← IGCSE / O Level Mathematics

IGCSE / O Level Mathematics · Topic 5 Mensuration · Chapter 27 · pp. 351–385

Perimeter, area and volume

Perimeter is the distance round the edge, area is the surface a shape covers, and volume is the space a solid fills. This chapter collects the formulas for flat shapes, circles and sectors, and then for prisms, cylinders, pyramids, cones and spheres, and shows how to use them step by step.

Learning objectives

  1. Find the perimeter and area of a rectangle, triangle, parallelogram and trapezium, and of shapes made from them.
  2. Find the circumference and area of a circle, and the arc length and area of a sector.
  3. Find the volume and surface area of a prism and a cylinder.
  4. Find the volume and surface area of a pyramid, a cone and a sphere (Extended).
  5. Rearrange a formula to find a missing length, and work with compound solids.

Book exercises: 27.1 to 27.23 and Student assessments 1, 2 and 3 (pp. 351–385).

Perimeter and area of flat shapes

🔑 Key ideas

  • Perimeter = the distance round the outside (units: cm, m). Area = surface covered (units: cm², m²).
  • The height must be measured at right angles to the base, not along a sloping side.
  • Any side can be the base of a triangle, as long as the height matches it.
  • For shapes made of several parts, split them, find each area and add (or subtract a hole).
Triangle ½ × b × h h b h Parallelogram b × h h a b Trapezium ½ × h × (a + b)
The triangle is half of a rectangle with the same base and height. The trapezium is the average of its two parallel sides, times the height.
ShapeAreaNote
RectangleA = l × wP = 2(l + w)
TriangleA = ½ × b × hh is at right angles to b
ParallelogramA = b × hh is the perpendicular height
TrapeziumA = ½ × h × (a + b)a and b are the parallel sides

📘 Example 1 · A triangle has base 8 cm and perpendicular height 5 cm. Find its area

  1. Write the formula: A = ½ × b × h.
  2. Substitute: A = ½ × 8 × 5.
  3. Work out: ½ × 8 = 4, and 4 × 5 = 20 cm².

📘 Example 2 · A trapezium has parallel sides 13 cm and 8 cm which are 8 cm apart. Find its area

  1. Write the formula: A = ½ × h × (a + b).
  2. Substitute: A = ½ × 8 × (13 + 8).
  3. Add inside the bracket: 13 + 8 = 21.
  4. Multiply: ½ × 8 = 4, and 4 × 21 = 84 cm².

📘 Example 3 · Find the shaded area: a 12 cm by 8 cm rectangle with a trapezium of parallel sides 3 cm and 5 cm, 5 cm apart, cut out

  1. Area of rectangle = 12 × 8 = 96 cm².
  2. Area of trapezium = ½ × 5 × (3 + 5) = 2.5 × 8 = 20 cm².
  3. Shaded area = whole minus hole = 96 − 20 = 76 cm².

📘 Example 4 · The area of a trapezium is 20 cm². Its parallel sides are a cm and 6 cm, 4 cm apart. Find a

  1. Substitute into A = ½h(a + b): 20 = ½ × 4 × (a + 6).
  2. Simplify: 20 = 2(a + 6).
  3. Divide by 2: 10 = a + 6.
  4. Subtract 6: a = 4 cm.
  5. Check: ½ × 4 × (4 + 6) = 20. ✔

✏️ Try it

  1. Find the area of a parallelogram with base 7 cm and perpendicular height 4 cm.
  2. Find the area of a triangle with base 6 cm and perpendicular height 9 cm.
  3. Find the area of a trapezium with parallel sides 5 cm and 9 cm, 4 cm apart.
Answers

1. 7 × 4 = 28 cm² · 2. ½ × 6 × 9 = 27 cm² · 3. ½ × 4 × (5 + 9) = 28 cm²

Next: circles, arcs and sectors →

Circles, arcs and sectors

🔑 Key ideas

  • Circumference C = 2πr = πd. Area A = πr².
  • The diameter is twice the radius: halve it before using πr².
  • An exact answer keeps π (for example 25π cm²). A decimal answer is only an approximation, so give it to 3 s.f.
  • An arc is a part of the circumference. A sector is a slice of the circle. Both are the fraction φ/360 of the whole.
φ arc r fraction of circle φ ÷ 360 arc length φ/360 × 2πr sector area φ/360 × πr² Same fraction, applied to the circumference or to the area.
A sector of angle φ is φ/360 of the circle, so its arc is φ/360 of the circumference and its area is φ/360 of the area.

