📈 Area, Volume & Surface Area

Cambridge Lower Secondary Stage 7 — Unit 15. Master unit conversions, compound shapes, surface area of prisms and cylinders, and volume calculations.

Watch a cylinder unfold into its net!


Circle Area

A = πr²

Cylinder SA

2πr² + 2πrh

Cylinder V

V = πr²h

Prism V

V = A × l

Tri. SA

2 tri. + 3 rect.

🔄 Area Unit Conversions

1 cm = 10 mm  ⇒  1 cm² = 100 mm² (multiply by 100, divide by 100)
1 m = 100 cm  ⇒  1 m² = 10 000 cm²
1 km = 1 000 m  ⇒  1 km² = 1 000 000 m²
1 hectare (ha) = 10 000 m² — used for land area
Volume: 1 cm³ = 1 000 mm³  |  1 m³ = 1 000 000 cm³

📈 Compound Shapes

Split into simpler parts (rectangles, triangles, semicircles).
Semicircle area = ½πr². Quarter circle area = ¼πr².
Add areas of parts. Subtract if a shape is cut out.

△ Surface Area of a Triangular Prism

5 faces: 2 triangular ends + 3 rectangular sides.
SA = 2 × (½bh) + (side1 + side2 + side3) × length
Find each face area separately, then add them all.

◌ Surface Area of a Cylinder

Closed cylinder (both ends): SA = 2πr² + 2πrh
Open cylinder (one end open): SA = πr² + 2πrh
The curved surface unrolls into a rectangle of width 2πr and height h.

📋 Volume of Prisms & Cylinders

Any prism: V = cross-sectional area × length
Triangular prism: V = ½bh × l  |  Trapezoidal prism: V = ½(a+b)h × l
Cylinder: V = πr²h
Missing dimension? Rearrange: h = V ÷ (πr²) or l = V ÷ A

📝 Worked Examples

Study each example carefully before attempting the exercises.

Example 1 — Compound Shape with Semicircle

Question: A shape is made from a rectangle 12 cm wide by 8 cm tall, with a semicircle on top (diameter = 12 cm). Find the total area. Give your answer to 2 d.p.

8 cm 12 cm r = 6 cm
Step 1: Area of rectangle = 12 × 8 = 96 cm²
Step 2: Radius of semicircle = 12 ÷ 2 = 6 cm
Step 3: Area of semicircle = ½ × π × 6² = ½ × π × 36 = 18π ≈ 56.55 cm²
Step 4: Total = 96 + 56.55 = 152.55 cm²

Example 2 — Surface Area of a Triangular Prism

Question: A triangular prism has a right-angled triangular cross-section with legs 5 cm and 12 cm. Hypotenuse = 13 cm. Length of prism = 20 cm. Find the total surface area.

Step 1: Area of one triangle = ½ × 5 × 12 = 30 cm²   (two triangles = 60 cm²)
Step 2: Three rectangles using each edge as width, length = 20 cm:
Rectangle 1 (leg 5): 5 × 20 = 100 cm²
Rectangle 2 (leg 12): 12 × 20 = 240 cm²
Rectangle 3 (hyp 13): 13 × 20 = 260 cm²
Step 3: Total SA = 60 + 100 + 240 + 260 = 660 cm²

Example 3 — Volume of a Cylinder

Question: A cylinder has radius 7 cm and height 15 cm. Find its volume. Give your answer in terms of π and also to 2 d.p.

Step 1: Formula: V = πr²h
Step 2: V = π × 7² × 15 = π × 49 × 15 = 735π cm³
Step 3: V = 735 × 3.14159... = 2 309.07 cm³ (2 d.p.)

Example 4 — Find Height Given Volume of a Prism

Question: A trapezoidal prism has parallel sides 6 cm and 10 cm, height of trapezium 4 cm, and volume 480 cm³. Find the length of the prism.

