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How do you use Pythagoras theorem to find a missing side in a right-angled triangle, including in three dimensions?

Using Pythagoras theorem to find a missing side of a right-angled triangle, including the converse to test for a right angle and applications in three dimensions.

A focused answer to the SQA National 5 Mathematics Pythagoras content, covering finding the hypotenuse and a shorter side of a right-angled triangle, using the converse to check for a right angle, and applying Pythagoras in three dimensions.

Generated by Claude Opus 4.89 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. The theorem
  3. Finding the hypotenuse
  4. Finding a shorter side
  5. The converse: testing for a right angle
  6. Pythagoras in three dimensions
  7. Examples in context
  8. Try this

What this dot point is asking

The SQA wants you to use Pythagoras theorem to find a missing side of a right-angled triangle, choosing correctly between finding the hypotenuse and finding a shorter side, to use the converse to test whether a triangle has a right angle, and to apply Pythagoras in three-dimensional shapes.

The theorem

In a right-angled triangle, the side opposite the right angle is the hypotenuse and is always the longest side. Pythagoras theorem links the three sides.

Finding the hypotenuse

When the unknown is the longest side, add the squares of the two known sides and take the square root.

Finding a shorter side

When the hypotenuse is known and a shorter side is unknown, subtract the square of the known shorter side from the square of the hypotenuse.

Deciding whether to add or subtract is the key skill: add when finding the hypotenuse, subtract when finding a shorter side.

The converse: testing for a right angle

The converse of Pythagoras lets you check whether a triangle contains a right angle. Square all three sides; if the two smaller squares add to the largest square, the angle opposite the longest side is a right angle.

Pythagoras in three dimensions

In a cuboid or pyramid, a space diagonal can be found by applying Pythagoras twice: first to a face to find a diagonal, then to a right-angled triangle that uses that diagonal and a perpendicular edge. Identify the right-angled triangle inside the solid, then proceed exactly as in two dimensions.

The skill is spotting the two right-angled triangles inside the solid and using the answer to the first as a side of the second.

Examples in context

Pythagoras measures distances that cannot be reached directly. A surveyor finds the straight-line distance across a rectangular field by treating it as the hypotenuse of the two sides. Builders check that a corner is square by measuring 33, 44 and 55 units along the walls and confirming 32+42=523^2 + 4^2 = 5^2. The same theorem gives the length of a roof rafter from its rise and run.

Try this

Q1. Find the hypotenuse of a triangle with shorter sides 99 cm and 1212 cm. [2 marks]

  • Cue. 81+144=225=15\sqrt{81 + 144} = \sqrt{225} = 15 cm.

Q2. A triangle has hypotenuse 1313 cm and one side 55 cm. Find the other side. [2 marks]

  • Cue. 16925=144=12\sqrt{169 - 25} = \sqrt{144} = 12 cm.

Q3. Is a triangle with sides 66, 88 and 1111 right-angled? [2 marks]

  • Cue. 36+64=10012136 + 64 = 100 \ne 121, so no.

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SQA National 5 20183 marksA right-angled triangle has shorter sides of 55 cm and 1212 cm. Calculate the length of the hypotenuse.
Show worked answer →

Use c2=a2+b2c^2 = a^2 + b^2 with the two shorter sides (1 mark): c2=52+122=25+144=169c^2 = 5^2 + 12^2 = 25 + 144 = 169 (1 mark). Take the square root: c=169=13c = \sqrt{169} = 13 cm (1 mark). Markers reward squaring and adding the shorter sides and the square root for the hypotenuse.

SQA National 5 20223 marksA ladder of length 55 m leans against a wall with its foot 1.51.5 m from the base. How far up the wall does it reach, to 2 decimal places?
Show worked answer →

The ladder is the hypotenuse, so find a shorter side: h2=521.52=252.25=22.75h^2 = 5^2 - 1.5^2 = 25 - 2.25 = 22.75 (2 marks). Take the square root: h=22.75=4.77h = \sqrt{22.75} = 4.77 m to 2 decimal places (1 mark). Markers reward subtracting to find the shorter side and the rounded height.

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