NCEA Level 1Pythagoras' theorem in 2D and 3DAS 91947

Pythagoras: 7 exam questions solved step by step

What this lesson covers

  • In a right-angled triangle, the square on the hypotenuse equals the sum of the squares on the other two sides.
  • A 3D length is found by using Pythagoras' theorem twice, in two right-angled triangles that share a side.
Key idea

7 questions, one theorem

  • 1. Find the right angle.
  • 2. Hypotenuse c: opposite it, the longest.
  • 3. Finding c: add the squares.
  • 4. A shorter side: subtract.
  • 5. Keep it exact; round at the end.
Your turn

Question 1 · Achieved

  • A right-angled triangle has shorter
  • sides 5 cm and 9 cm.
  • Find the hypotenuse c, to 1 dp.
Question 1 · Achieved
Worked example

Q1: find the hypotenuse

  • c is opposite the 90°.
  • Add the squares.
  1. c² = 5² + 9²Pythagoras
  2. c² = 25 + 81 = 106
  3. c = √106square root
  4. √106 = 10.2956…calculator
  5. c ≈ 10.3 cm1 dp
  6. 10.3 > 9 and 10.3 > 5
Your turn

Question 2 · Achieved

  • A right-angled triangle has
  • hypotenuse 12 cm and a side of 8 cm.
  • Find the other side a, to 1 dp.
  • Careful: 12 cm is the longest side.
Question 2 · Achieved
Worked example

Q2: find a shorter side

  • a is a shorter side.
  • So subtract.
  1. a² + 8² = 12²Pythagoras
  2. a² = 12² − 8²−8² both sides
  3. a² = 144 − 64 = 80
  4. a = √80
  5. √80 = 8.9442…calculator
  6. a ≈ 8.9 cm1 dp
  7. 144 + 64 = 208, √208 ≈ 14.4
  8. 14.4 > 12: impossiblelonger than c
  9. 8.9 < 12
Your turn

Question 3 · Merit, in context

  • A 4.5 m ladder leans on a wall.
  • Its foot is 1.2 m from the wall.
  • How high up the wall does it reach?
  • Give your answer to 2 dp.
Question 3 · Merit, in context
Worked example

Q3: the ladder

  • Ladder is the hypotenuse.
  • Subtract, then root.
  1. h² + 1.2² = 4.5²Pythagoras
  2. h² = 4.5² − 1.2²−1.2² both sides
  3. h² = 20.25 − 1.44 = 18.81
  4. h = √18.81
  5. √18.81 = 4.3370…calculator
  6. h ≈ 4.34 m2 dp
  7. 4.34 < 4.5shorter than ladder
  8. It reaches about 4.34 m up
Your turn

Question 4 · Merit, in context

  • A builder marks out a deck corner.
  • Sides 2.4 m and 3.2 m, diagonal 4.1 m.
  • Is the corner a right angle? Explain.
Worked example

Q4: is it a right angle?

  • Use the converse.
  • Compare, then explain.
  1. 2.4² + 3.2² = 5.76 + 10.24
  2. = 16
  3. 4.1² = 16.81longest side
  4. 16 ≠ 16.81
  5. Not a right anglesquares do not match
  6. For 90°: diagonal = √16 = 4 m
Your turn

Question 5 · Merit, in context

  • A kayak paddles 1.2 km due east,
  • then 500 m due north.
  • How far is it from its start, in m?
Question 5 · Merit, in context
Worked example

Q5: the kayak

  • East, then north: 90°.
  • Same units first.
  1. √(1.2² + 500²) ≈ 500mixed units
  2. 1.2 km = 1200 m× 1000
  3. d² = 1200² + 500²Pythagoras
  4. d² = 1 440 000 + 250 000
  5. = 1 690 000
  6. d = √1 690 000 = 1300
  7. 1300 m from the start
  8. 1.2² + 0.5² = 1.69 = 1.3²in km
Your turn

Question 6 · Excellence

  • A box is 80 cm by 50 cm by 30 cm.
  • Will a 1 m fishing rod fit inside,
  • lying straight? Explain.
Question 6 · Excellence
Worked example

Q6: two triangles

  • Longest: space diagonal.
  • Floor first, keep exact.
  1. f² = 80² + 50²floor triangle
  2. f² = 6400 + 2500 = 8900
  3. f = √8900keep exact
  4. Triangle 2: sides f, 30, d
  5. d² = f² + 30²height at 90° to floor
  6. d² = 8900 + 900 = 9800
Worked example

Q6: does the rod fit?

  • Round only at the end.
  • Compare in one unit.
  1. d² = 9800
  2. d = √9800
  3. √9800 = 98.9949…calculator
  4. d ≈ 98.99 cm2 dp
  5. 1 m = 100 cm
  6. 98.99 cm < 100 cm
  7. No, the 1 m rod will not fit
Your turn

Question 7 · Excellence, in context

  • A tent is a square-based pyramid.
  • Its base is 6 m by 6 m.
  • Each sloping edge is 5 m long.
  • How tall is the centre pole? (2 dp)
Question 7 · Excellence, in context
Worked example

Q7: centre to corner first

  • Pole meets base at 90°.
  • Base triangle first.
  1. k = centre to a corner
  2. k² = 3² + 3²half of 6 each way
  3. k² = 9 + 9 = 18
  4. k = √18keep exact
  5. Triangle 2: pole h, k, edge 5
  6. h² + k² = 5²edge is hypotenuse
  7. h² + 18 = 25k² = 18
Worked example

Q7: the height of the pole

  • Subtract, then root.
  • Round at the end.
  1. h² + 18 = 25
  2. h² = 25 − 18 = 7−18 both sides
  3. h = √7
  4. √7 = 2.6457…calculator
  5. h ≈ 2.65 m2 dp
  6. 2.65 < 5shorter than edge
  7. The pole is about 2.65 m tall
Summary

The method that solved all seven

  • Right angle, hypotenuse, square, root, then round.
StepHow
Right anglefind it before anything else
Hypotenuseopposite 90°, the longest
Short sidesubtract, never add
Same unitsconvert before you square
Conversesquares match: a right angle
3D lengthstwo triangles, keep it exact
Roundingonly at the end, with units

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