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Pythagoras' theorem in practical problems: QCE Essential Mathematics Unit 3

Syllabus dot point

“Identify right-angled triangles in practical situations and use Pythagoras' theorem to find an unknown side length, including ladders, ramps, roofs, diagonals and checking for square corners”

QCEEssential MathematicsUnit 3: Measurement, scales and chance6 min read

Quick answer

Pythagoras' theorem, c2=a2+b2c^2 = a^2 + b^2, links the sides of a right-angled triangle, where cc is the hypotenuse. Add squares to find the hypotenuse and subtract to find a shorter side. Spot the right-angled triangle in ladders, diagonals, roofs and ramps, and use the 3-4-5 rule to check square corners.

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  1. What this dot point is asking
  2. The answer
  3. Practice questions

What this dot point is asking

Many practical measurement problems hide a right-angled triangle: a ladder against a wall, a roof rafter, the diagonal of a room or screen, a ramp. You need to find the triangle, identify the hypotenuse, and use Pythagoras' theorem to find a missing length.

The answer

Pythagoras' theorem

In a right-angled triangle with shorter sides aa and bb and hypotenuse cc (the longest side, opposite the right angle):

c2=a2+b2c^2 = a^2 + b^2

Two cases
  • Finding the hypotenuse (longest side): add the squares, then take the square root. c=a2+b2c = \sqrt{a^2 + b^2}
  • Finding a shorter side: subtract the squares, then take the square root. a=c2−b2a = \sqrt{c^2 - b^2}

Always check that your answer makes sense: the hypotenuse must be the longest side.

Spotting the triangle

  • Ladders: ladder = hypotenuse, wall height and distance from the wall = shorter sides.
  • Diagonals of rectangles (screens, rooms, gates): the diagonal is the hypotenuse.
  • Roofs: the rafter is the hypotenuse; half the span and the roof rise are the shorter sides.
  • Ramps: the ramp surface is the hypotenuse; rise and horizontal run are the shorter sides.

Checking for square corners

If a2+b2=c2a^2 + b^2 = c^2, the triangle has a right angle. Builders use the 3-4-5 rule: mark 3 m along one wall and 4 m along the other; if the diagonal between the marks is exactly 5 m, the corner is square. Multiples (6-8-10, 1.5-2-2.5) work too.

Worked example

A gable roof spans 7.2 m and rises 1.5 m at the centre. Find the length of each rafter to the nearest centimetre (ignore overhang).

  1. Half the span = 3.6 m (one shorter side). Rise = 1.5 m (other shorter side).
  2. r2=3.62+1.52=12.96+2.25=15.21r^2 = 3.6^2 + 1.5^2 = 12.96 + 2.25 = 15.21
  3. r=15.21=3.9r = \sqrt{15.21} = 3.9 m.
  4. Each rafter is 3.90 m.
Common traps
Adding when you should subtract
If you know the hypotenuse, subtract.
Using the whole span for a roof
Each rafter covers half the span.
Forgetting the square root
c2=100c^2 = 100 means c=10c = 10, not 100.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation3 marks
A rectangular TV screen is 88 cm wide and 50 cm high. Find the length of its diagonal, to the nearest centimetre.
Show worked solution →

The diagonal is the hypotenuse.

c2=882+502=7744+2500=10 244c^2 = 88^2 + 50^2 = 7744 + 2500 = 10\,244

c=10 244=101.2c = \sqrt{10\,244} = 101.2, so the diagonal is 101 cm.

Marking guide: 1 mark for identifying the hypotenuse, 1 mark for the substitution, 1 mark for the answer.

core3 marks
A 4.5 m ladder leans against a wall with its foot 1.2 m from the wall. How high up the wall does it reach, to the nearest centimetre?
Show worked solution →

The ladder is the hypotenuse, so find a shorter side:

h2=4.52−1.22=20.25−1.44=18.81h^2 = 4.5^2 - 1.2^2 = 20.25 - 1.44 = 18.81

h=18.81=4.337h = \sqrt{18.81} = 4.337 m, so the ladder reaches 4.34 m (434 cm).

Marking guide: 1 mark for subtracting (shorter side), 1 mark for the working, 1 mark for the answer with units.

exam5 marks
A wheelchair ramp must rise 0.6 m to reach a doorway. Australian guidance for this kind of ramp is a gradient of no more than 1 in 14 (for every 1 m of rise, at least 14 m of horizontal run). Find the minimum horizontal run and the length of the ramp surface, and decide whether a 9 m ramp would meet the guidance.
Show worked solution →

Minimum horizontal run = 0.6 × 14 = 8.4 m.

Ramp surface length (hypotenuse):

L2=8.42+0.62=70.56+0.36=70.92L^2 = 8.4^2 + 0.6^2 = 70.56 + 0.36 = 70.92

L=70.92=8.42L = \sqrt{70.92} = 8.42 m.

9 m ramp. A ramp surface of 9 m with a 0.6 m rise has horizontal run 92−0.62=80.64=8.98\sqrt{9^2 - 0.6^2} = \sqrt{80.64} = 8.98 m. Since 8.98 m is more than 8.4 m, the gradient is gentler than 1 in 14, so it meets the guidance.

Marking guide: 1 mark for the run, 2 marks for the ramp length, 1 mark for the 9 m ramp's run, 1 mark for the justified decision.

Practise this

Sources & how we know this

Based on the official QCAA syllabus. Not yet individually reviewed by a person; always check QCAA and your school's materials for current rules.

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