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Pythagoras' Theorem0%

Measurement & Geometry · Topic 8 of 9

Pythagoras' Theorem

Video coming soon3 worked examples

Theory

The Golden Rule: In National 5 Applications of Mathematics, Pythagoras questions are strictly designed to be two-stage calculations. If you calculate a single missing side using a2=b2+c2a^2 = b^2 + c^2 and stop, you have only completed half the problem. You must always use your new measurement to complete a second step, such as finding a second missing side, calculating a total perimeter, or finding the depth of a liquid.

1. The Standard Formulae

  • Finding the hypotenuse (the longest side): Square the two shorter sides and add them together, then find the square root (c2=a2+b2c^2 = a^2 + b^2).
  • Finding a shorter side: Square the hypotenuse and the other shorter side and subtract the smaller from the larger, then find the square root (a2=c2−b2a^2 = c^2 - b^2).

2. Two-Stage Triangles

You will frequently be presented with two right-angled triangles joined together. You must use Pythagoras in the first triangle to find the length of the shared "hidden" side. You then use that newly calculated length as a dimension in the second triangle to find the final missing side.

3. Pythagoras in a Circle (Chords & Depths)

This is a classic context. You will be given a circular cross-section (like an oil tank or a pipe) and asked to find the depth of a liquid or the width of a table.

  • Step 1: Draw a straight vertical line down from the centre of the circle to the chord (the liquid surface), creating a right-angled triangle.
  • Step 2: Halve the length of the chord to find the base of your triangle.
  • Step 3: Use the circle's radius as the hypotenuse of your triangle.
  • Step 4: Calculate the missing vertical height. Finally, add or subtract this height from the full radius to find the final depth.

4. Significant Figures & Unrounded Answers

Pythagoras questions almost always feature an instruction to round your final answer to a specific number of significant figures. You must write down the full, unrounded decimal from your calculator display before you write your rounded answer.

⚠️ Common Examiner Traps

  • The "Wrong Dimensions" Trap: a common way to lose marks is knowing to use Pythagoras but failing to identify the correct dimensions from the diagram. When a boat sails from A to B, you must add the distances together to find the full length of the new triangle's base before calculating the hypotenuse.
  • The "Premature Rounding" Trap: When finding a shared side between two triangles, do not round the length of the shared side to 1 or 2 decimal places. Keep the exact square root in your calculator for the second stage to prevent compounding rounding errors.
  • The "Forgotten Final Step" Trap: In circle questions, candidates frequently calculate the height of the right-angled triangle and underline it as their final answer, forgetting that the question asked for the total depth of the water, which requires adding or subtracting that height from the radius.

Worked examples

Example 1

A boat sails due East from a harbour. A lighthouse is 400 metres due North of the harbour. When the boat is at position A, it is exactly 500 metres away from the lighthouse. The boat then sails a further 450 metres East to position B.

Calculate the direct distance from position B to the lighthouse. (4 marks)

Step 1: Focus on the first right-angled triangle (Harbour, Lighthouse, Position A). The hypotenuse is 500m and the vertical side is 400m. Calculate the horizontal distance from the harbour to A (subtracting because we are finding a shorter side).

x2=5002−4002=250,000−160,000=90,000x^2 = 500^2 - 400^2 = 250,000 - 160,000 = 90,000
x=90,000=300 metresx = \sqrt{90,000} = 300 \text{ metres}

Step 2: Calculate the total horizontal base for the second triangle (Harbour to Position B). Avoid the trap: you must add the new distance to the distance you just found!

300m+450m=750 metres300\text{m} + 450\text{m} = 750 \text{ metres}

Step 3: Focus on the large right-angled triangle (Harbour, Lighthouse, Position B). The vertical side is 400m and the horizontal base is 750m. Calculate the new hypotenuse (adding).

y2=4002+7502=160,000+562,500=722,500y^2 = 400^2 + 750^2 = 160,000 + 562,500 = 722,500
y=722,500=850 metresy = \sqrt{722,500} = 850 \text{ metres}

Final Answer: The direct distance from position B to the lighthouse is 850 metres.

Example 2

The diagram shows the circular cross-section of a drainage pipe.

  • The centre of the circle is O.
  • The surface of the water forms a chord AB measuring 1.6 metres.
  • The radius of the pipe is 1.0 metre.

Calculate the maximum depth, d, of the water. (4 marks)

Step 1: Create a right-angled triangle by drawing a line from O to the chord. Halve the chord to find the base of the triangle.

1.6÷2=0.8 metres1.6 \div 2 = 0.8 \text{ metres}

Step 2: The hypotenuse of this triangle is the radius of the pipe (1.0m). Calculate the missing vertical height, x, of the triangle.

x2=1.02−0.82=1.0−0.64=0.36x^2 = 1.0^2 - 0.8^2 = 1.0 - 0.64 = 0.36
x=0.36=0.6 metresx = \sqrt{0.36} = 0.6 \text{ metres}

Step 3: Calculate the final depth. The full radius from the centre to the bottom of the pipe is 1.0m. The water level is 0.6m below the centre.

d=1.0−0.6=0.4 metresd = 1.0 - 0.6 = 0.4 \text{ metres}

Final Answer: The depth of the water is 0.4 metres.

Example 3

A park features a concrete patio and a grass lawn.

  • The patio is in the shape of a right-angled triangle with a base of 6m and a height of 8m.
  • The longest side of the patio forms the base of the grass lawn.
  • The grass lawn is also a right-angled triangle, with a vertical height of 24m.

Calculate the length of the longest side of the grass lawn. Give your answer to 3 significant figures. (4 marks)

Step 1: Calculate the hypotenuse of the patio triangle (which is the shared side).

h2=62+82=36+64=100h^2 = 6^2 + 8^2 = 36 + 64 = 100
h=100=10 metresh = \sqrt{100} = 10 \text{ metres}

Step 2: Use this 10m shared side as the base for the grass lawn triangle. Calculate the new hypotenuse.

L2=102+242=100+576=676L^2 = 10^2 + 24^2 = 100 + 576 = 676
L=676=26 metresL = \sqrt{676} = 26 \text{ metres}

Step 3: Write the unrounded answer, then round to 3 significant figures.

Unrounded: 26.000...\text{Unrounded: } 26.000...
Rounded: 26.0 metres\text{Rounded: } 26.0 \text{ metres}

(Adding the .0 explicitly demonstrates 3 significant figures).

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9 Pythagoras Theorem questions with answers and video solutions