Year 9 - Pythagoras (applied) - Free practice

Year 9 Pythagoras & Applied Geometry Worksheets

Combine Pythagoras with similarity, ratio and trig.

Quick answer

How do you apply Pythagoras' theorem to real-world and 3D problems in Year 9?

Identify a right-angled triangle within the situation, label the two shorter sides and the hypotenuse, then use a² + b² = c². In 3D problems such as finding a box's space diagonal, apply Pythagoras twice: first across the base to get the base diagonal, then use that result as one leg of a second triangle with the height.

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Six questions across Foundation, Standard and Extension, with full worked solutions, written to the Australian Curriculum.

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Worked examples

Year 9 pythagoras (applied): worked examples

Three examples, from a standard question to a worded problem. Have a go at each one before you open the solution.

  1. Example 1 · Diagonal of a rectangular prism

    A rectangular box has length 1212 cm, width 55 cm and height 88 cm. Find the length of the longest diagonal of the box, correct to one decimal place.

    Show worked solution

    Key idea: The longest diagonal of a rectangular prism goes from one corner to the opposite corner. Use Pythagoras twice: first on the base, then use that result with the height.

    1. Find the diagonal dd of the base using Pythagoras:
      d2=122+52=144+25=169d=13 cm\begin{gathered}d^2 = 12^2 + 5^2 = 144 + 25 = 169 \\ d = 13 \text{ cm}\end{gathered}
    2. Use dd and the height to find the space diagonal DD:
      D2=d2+82=169+64=233D=233≈15.3 cm\begin{gathered}D^2 = d^2 + 8^2 = 169 + 64 = 233 \\ D = \sqrt{233} \approx 15.3 \text{ cm}\end{gathered}

    Answer: The longest diagonal is approximately 15.315.3 cm.

  2. Example 2 · Similarity and Pythagoras to find a missing length

    Two right-angled triangles are similar. In the smaller triangle, the two shorter sides are 66 cm and 88 cm. The longer of those two sides in the larger triangle is 2020 cm. Find the hypotenuse of the larger triangle.

    Show worked solution

    Key idea: Use Pythagoras to find the hypotenuse of the smaller triangle, then use the similarity ratio to scale it up.

    1. Find the hypotenuse of the smaller triangle:
      c2=62+82=36+64=100  ⟹  c=10 cmc^2 = 6^2 + 8^2 = 36 + 64 = 100 \implies c = 10 \text{ cm}
    2. The side of length 88 cm in the smaller triangle corresponds to the side of length 2020 cm in the larger triangle. Find the scale factor:
      k=208=2.5k = \frac{20}{8} = 2.5
    3. Multiply the smaller hypotenuse by the scale factor:
      Larger hypotenuse=10×2.5=25 cm\text{Larger hypotenuse} = 10 \times 2.5 = 25 \text{ cm}

    Answer: The hypotenuse of the larger triangle is 2525 cm.

  3. Example 3 · Real-life problem: ladder against a wall

    A 6.56.5 m ladder leans against a vertical wall. The base of the ladder is on horizontal ground and sits 2.52.5 m from the base of the wall. A safety rule says the angle the ladder makes with the ground must be between 70°70° and 80°80°. Use trigonometry to find the angle the ladder makes with the ground, correct to one decimal place, and state whether the ladder is positioned safely.

    Show worked solution

    Key idea: The ladder, wall and ground form a right-angled triangle. Use the cosine ratio, since the adjacent side (ground distance) and hypotenuse (ladder) are known, to find the angle at the base.

    1. Identify the sides relative to the angle at the ground: adjacent =2.5= 2.5 m, hypotenuse =6.5= 6.5 m.
    2. Write the cosine ratio:
      cos⁡(θ)=2.56.5\cos(\theta) = \frac{2.5}{6.5}
    3. Solve for θ\theta:
      θ=cos⁡−1 ⁣(2.56.5)≈cos⁡−1(0.3846)≈67.4°\theta = \cos^{-1}\!\left(\frac{2.5}{6.5}\right) \approx \cos^{-1}(0.3846) \approx 67.4°
    4. Compare with the safe range of 70°70° to 80°80°: 67.4°<70°67.4° < 70°, so the ladder is outside the safe range.

    Answer: The ladder makes an angle of approximately 67.4°67.4° with the ground, which is not safe.

Common mistakes

Where students usually lose marks

Forgetting to apply Pythagoras twice in 3D

Student writes D2=122+52+82D^2 = 12^2 + 5^2 + 8^2 and jumps straight to D=233D = \sqrt{233} without justifying the formula.

Always show both steps clearly: find the base diagonal first using d2=l2+w2d^2 = l^2 + w^2, then find the space diagonal using D2=d2+h2D^2 = d^2 + h^2. Skipping steps can lead to errors and loses marks.

Using the wrong scale factor in similarity

Student matches the 66 cm side (not the 88 cm side) to 2020 cm, giving k=20/6k = 20/6 and an incorrect hypotenuse.

Always match corresponding sides carefully before forming the ratio. Label which side in the smaller triangle lines up with which side in the larger triangle before calculating the scale factor.

Mixing up sin, cos and tan

Student writes sin⁡(θ)=2.5/6.5\sin(\theta) = 2.5/6.5 when the known sides are the adjacent and hypotenuse, giving the wrong angle.

Use SOH-CAH-TOA every time: label the sides as opposite, adjacent and hypotenuse relative to the angle you want, then choose the correct ratio. Writing the labels on a sketch helps avoid this.

Rounding too early in a multi-step problem

Student rounds d=169=13d = \sqrt{169} = 13 cm correctly, but in a different problem rounds a decimal base diagonal early, causing the final answer to be inaccurate.

Keep full calculator accuracy in every intermediate step and only round at the very last step. Store intermediate values in your calculator's memory rather than writing down rounded decimals.

Try one yourself

Your turn

A rectangular box has a length of 99 cm, a width of 66 cm and a height of 66 cm. Find the length of the space diagonal of the box, correct to 2 decimal places. Give your answer in cm.

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About this worksheet

Why we built it

Year 9 students are expected to combine Pythagoras with similarity, scale and angle properties to solve more complex spatial problems. These worksheets push past the basic "find the hypotenuse" stage into multi-step applied problems.

What's covered

Sub-skills your student will practise

  • ✓Pythagoras combined with angle-property reasoning
  • ✓Pythagoras in 3D problems (diagonals of boxes)
  • ✓Similarity + Pythagoras to find missing lengths
  • ✓Scale drawings and ratio-based problems
  • ✓Applied problems mixing Pythagoras and basic trig

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Frequently asked

Questions parents ask about pythagoras (applied)

How is Year 9 Pythagoras different from Year 8?

Year 9 combines Pythagoras with similarity, scale, ratio, and angle properties - so problems are multi-step rather than 'find the hypotenuse'. Many Year 9 problems also feed into trig.

How is Pythagoras used in 3D problems?

Find the diagonal of a face first (using 2D Pythagoras), then use that diagonal as one of the shorter sides for a second Pythagoras on the body diagonal. So for a cuboid, body diagonal = sqrt(l^2 + w^2 + h^2).

What's an exact answer vs a decimal answer?

Exact answers leave surds in (sqrt(2), sqrt(5)) without converting to decimals. Decimal answers give an approximation (1.414, 2.236). Exam questions usually specify which is required.

Why is Pythagoras such a focus across Years 8-10?

It's the most-used theorem in school geometry. It underlies trigonometry, coordinate geometry (distance formula), 3D measurement, and circle geometry. Every subsequent topic assumes fluency with it.

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