x 2 + y 2 = z 2
One of the most important rules you will meet in mathematics – and it’s 2500 years old…..

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Briefly revisit some angle problems and the idea of similar triangles
Understand a special connection between the lengths of the sides of some triangles;
Know that this connection is called
Pythagoras’s Theorem
Use the Theorem in some problems..

1.
2.
3.
What is the total (sum) of all the exterior angles of ANY polygon?
What name do we give to two angles between two parallel lines which form a ‘Z’ shape?
What is the sum of all the angles inside a quadrilateral?

4.
The diagram shows part of a regular polygon that has 8 sides
(an octagon). What are the sizes of the angles x and y?
y x

5.
Look at the diagrams below. Find the sizes of angles a, b, c and d.
73
28 b a d c
6. What is the name given to angles 73 and c in the diagram above?

B
6
7
8
C
9
A
5 1
2
F
3
4
10
11
D
12
E
7.Give the three-letter code for angle 3.

1.
2.
3.
What is the total (sum) of all the exterior angles of ANY polygon?
(360)
What name do we give to two angles between two parallel lines which form a ‘Z’ shape? (Alternating)
What is the sum of all the angles inside a quadrilateral? (360)

4.
The diagram shows part of a regular polygon that has 8 sides
(an octagon). What are the sizes of the angles x and y?
y x
X = 360 ÷ 8 = 45 ° and y = 180 – 45 = 135 °

5.
a = b = 76 °
Look at the diagrams below. Find the sizes of angles a, b, c and d.
73 c=73
28 d=107 b a d c
6. What is the name given to angles 73 and c in the diagram above?
F-shape, so CORRESPONDING

B
6
7
8
C
9
10
11
D
2
3
5 1
4
A
F
6.Give the three-letter code for angle 3.
12
E
Angle 3 = CFD or DFC

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Shapes are similar when one is an
ENLARGEMENT of the other
The enlargement must be achieved by multiplying all the sides by the same amount, called the scale factor
Two shapes are NOT similar if we just add the same amount onto all the sides.
If the two shapes are identical size, they are called CONGRUENT instead.

1. Below are two SIMILAR triangles. Work out the length of the sides marked x and y, and the angle a.
4.5cm
y
52 °
4cm x
18cm a
12cm

1. Below are two SIMILAR triangles. Work out the length of the sides marked x and y, and the angle a.
4.5cm
y
52 °
4cm x
18cm a
12cm
The 4cm is enlarged to 12cm, so the scale factor is 3 . So x = 4.5 x 3
= 13.5 cm and y is 18 ÷ 3 = 6 cm. Angles never change, so a = 52 °

KEY QUESTION: Quite a hard GCSE-type problem:
Timpkins Builders make wooden frames for roofs on new houses.
In the diagram of the wooden frame shown below,
PQ is parallel to BC.
A
200 cm
P Q
NOT TO SCALE
400 cm
B
600 cm
C
Calculate length PQ using similar triangles.

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
For the following triangles, use a compass and a ruler to draw them accurately, putting the middle sized length as the base;
Then, for each triangle , multiply each side length by itself (square it), writing the three answers you get inside the triangle you have drawn.
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Do you notice anything about four of the triangles and the values you work out?
TRIANGLE 1: 3cm, 4cm, 5cm
TRIANGLE 2 : 5cm, 12cm, 13cm
TRIANGLE 3: 10cm, 8cm, 6cm
WHICH OF
THESE
TRIANGLES
IS THE ODD
ONE OUT
TRIANGLE 4: 3.5cm, 12cm, 12.5cm
TRIANGLE 5: 4cm, 7cm, 9.5cm
?

What you
have found:
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For any right-angled triangle the longest side (called the HYPOTENUSE ) squared equals the total of the other two sides squared!!
This rule is called PYTHAGORAS’S
THEOREM
It won’t work if the triangle has not got a right angle…..Like number 5!

