NCEA Level 1Surface areaAS 91947

Surface area: 7 exam questions solved step by step

What this lesson covers

  • Surface area is the total area of every face, so a net shows exactly what to add up.
  • Cones and spheres have curved surfaces with their own area formulas; the skill is knowing which pieces a solid has, and which faces are hidden.
Key idea

7 questions, one method

  • 1. Picture the net: list every face.
  • 2. Slant heights from Pythagoras.
  • 3. Leave out hidden faces.
  • 4. Add, keep π, round at the end.
Your turn

Question 1 · Achieved

  • A triangular prism is 20 cm long.
  • Its ends are right-angled triangles,
  • sides 5 cm, 12 cm and 13 cm.
  • Find its total surface area.
Question 1 · Achieved
Worked example

Q1: triangular prism, 20 cm

  • List the five faces.
  • Triangle: the right angle.
  1. 2 triangles, 3 rectangles
  2. Ends: 2 × ½ × 5 × 12 = 60sides at right angle
  3. 5 × 20 = 100rectangle
  4. 12 × 20 = 240rectangle
  5. 13 × 20 = 260rectangle
  6. SA = 60 + 100 + 240 + 260
  7. SA = 660 cm²
  8. 30 × 20 + 60 = 660perimeter × length
Your turn

Question 2 · Achieved

  • A closed cylinder has radius 3 cm
  • and height 10 cm.
  • Find its total surface area, to 1 d.p.
Worked example

Q2: cylinder, r = 3, h = 10

  • Closed: two circles.
  • Keep π until the end.
  1. SA = 2πr² + 2πrhcircles + curved side
  2. 2 × π × 3² = 18πtwo circles
  3. 2 × π × 3 × 10 = 60πcurved side
  4. SA = 18π + 60π = 78π cm²
  5. 78π = 245.044...
  6. SA ≈ 245.0 cm² (1 d.p.)
Your turn

Question 3 · Merit

  • A square-based pyramid has a base
  • 16 cm by 16 cm, vertical height 6 cm.
  • Find its total surface area.
  • Careful: 6 cm is not up the triangles.
Question 3 · Merit
Worked example

Q3: square-based pyramid

  • Slant height first.
  • Then base + 4 triangles.
  1. l² = 6² + 8²Pythagoras, half of 16
  2. l² = 36 + 64 = 100
  3. l = 10 cm
  4. Base: 16 × 16 = 256
  5. Triangle: ½ × 16 × 10 = 80
  6. 4 × 80 = 320four triangles
  7. SA = 256 + 320
  8. SA = 576 cm²
  9. Triangles with 6: SA = 448used 6, not l
Your turn

Question 4 · Merit

  • A solid cone has base radius 8 cm
  • and vertical height 6 cm.
  • Find its total surface area, to 1 d.p.
Worked example

Q4: cone, r = 8, h = 6

  • πrl needs the slant l.
  • Base + curved surface.
  1. l² = 8² + 6² = 64 + 36Pythagoras
  2. l = √100 = 10 cm
  3. SA = πr² + πrlbase + curved surface
  4. π × 8² = 64πbase
  5. π × 8 × 10 = 80πcurved surface
  6. SA = 64π + 80π = 144π cm²
  7. 144π = 452.389...
  8. SA ≈ 452.4 cm² (1 d.p.)
Your turn

Question 5 · Merit, in context

  • A climbing dome: a hemisphere, 6 m across.
  • Only the curved outside is painted.
  • 1 L of paint covers 8 m².
  • Paint comes in 1 L tins. How many tins?
Worked example

Q5: painting the dome

  • 6 m is the diameter.
  • Round the tins up.
  1. r = 6 ÷ 2 = 3 m6 m is the diameter
  2. Curved SA = ½ × 4πr² = 2πr²no floor
  3. 2 × π × 3² = 18π m²
  4. 18π = 56.548... m²
  5. 18π ÷ 8 = 7.068... L8 m² per litre
  6. 7 tinscover only 56 m²
  7. 8 tins of paintround up
Your turn

Question 6 · Excellence, in context

  • A water tank: a cylinder, radius 3 m,
  • height 5 m, with a cone roof on top,
  • vertical height 4 m.
  • Find its total surface area, to 1 d.p.
Worked example

Q6: which faces count?

  • Count what paint reaches.
  • Joined faces are hidden.
  1. Base circle: outside
  2. Cylinder side: outside
  3. Roof curve: outside
  4. Cylinder top, roof basejoined, inside
  5. l² = 3² + 4² = 9 + 16 = 25Pythagoras
  6. l = 5 m
Worked example

Q6: add the three pieces

  • Keep π in.
  • Round once, in m².
  1. l = 5 m
  2. Base: π × 3² = 9ππr²
  3. Side: 2 × π × 3 × 5 = 30π2πrh
  4. Roof: π × 3 × 5 = 15ππrl
  5. SA = 9π + 30π + 15π
  6. = 54π m²
  7. 54π = 169.646...
  8. SA ≈ 169.6 m² (1 d.p.)
Your turn

Question 7 · Excellence

  • A closed cylinder has height = diameter.
  • Its total surface area is 150π cm².
  • Find its radius and its height.
Worked example

Q7: form the equation

  • Name the unknown: r.
  • Height = diameter = 2r.
  1. Let radius = r, so h = 2r
  2. SA = 2πr² + 2πrh
  3. = 2πr² + 2πr(2r)h = 2r
  4. = 2πr² + 4πr²
  5. = 6πr²collect
  6. 6πr² = 150πthe given area
Worked example

Q7: solve, reject, check

  • A length cannot be negative.
  • Answer in words, with units.
  1. 6πr² = 150π
  2. r² = 25÷ 6π both sides
  3. r = 5 or r = −5
  4. r = −5a radius is positive
  5. h = 2 × 5 = 10 cm
  6. 2π × 5² = 50π2πr²
  7. 2π × 5 × 10 = 100π2πrh
  8. 50π + 100π = 150π
  9. Radius 5 cm, height 10 cm
Summary

The method that solved all seven

  • List every face, add, then round last.
StepHow
Every facepicture the net, list them
Slant heightPythagoras, not the height
Hidden facejoined faces do not count
Keep π inadd the π terms together
Round lastonce, in cm² or m²

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