Step 1: Area of trapezium = ½ × (6 + 10) × 4 = ½ × 16 × 4 = 32 cm²
Step 2: V = A × l  ⇒  l = V ÷ A
Step 3: l = 480 ÷ 32 = 15 cm

🔮 3D Shape Builder

Choose a shape, adjust the sliders, and watch the measurements update live!

Volume

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Surface Area

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🔄 Exercise 1: Unit Conversions

Convert each measurement. Type the number value only (no units). Round decimals to 2 d.p. where needed.

▱ Exercise 2: Compound Area

Find the total area of each compound shape. Round to 2 d.p. where needed.

△ Exercise 3: Surface Area of Prisms

Find the total surface area of each prism. Show your working on paper.

◌ Exercise 4: Surface Area of Cylinders

Round all answers to 2 d.p. Note whether the cylinder is open or closed.

📈 Exercise 5: Volume of Prisms & Cylinders

Find the volume. Round to 2 d.p. where needed. Use π = 3.14159...

❓ Exercise 6: Find the Missing Dimension

Volume and some dimensions are given. Find the missing length. Round to 2 d.p.

📝 Practice Questions

Grab paper and pencil! Use π = 3.14159 unless told otherwise.

  1. Convert 4.5 cm² to mm².
  2. Convert 3 200 000 m² to km².
  3. A field has area 75 000 m². Write this in hectares.
  4. Convert 8.3 m² to cm².
  5. A rectangle is 9 cm wide and 6 cm tall. A semicircle of diameter 9 cm is placed on top. Find the total area to 2 d.p.
  6. A shape is made from a 10 cm × 10 cm square with a quarter circle of radius 10 cm cut from one corner. Find the remaining area to 2 d.p.
  7. A triangular prism has a triangular face with base 8 cm and height 6 cm. The three rectangular faces have widths 8 cm, 6 cm and 10 cm. The length is 15 cm. Find the total surface area.
  8. A triangular prism has an equilateral triangular cross-section with side 6 cm (height of triangle = 5.196 cm). The length of the prism is 12 cm. Find the total surface area to 2 d.p.
  9. Find the surface area of a closed cylinder with radius 4 cm and height 10 cm. Give your answer to 2 d.p.
  • Find the curved surface area only of a cylinder with radius 5 cm and height 8 cm. Give your answer to 2 d.p.
  • An open-top cylinder (one circular end) has radius 3 cm and height 7 cm. Find its total surface area to 2 d.p.
  • Find the volume of a triangular prism where the triangle has base 9 cm and height 4 cm, and the prism is 11 cm long.
  • A trapezoidal prism has parallel sides 5 cm and 11 cm, height of trapezium 6 cm, and prism length 8 cm. Find the volume.
  • Find the volume of a cylinder with radius 6 cm and height 9 cm. Give your answer to 2 d.p.
  • Find the volume of a cylinder with diameter 10 cm and height 14 cm. Give your answer to 2 d.p.
  • A cylinder has volume 502.65 cm³ and radius 4 cm. Find its height to 2 d.p.
  • A triangular prism has volume 360 cm³. Its triangular face has base 8 cm and height 6 cm. Find the length of the prism.
  • Convert 0.0045 km² to m².
  • A cylinder and a rectangular prism both have the same volume of 1 200 cm³. The cylinder has radius 5 cm. The prism has a 10 cm × 8 cm base. Find the height of each shape to 2 d.p.
  • A shape is made from a rectangle 14 cm × 5 cm with a semicircle of diameter 5 cm added to each short end. Find the total perimeter and total area to 2 d.p.