EXAMPLE : Work out the length marked x in this triangle: x cm
9 cm
40 cm

EXAMPLE 2 : Work out the length marked x in this triangle (give your answer to 1 d.p) x cm
7.3 cm
12.8 cm

7.3 cm
Here’s my special method:
Step 1: Write the three sides in order of size – like this:
7.3
12.8
x (miss out cm)
Step 2: put ‘squares’ onto each number – like this:
7.3
2 12.8
2 x 2
Step 3: put a + and an = in the two gaps – like this:
7.3
2 + 12.8
2 = x 2
12.8 cm x cm
HYPOTENUSE
(make sure it’s at the end)

7.3 cm
Here’s my special method:
Step 4: Work out the two parts you can – like this:
53.29 + 163.84 = x 2
Step 5: Now add the first two answers – like this:
217.13 = x 2
Step 6: Now we need to know what number squared actually gives 217.13. For this we need the square root key – it looks like this: √ x cm x = √217.13
x = 14.7 cm to 1dp
12.8 cm

7.3 cm
….and finally (but crucial!):
Always CHECK sides….
your answer looks right. It has
got to be bigger (longer) than the other two
WHY????
Because it’s supposed to be the HYPOTENUSE which is the longest side!!
So x = 14.7 cm is probably OK.
x cm
12.8 cm

Question 3: Work out the length marked x in this triangle (give your answer to 1 d.p) x cm
6.6 cm
11.9 cm

6.6
6.6
2
11.9
11.9
2 x x 2
6.6
2 + 11.9
2 = x 2
43.56 + 141.61
= x 2
185.17
= x 2 x = √185.17
x = 13.6 cm to 1dp x cm
CHECK : Does it look right?
6.6 cm
11.9 cm

Questions: Work out the length marked x in these triangles (give your answer to 1 d.p)
Q1
Q2
3.7 cm
15 cm x cm
Q3 x cm
6.6 cm d cm
6.1 cm
4.5 cm
7.4 cm

Q1: x = √63.81 = 8.0 cm (to 1 dp)
Q2: x = √238.69 = 15.4 cm (to 1 dp)
Q3: d = √91.97 = 9.6 cm (to 1 dp)

EXAMPLE Work out the length marked x in this triangle (give your answer to 1 d.p)
Well, we stick with the same method as before!!
x cm
17.3 cm
13.6 cm

x cm
So now for the special method:
Step 1: Write the three sides in order of size – like this: x 13.6
17.3
Step 2: put ‘squares’ onto each number – like this: x 2 13.6
2 17.3
2
Step 3: put a + and an = in the two gaps – like this: x 2 + 13.6
2 = 17.3
2
13.6 cm
17.3 cm
HYPOTENUSE
(Again, it’s at the end)

x cm
Here’s my special method:
Step 4: Work out the two parts you can – like this: x 2 + 184.96
= 299.29
Step 5: Now subtract these two answers – like this: x 2 = 299.29 – 184.96
x 2 = 114.33
Step 6: Now we need to know what number squared actually gives 114.33. For this we need the square root key – it looks like this: √ x = √114.33
17.3 cm x = 10.7 cm to 1dp
13.6 cm

x cm
….and finally (but crucial!):
Now again CHECK your answer looks right. It has got to be smaller than the hypotenuse…
WHY????
Because the HYPOTENUSE is the longest side!!
So x = 10.7 cm is probably OK.
17.3 cm
13.6 cm

Question 3: Work out the length marked x in this triangle (give your answer to 1 d.p)
12.5 cm x cm
8.9 cm

x x 2 x 2 x 2
8.9
8.9
2
+ 8.9
2
+ 79.21
x 2 x 2
12.5
12.5
2
= 12.5
2
= 156.25
= 156.25 – 79.21
= 77.04
x = √77.04
x = 8.8 cm to 1dp
CHECK : Does it look right?
x cm
12.5 cm
8.9 cm

Questions: Work out the length marked x in these triangles (give your answer to 1 d.p)
Q1
Q2
4.1 cm x cm
13.7cm
Q3
14.5 cm
9.6 cm
24 cm t cm x cm
17 cm

Q1: x = √118.09 = 10.9 cm (to 1 dp)
Q2: x = √170.88 = 13.1 cm (to 1 dp)
Q3: d = √287 = 16.9 cm (to 1 dp)

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Look at the problem on the next slide
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It’s like what you could get at
Foundation level …….
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And shows how you might be asked to apply Pythagoras’s theorem

A boat leaves a harbour and sails due North for
18Km , then turns East and sails for a distance of 25km . How far is the direct route back to the Harbour?
25km
18km x km

Solution
25km
18km
Does the answer
LOOK right??
x km
18 25 = x
18 2 25 2 = x 2
18 2 + 25 2 = x 2
324 + 625 = x 2
949 = x 2 x = 30.8km