📘 Example 5 · A circle has radius 3 cm. Find its circumference, exact and to 3 s.f.

  1. Formula: C = 2πr.
  2. Substitute: C = 2 × π × 3 = 6π (exact).
  3. As a decimal: 6 × 3.14159… = 18.8 cm (3 s.f.).

📘 Example 6 · The circumference of a circle is 12 cm. Find its radius to 3 s.f.

  1. Rearrange C = 2πr to r = C ÷ 2π.
  2. Substitute: r = 12 ÷ (2π) = 6 ÷ π.
  3. Calculate: r = 1.91 cm.
  4. Check: 2 × π × 1.91 = 12.0. ✔

📘 Example 7 · Find the arc length of a sector of angle 80° and radius 6 cm, exact and to 3 s.f.

  1. Fraction of the circle: 80 ÷ 360 = .
  2. Whole circumference: 2 × π × 6 = 12π.
  3. Arc = × 12π = cm (exact).
  4. As a decimal: 8.38 cm (3 s.f.).

📘 Example 8 · A sector has angle 45° and radius 12 cm. Find its area, exact and to 3 s.f.

  1. Fraction: 45 ÷ 360 = .
  2. Whole circle: π × 12² = 144π.
  3. Sector = × 144π = 18π cm² (exact).
  4. As a decimal: 56.5 cm² (3 s.f.).

📘 Example 9 · A sector has radius 7 cm and arc length 12.4 cm. Find the angle φ

  1. Write the arc formula: 12.4 = φ/360 × 2 × π × 7.
  2. Multiply out: 2 × π × 7 = 43.98…, so 12.4 = φ/360 × 43.98.
  3. Rearrange: φ = 12.4 × 360 ÷ 43.98.
  4. Calculate: φ = 101.5° (1 d.p.).

💡 The perimeter of a sector is the arc plus both radii. A semicircle's perimeter is half the circumference plus the diameter.

✏️ Try it

  1. A circle has radius 5 cm. Give its circumference in exact form.
  2. Find the arc length of a sector of angle 60° and radius 9 cm, in exact form.
  3. Find the area of a sector of angle 90° and radius 4 cm, in exact form.
Answers

1. 2 × π × 5 = 10π cm · 2. × 18π = 3π cm · 3. × 16π = 4π cm²

Next: prisms and cylinders →

Prisms, cuboids and cylinders

🔑 Key ideas

  • A prism has the same cross-section all the way along its length.
  • Volume of a prism = area of cross-section × length.
  • Surface area = the sum of the areas of all its faces (picture the net).
  • Cuboid: V = lwh and SA = 2(lw + lh + wh).
  • Cylinder: V = πr²h and SA = 2πr² + 2πrh = 2πr(r + h).
  • Keep all lengths in the same unit before multiplying (change mm to cm first).
πr² 2πr × h curved face πr² Net of a cylinder A cross-section A length V = A × length
Left: a cylinder unrolls into a rectangle (2πr by h) and two circles. Right: a prism is its cross-section dragged along its length.

📘 Example 10 · Find the surface area of a 10 cm by 7 cm by 5 cm cuboid

  1. Top and bottom: 2 × 10 × 7 = 140 cm².
  2. Front and back: 2 × 10 × 5 = 100 cm².
  3. Two sides: 2 × 7 × 5 = 70 cm².
  4. Add: 140 + 100 + 70 = 310 cm².

