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Unit F7 Section 1
Pythagoras' Theorem

Pythagoras' Theorem gives a relationship between the lengths of the sides of a right angled triangle.

Pythagoras' Theorem states that:

In any right angled triangle, the area of the square on the hypotenuse (the side opposite the right angle) is equal to the sum of the areas of the squares on the other two sides (the two sides that meet at the right angle).

For the triangle shown opossite,

a^2 = b^2 + c^2

Note

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

Proof

Draw a square of side b + c , as shown opposite. Join up the points PQ, QR, RS, SP as shown, to give a quadrilateral, PQRS.

In fact, PQRS is a square as each side is equal to a (as the four triangles are congruent) and at the point P,

x + angle SPQ + y = 180°

But we know that x + y = 90° , so

angle SPQ = 90°

Similarly for the other three angles in PQRS. Thus PQRS is a square, and equating areas,

a^2 + 4 × ⎛
⎝bc⎞
⎠ = (b + c)^2

a^2 + 2bc = b^2 + 2bc + c^2

Hence

a^2 = c^2 + b^2

Worked Examples

1

Find the length of the hypotenuse of the triangle shown in the diagram. Give your answer correct to 2 decimal places.

As this is a right angled triangle, Pythagoras' Theorem can be used. If the length of the hypotenuse is a, then b = 4 and c = 6.

So

a2 = b^2 + c^2
a2 = 4^2 + 6^2
a2 = 16 + 36
a2 = 52
a =
= 7.2 cm (to one decimal place)
2

Find the length of the side of the triangle marked x in the diagram.

As this is a right angled triangle, Pythagoras' Theorem can be used. Here the length of the hypotenuse is 6 cm, so writing a = 6 cm and c = 3 cm with b = x , we have

a2 = b^2 + c^2
62 = x^2 + 3^2
36 = x^2 + 9
36 − 9 = x^2
27 = x^2
= x
x = 5.2 cm (to one decimal place)

Exercises

Find the length of the side marked x in each triangle.

(a)
x = m
(b)
x = cm
(c)
x = cm
(d)
x = m
(e)
x = m
(f)
x = cm
(g)
x = m
(h)
x = cm

Find the length of the side marked x in each triangle. Give your answers correct to 2 decimal places.

(a)
x = cm
(b)
x = cm
(c)
x = cm
(d)
x = m
(e)
x = cm
(f)
x = cm
(g)
x = m
(h)
x = m
(i)
x = cm
(j)
x = m
(k)
x = m
(l)
x = m
(m)
x = cm
(n)
x = m

Andre runs diagonally across a school field, while Rakeif runs around the edge.

(a)
How far does Rakeif run? m
(b)
How far does Andre run? m
(c)

How much further does Rakeif run than Andre?

m

Daisy is 1.4 metres tall. At a certain time her shadow is 2 metres long. What is the distance from the top of her head to the tip of her shadow?

m

A rope of length 10 metres is stretched from the top of a pole 3 metres high until it reaches ground level. How far is the end of the line from the base of the pole?

m

A rope is fixed between two trees that are 10 metres apart. When a child hangs on to the centre of the rope, it sags so that the centre is 2 metres below the level of the ends. Find the length of the rope.

m

The picture shows a shed. Find the length, AB, of the roof.

m

Rohan walks 3 km east and then 10 km north.

(a)

How far is he from his starting point?

km

(b)

He then walks east until he is 20 km from his starting point. How much further east has he walked?

km

Information

The Greeks, (in their analysis of the arcs of circles) were the first to establish the relationships or ratios between the sides and the angles of a right angled triangle.
The Chinese also recognised the ratios of sides in a right angled triangle and some survey problems involving such ratios were quoted in Zhou Bi Suan Jing. It is interesting to note that sound waves are related to the sine curve. This discovery by Joseph Fourier, a French mathematician, is the essence of the electronic musical instrument developments today.