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Home > Maths I > Measure and Shape > Pythagoras

Maths I

Pythagoras

Calculating the length of a shorter side

How can we use Pythagoras' Theorem to calculate the length of one of the shorter sides?

Relax, all you will need to do is 1 extra line of working.

Example

Calculate the length of the side marked a.

Give your answer to 2 decimal places.

A right angled triangle. The hypotenuse is labelled 12, the shortest edge is 8, and the medium edge is a.
  • Write the equation: 12^2 = a^2 + 8^2

  • Organise the equation: a^2 = 12^2 - 8^2

  • Square the lengths you know: a^2 = 144 - 64

  • Do the subtraction: a^2 = 80

  • Find the square root: a = \surd80

  • a = 8.94 (to 2 decimal places)

Now you try this type of question.

Question
A right angled triangle. The hypotenuse is labelled as 14, the shortest edge is 9, and the medium edge is labelled a.
Answer
  • 14^2 = a^2 + 9^2

  • a^2 = 14^2 - 9^2

  • a^2 = 196 - 81

  • a^2 = 115

  • a = \surd115

  • = 10.72 (to 2 decimal places)

Question
A right angled triangle. The hypotenuse is 20, the shortest edge is 11, and the medium edge is r.
Answer
  • 20^2 = r^2 + 11^2

  • r^2 = 20^2 - 11^2

  • r^2 = 400 - 121

  • r^2 = 279

  • r = \surd279

  • = 16.70 (to 2 decimal places)

Question
A right angled triangle. The hypotenuse is 3.8, the shortest edge is 2.4, and the medium edge is e.
Answer
  • 3.8^2 = e^2 + 2.4^2

  • e^2 = 3.8^2 - 2.4^2

  • e^2 = 14.44 - 5.76

  • e^2 = 8.68

  • e = \surd8.68

  • = 2.95 (to 2 decimal places)

Well done if you got the correct answers. You remembered to organise the equation. You squared the lengths you knew. You found the square root correctly.

If you got the wrong answers -

  • Your first equation must start with the hypotenuse squared.

  • Always remember the extra line and to organise the equation!

  • The hypotenuse must always be the longest side.

Now we can go on to solving problems using Pythagoras' Theorem.

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