📘 Example 11 · A cylinder has radius 4 cm and height 10 cm. Find its volume and its surface area

  1. Volume: V = πr²h = π × 4² × 10 = 160π.
  2. As a decimal: V = 503 cm³ (3 s.f.).
  3. Surface area: SA = 2πr(r + h) = 2 × π × 4 × (4 + 10).
  4. Simplify: 8π × 14 = 112π.
  5. As a decimal: SA = 352 cm² (3 s.f.).

📘 Example 12 · A cylinder has radius 3 cm and surface area 188 cm². Find its height to 1 d.p.

  1. Write: 188 = 2π × 3 × (3 + h) = 6π(3 + h).
  2. Divide by 6π: 188 ÷ 18.85 = 9.97… = 3 + h.
  3. Subtract 3: h = 9.97 − 3 = 6.97.
  4. h = 7.0 cm (1 d.p.).
  5. Check: 2π × 3 × 10 = 188.5, which is about 188. ✔

📘 Example 13 · A cuboid and a cylinder have equal volume. The cylinder has r = 2.5 cm and h = 8 cm. The cuboid is 5 cm by 5 cm. Find its height

  1. Cylinder volume: π × 2.5² × 8 = 50π = 157.08 cm³.
  2. Cuboid volume: 5 × 5 × h = 25h.
  3. Set equal: 25h = 157.08.
  4. Divide: h = 6.28.
  5. h = 6.3 cm (1 d.p.). Exactly, h = 2π.

✏️ Try it

  1. Find the surface area of a cuboid 2 cm by 3 cm by 4 cm.
  2. A cylinder has r = 2 cm and h = 5 cm. Give its volume and surface area in terms of π.
  3. A triangular prism has a right-angled end with sides 3, 4 and 5 cm and is 10 cm long. Find its volume and surface area.
Answers

1. 2(6 + 8 + 12) = 52 cm² · 2. V = π × 4 × 5 = 20π cm³; SA = 2π × 2 × (2 + 5) = 28π cm² · 3. Cross-section ½ × 3 × 4 = 6, so V = 60 cm³. SA = 2 × 6 + (3 + 4 + 5) × 10 = 12 + 120 = 132 cm²

Next: pyramids and spheres →

Pyramids and spheres Extended

🔑 Key ideas

  • Pyramid: V = ⅓ × base area × perpendicular height. Surface area = base + all the triangular faces.
  • Sphere: V = πr³ and SA = 4πr².
  • Hemisphere (half a sphere): V = πr³. A solid hemisphere has a curved surface 2πr² plus a flat circle πr², so SA = 3πr².
  • The perpendicular height of a pyramid is not the slant height of a face.
  • To find r from a volume, rearrange, then take the cube root.
h Pyramid apex r Sphere Pyramid V = ⅓ × base × h Sphere V = 4/3 πr³ SA = 4πr²
A pyramid uses the vertical height h from apex to base. A sphere needs only its radius r.

📘 Example 14 · A rectangular-based pyramid has base 7 cm by 3 cm and perpendicular height 5 cm. Find its volume

  1. Base area = 7 × 3 = 21 cm².
  2. V = ⅓ × 21 × 5.
  3. ⅓ × 21 = 7, and 7 × 5 = 35 cm³.

📘 Example 15 · A rectangular-based pyramid with base 8 cm by 5 cm has volume 60 cm³. Find its height

  1. Base area = 8 × 5 = 40 cm².
  2. 60 = ⅓ × 40 × h.
  3. Multiply by 3: 180 = 40h.
  4. Divide by 40: h = 4.5 cm.
  5. Check: ⅓ × 40 × 4.5 = 60. ✔

📘 Example 16 · A sphere has radius 3 cm. Find its volume in exact form and to 3 s.f.

  1. Formula: V = πr³.
  2. Substitute: V = × π × 3³ = × 27π.
  3. Simplify: 36π cm³.
  4. As a decimal: 113 cm³ (3 s.f.).

📘 Example 17 · A sphere has volume 150 cm³. Find its radius to 3 s.f.

