywhich one it must be and why the other isobviously wrong.

4. The tip of this pencil is in the shape of anisosceles triangle.

Calculate the width of the pencil (w).

5. This Scottish Flag is 2·35 metres long and 1·86metres wide.

What length must each diagonal strip be?

6. A cannon ball was fired and flew in a straightline for 450 metres where it exploded 85 metres above the enemy lines.

Calculate the distance (d m) from the cannon to the enemy soldiers.

14·0 cm

y cm

7·2 cmy = 9·6 cm ?

y = 15·7 cm ?

7. This warning sign isin the shape of an isosceles triangle.

Calculate the height of the sign.

8. A ladder was leaning against a wall. It began to slide away from the wall, but it stopped when its base came to rest against a smaller wall.

(a) Calculate the original height (H) of the top of the ladder above the ground.

(b) Calculate the new height (h) of the top of the ladder.

(c) By how many metres had the top of the ladder slipped ?

9. An orienteering competition was held over a triangular course.

From the start, the participants walk East to the 1st checkpoint, North to the 2nd one and then race back to the finishing line.

Calculate the overall distance of the event.

10. A lamppost fell over during a storm and cameto rest with its top resting against the top of a wall.

Calculate the height of the wall.

2·35 m

1·86 m

44 cm

48 cm

DANGER

3·4 m

6·3 m ? mwall

450 m85 m

d m

12·5 mm

11·3 mm w

2nd checkpoint

StartFinish 3·2 km

1·9 km

1st checkpoint•

•

•

7·2 m

2·9 m

before

H 7·2 m

after

h

6·1 m

AB2 = AP2 + PB2

AB2 = 82 + 62

AB2 = 64 + 36 = 100

AB = √100 = 10 boxes


Distances Between Coordinate Points

Consider the two coordinate points A(–3, –1) and B(5, 5).

They are plotted on the coordinate diagram opposite.

To calculate the distance from A to B :-

• draw in the 2 dotted lines to make a rightangled triangle APB.

• write down the lengths of the two sides AP and BP.

• use Pythagoras’ Theorem to calculate length of AB.

x

y

•

•

A

B

P

Exercise 2·5

1. (a) Make a copy of this coordinate diagram, showing the 2 points P(1, –1) and Q(4, 3).

(b) By drawing in the 2 dotted lines, create a right angled triangle and use it to calculate the length of the line PQ.

2. (a) Draw a new coordinate diagram and plot the 2 points M(– 4, 2) and N(8, 7)

(b) Create a right angle triangle in your figureand determine the length of the line MN.

3. Calculate the distance between the 2 points :- R(–2, 0) and S(5, 4),

giving your answer correct to 2 decimal places.

x

y

•

•

P

Q

4. For each pair of points below, calculate the length of the line joining them, giving your answer to 2 decimal places each time.

(a) F(2, – 4) and G(–1, 5)

(b) U(6, –2) and V(0, 4).

5. Terry thinks triangle AST below is isosceles.

To prove it is, he has to find the lengths of the2 lines AS and AT and show they are equal.

(a) Write down the length of the line AT.

(b) Calculate the length of the line AS.

(c) Was Terry correct ?

6. Prove that triangle LMN is isosceles whereL(–2, 2), M(6, 8) and N(4, –6).

x

y

•

T

A

S••


Exercise 2·6

1. Check if this triangle is right angled at Q.

Copy and complete :-

• PQ2 = 182 = 324,

• QR2 = 7·52 = .....

• PR2 = ....2 = ....

• PQ2 + QR2 = 324 + .... = ...... = PR2

• by the Converse of Pythagoras’ Theorem, triangle PQR must be r....... a...... at Q

2. Show that this triangle is NOT right angled.

i.e. (Show that UW2 + VW2 ≠ UV2)

The CONVERSE of Pythagoras’ Theorem

Pythagoras’ Theorem only works on a right angled triangle.

We can use Pythagoras Theorem “in reverse” to actually prove that a triangle is right angled.

Example :-

Look at triangle ABC opposite

We can prove it is right angled as follows :-

• Write down the 3 sides :- AB = 5·2, AC = 3·9, BC = 6·5.

• Square each side :- AB2 = 27·04, AC2 = 15·21, BC2 = 42·25.

• Add the two smaller squares together :- AB2 + AC2 = 27·04 + 15·21 = 42·25.

• Check if this is the same value as the largest square :- AB2 + AC2 = 42·25 = BC2.

• We say that, by the CONVERSE of Pythagoras’ Theorem, the triangle is proven to be right angled at A.

BC

A

6·5 cm

5·2 cm3·9 cm

18 cm

19·5 cm

7·5 cmP

Q

R

6·6 cm

11·1 cm

8·8 cmU

W

V

3. Decide which of the following is/are right angled triangles, and which is/are not :-

(a) (b)

4. A groundsman wishes to make sure the footballpitch is “square” (its corners are at 90°).

To check, he measures the diagonal length.Is the pitch “square” ?

5. Has this flagpole beenerected correctly, so that it is vertical ?

84 mm

91 mm

35 mm9·6 m

20·4 m

18·0 m

105 m

84 m

63 m

13·5 m 10·8 m

8·1 m


1. Calculate the lengths of the missing sides in thefollowing right angled triangles :-

2. Shown is an isosceles triangle.

(a) Calculate the heightof the triangle.

(b) Now calculate its area.

3. Calculate the area of the following rectangle :-

4. Calculate the perimeter of this right angled triangle :-

5. Calculate the value of x, which indicates the length of the slopingside of this trapezium.

7 cm

x cm7 cm

23 cm

18 cm

y cm

2·9 mm 3·1 mm

w mm

5·6 m

6·8 mz m

30 cm

39 cm

48 cm

50 cm

40 cm

9 cm

6·2 m

8·7 m

13·5 m

x m

6. This shape consists of a rectangle with an isosceles triangle attached to its end.

(a) Calculate the total length (L) of the figure.

(b) Now calculate its area.

7. Shown are the points F(–5, –2) and G(4, 3).

Draw a coordinate diagram, plot the two points and calculate the length of the line FG.

8. Draw a new set of axes, plot the 2 pointsM(–2, 6) and N(5, –3) and calculate the length of the line MN.

9. Prove that one of the following IS a right angled triangle and the other is NOT.

10. Prove that PQRS is a rectangle :-

22 cm

35 cm

24 cm

L cm

5·1 cm

6·8 cm

8·3 cm

20·8 cm

19·2 cm8·0 cm

16·5 cm

18·7 cm8·8 cm

P

S R

Q

Remember Remember..... ?Remember Remember..... ? Topic in aNutshell

A B

x

y

•

•

F

G

chapter 2chapter 2 pythagoraspythagoras - …€¦ · chapter 2 this is page 19 pythagoras 11 cm 7...

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