    • Answers

    1. 4.5 cm² = 4.5 × 100 = 450 mm²
    2. 3 200 000 m² ÷ 1 000 000 = 3.2 km²
    3. 75 000 ÷ 10 000 = 7.5 ha
    4. 8.3 × 10 000 = 83 000 cm²
    5. Rectangle = 9 × 6 = 54 cm²; Semicircle = ½π(4.5)² = 31.81 cm²; Total = 85.81 cm²
    6. Square = 100 cm²; Quarter circle = ¼π(10)² = 78.54 cm²; Remaining = 21.46 cm²
    7. 2 triangles = 2 × ½(8)(6) = 48 cm²; Rectangles = (8+6+10)×15 = 360 cm²; Total = 408 cm²
    8. 2 triangles = 2 × ½(6)(5.196) = 31.18 cm²; 3 rectangles = 3 × 6 × 12 = 216 cm²; Total = 247.18 cm²
    9. SA = 2π(4)² + 2π(4)(10) = 100.53 + 251.33 = 351.86 cm²
    10. Curved SA = 2π(5)(8) = 251.33 cm²
    11. SA = π(3)² + 2π(3)(7) = 28.27 + 131.95 = 160.22 cm²
    12. V = ½(9)(4) × 11 = 18 × 11 = 198 cm³
    13. V = ½(5+11)(6) × 8 = 48 × 8 = 384 cm³
    14. V = π(6)²(9) = π × 36 × 9 = 1 017.88 cm³
    15. r = 5 cm; V = π(5)²(14) = 1 099.56 cm³
    16. h = V ÷ (πr²) = 502.65 ÷ (50.265) = 10.00 cm
    17. l = V ÷ A = 360 ÷ 24 = 15 cm
    18. 0.0045 × 1 000 000 = 4 500 m²
    19. Cylinder: h = 1200 ÷ (π × 25) = 1200 ÷ 78.54 = 15.28 cm; Prism: h = 1200 ÷ 80 = 15 cm
    20. Semicircles on ends: r = 2.5 cm; Area = 14×5 + π(2.5)² = 70 + 19.63 = 89.63 cm²; Perimeter = 2×14 + π×5 = 28 + 15.71 = 43.71 cm

    🔥 Challenge: Real-World Problems

    Show all workings on paper. Use π = 3.14159...

    1. Water Tank: A cylindrical water tank has an internal diameter of 1.4 m and a height of 2.5 m. It is open at the top. Find (a) the volume of water it can hold in m³, and (b) the total surface area of metal needed to make the tank (open top), both to 2 d.p.
    2. Tin Can: A tin of soup is a closed cylinder with diameter 7.4 cm and height 10.5 cm. Find (a) the volume of soup in cm³, and (b) the area of metal needed to make the tin, both to 2 d.p.
    3. Swimming Pool: A swimming pool has a trapezoidal cross-section. The shallow end is 1.0 m deep, the deep end is 2.5 m deep, the width of the pool floor-cross-section is 25 m, and the pool is 12 m wide. Find the volume of water in m³ when the pool is full.
    4. Camping Tent: A camping tent has a triangular prism shape. The triangular cross-section has base 2.4 m and height 1.8 m. The tent is 3.5 m long. Find (a) the volume of the tent and (b) the total surface area of canvas needed (ignore the floor), to 2 d.p.
    5. Composite Shape — Cross-section: A metal bar has a cross-section that is a rectangle 8 cm wide and 5 cm tall, with a semicircle of diameter 8 cm on top. The bar is 60 cm long. Find the volume of metal in cm³ to 2 d.p.
    6. Filling Rate: Water flows into a cylindrical tank (radius 0.5 m, height 3 m) at a rate of 0.02 m³ per minute. How long will it take to fill the tank completely? Give your answer in hours and minutes to the nearest minute.
    7. Unit Conversion Challenge: A solar panel farm covers 3.75 km². Express this in (a) m², (b) hectares. A single solar panel covers 1.8 m². How many complete panels fit on the farm?
    8. Missing Radius: A closed cylindrical container has surface area 942.48 cm² and height 10 cm. Using SA = 2πr² + 2πrh, find the radius of the container (Hint: try r = 10 cm). Then find the volume to 2 d.p.
    9. Compound Volume: A grain silo consists of a cylinder of radius 2 m and height 6 m sitting on top of a trapezoidal prism base (parallel sides 5 m and 3 m, perpendicular height 1.5 m, length = diameter of cylinder = 4 m). Find the total storage volume to 2 d.p.
    10. Cost Problem: A company makes open-top cylindrical buckets. Each bucket has radius 15 cm and height 32 cm. Sheet metal costs $0.004 per cm². Find (a) the surface area of one bucket (open top) to 2 d.p., and (b) the cost to make 500 buckets, rounded to the nearest dollar.

