Pythagoras
Wackford Squeers
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Pythagoras’ Theorem State and apply Pythagoras’ Theorem to the solution of problems involving right-angle triangles. Perhaps the most famous Theorem in Mathematics – or at least the one everyone can remember! It states: “In a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides” So we can see what this means we have a right-angled triangle as follows:
LEARN: the side c is opposite the right-angle and is called the HYPOTENUSE
c
a
b The theorem tells us that:
a2 + b2 = c2 Illustration 1
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c
LEARN: in any triangles – the longest side is opposite the largest angle
4 We know that:
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c2 = 32 + 42 = 9 + 16 = 25 So c = 5 and that’s it! NOTE: To illustrate how the theorem works we have chosen ‘easy’ numbers – i.e. numbers which produce a whole number (integer) answer. This will not always be the case!
Illustration 2
Find the length of the hypotenuse in the following triangle (give your answer correct to one decimal place):
c
5
7 We know (from the theorem) that: C2 = 72 + 52 = 49 + 25 = 74 So
c
74 8.6023 8.6 (correct to 1 decimal place)
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A Different Kind of Problem
Sometimes you are asked to work out the length of another side (other than the hypotenuse). This means rearranging the equation to make that side the ‘subject’ of the equation (rather than the hypotenuse which is usually the subject) Look at the diagram:
c
a
b We know that the ‘normal’ equation is:
c2 = a2 + b2 and suppose we want to find the length of the side b The equation can be rewritten:
b2 = c 2 – a 2 [we subtracted b2 from both sides] The rearranged equation to find the other side will be:
a2 = c 2 – b 2
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Illustration Find the length of the side marked
a
a
11
8 a2 = 112 – 82 = 121 – 64 = 57
a
57 = 7.55 (to two decimal places)
Go the next page and do the exercises:
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Exercises Find the length of the side denoted by a letter (give your answers correct to one decimal place where appropriate):
a
6m
4.5m a
7cm 5cm
c
3m
b
8m
8.2cm a
e
4.2cm a
d
9m
7.5cm a
11cm f
12m 16.1cm a
4.3m a 22m
g
6.1m j h 3.9m
22.3cm
0.8m a
k
0.6m
16.2 m
44 m
m 8.7m
n 32m
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Extension Sailing Along 1
Jonah sets sail from his base B and travels due East for 8 miles. He then travels 6 miles due North. How far is he now from his base?
Sailing Along 2
Joanne sets sail form her base C and sails for 12 miles due West. She then sails 7 miles due South. How far away from her base is she?
Sailing Along 3
Peter sets sail from his base D and sails NorthWest for 10 miles. At the same time his fried William sails 4 miles due West (from the same point). How far apart are they?
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