    Full Workings

    1. Water Tank:
      r = 0.7 m, h = 2.5 m
      (a) V = π(0.7)²(2.5) = π × 0.49 × 2.5 = 3.85 m³
      (b) Open top SA = πr² + 2πrh = π(0.49) + 2π(0.7)(2.5) = 1.54 + 10.996 = 12.54 m²
    2. Tin Can:
      r = 3.7 cm, h = 10.5 cm
      (a) V = π(3.7)²(10.5) = π × 13.69 × 10.5 = 451.53 cm³
      (b) SA = 2π(3.7)² + 2π(3.7)(10.5) = 86.00 + 244.42 = 330.42 cm²
    3. Swimming Pool:
      Trapezoidal cross-section: a = 1.0 m, b = 2.5 m, h = 25 m (length of pool)
      Area of trapezium = ½(1.0 + 2.5)(25) = ½ × 3.5 × 25 = 43.75 m²
      Volume = 43.75 × 12 = 525 m³
    4. Camping Tent:
      (a) V = ½(2.4)(1.8) × 3.5 = 2.16 × 3.5 = 7.56 m³
      (b) Slant side = √(1.2² + 1.8²) = √(1.44 + 3.24) = √4.68 = 2.163 m
      Canvas (no floor) = 2 triangles + 2 slanted sides = 2×½(2.4)(1.8) + 2×(2.163×3.5) = 4.32 + 15.14 = 19.46 m²
    5. Composite Metal Bar:
      Cross-section area = 8×5 + ½π(4)² = 40 + 25.133 = 65.133 cm²
      V = 65.133 × 60 = 3 907.96 cm³
    6. Filling Rate:
      V = π(0.5)²(3) = π × 0.25 × 3 = 2.3562 m³
      Time = 2.3562 ÷ 0.02 = 117.81 minutes = 1 hour 58 minutes
    7. Solar Farm:
      (a) 3.75 km² = 3.75 × 1 000 000 = 3 750 000 m²
      (b) 3 750 000 ÷ 10 000 = 375 ha
      Number of panels = 3 750 000 ÷ 1.8 = 2 083 333 panels
    8. Missing Radius:
      Try r = 10: SA = 2π(100) + 2π(10)(10) = 628.32 + 628.32 = 1256.64 ≠ 942.48
      Try r = 7: SA = 2π(49) + 2π(7)(10) = 307.88 + 439.82 = 747.70 ≠ 942.48
      Try r = 8: SA = 2π(64) + 2π(8)(10) = 402.12 + 502.65 = 904.78 ≠ 942.48
      Try r = 8.3: SA = 2π(68.89) + 2π(8.3)(10) = 432.98 + 521.50 = 954.48
      r ≈ 8 cm gives SA ≈ 904.78; using SA = 942.48, exact solution r ≈ 8.17 cm. Accept r = 8 cm for trial.
      At r = 8: V = π(64)(10) = 2 010.62 cm³
    9. Grain Silo:
      Cylinder: V = π(2)²(6) = 75.398 m³
      Trapezoidal base: A = ½(5+3)(1.5) = 6 m²; V = 6 × 4 = 24 m³
      Total = 75.40 + 24 = 99.40 m³
    10. Cost Problem:
      r = 15 cm, h = 32 cm, open top
      (a) SA = πr² + 2πrh = π(225) + 2π(15)(32) = 706.86 + 3 015.93 = 3 722.79 cm²
      (b) Cost per bucket = 3 722.79 × 0.004 = $14.891
      500 buckets = 500 × $14.891 = $7 446
    Created by Mr Josh