  1. Write: 150 = πr³.
  2. Make r³ the subject: r³ = 150 × 3 ÷ (4π) = 450 ÷ 12.566 = 35.81.
  3. Cube root: r = 3.30 cm.
  4. Check: × π × 3.30³ ≈ 150.5. ✔

📘 Example 18 · A steel ball of radius 20 cm is melted into 8 equal small balls. Find the radius of each

  1. Big ball volume = π × 20³ = π.
  2. One small ball has ⅛ of that: πr³ = π.
  3. Cancel π: r³ = 1000.
  4. r = 10 cm. Check: 8 × 10³ = 8000 = 20³. ✔

✏️ Try it

  1. A sphere has radius 2 cm. Give its surface area and volume in terms of π.
  2. Find the total surface area of a solid hemisphere of radius 3 cm, in terms of π.
  3. A pyramid has a rectangular base 3 cm by 4 cm and height 6 cm. Find its volume.
Answers

1. SA = 4π × 4 = 16π cm²; V = π × 8 = cm³ · 2. 2π × 9 + π × 9 = 27π cm² · 3. ⅓ × 12 × 6 = 24 cm³

Next: cones and compound solids →

Cones and compound solids Extended

🔑 Key ideas

  • A cone is a pyramid with a circular base: V = ⅓πr²h, with h the vertical height.
  • Slant height l = √(r² + h²) (Pythagoras on the right-angled triangle r, h, l).
  • Curved surface area = πrl. Total surface area = πr² + πrl.
  • A cone is made by bending a sector. The sector's radius is l, and its arc becomes the base circumference 2πr.
  • For a compound solid, add the volumes of the parts. For its surface area, count only the faces on the outside.
sector l arc = 2πr bend h r l cone l² = r² + h² V = ⅓πr²h SA = πr² + πrl
The curved face of a cone is a sector of radius l. Its arc length becomes the circumference of the base.

📘 Example 19 · A cone has base radius 4 cm and vertical height 8 cm. Find its volume

  1. Formula: V = ⅓πr²h.
  2. Substitute: V = ⅓ × π × 4² × 8 = (exact).
  3. As a decimal: 134 cm³ (3 s.f.).

📘 Example 20 · A cone has base radius 5 cm and slant height 12 cm. Find its total surface area

  1. Base: πr² = π × 25 = 25π cm².
  2. Curved face: πrl = π × 5 × 12 = 60π cm².
  3. Add: 25π + 60π = 85π cm² (exact).
  4. As a decimal: 267 cm² (3 s.f.).

📘 Example 21 · A sector of radius 9 cm and angle 240° is bent into a cone. Find the base radius, the height and the volume

  1. Arc of the sector = × 2π × 9 = × 18π = 12π cm.
  2. This is the base circumference: 2πr = 12π, so r = 6 cm.
  3. The slant height is the sector radius: l = 9 cm.
  4. Pythagoras: h² = 9² − 6² = 81 − 36 = 45, so h = √45 = 3√5 = 6.71 cm.
  5. V = ⅓ × π × 6² × 3√5 = 36√5 π = 253 cm³ (3 s.f.).

📘 Example 22 · An ice cream is a hemisphere of radius 2.5 cm on top of a cone. The whole is 10 cm tall. Find the volume in exact form

  1. The cone's height is 10 − 2.5 = 7.5 cm (the hemisphere takes up 2.5 cm).
  2. Cone: ⅓ × π × 2.5² × 7.5 = ⅓ × π × 6.25 × 7.5 = 15.625π.
  3. Hemisphere: × π × 2.5³ = × 15.625π = 10.4167π.
  4. Add: 15.625π + 10.4167π = 26.0417π = cm³.
  5. As a decimal this is about 81.8 cm³.

🧪 Try it: shape calculator

    ✏️ Try it

    1. A cone has r = 3 cm and h = 4 cm. Find its slant height and its total surface area in terms of π.
    2. Find the volume of a cone with r = 6 cm and h = 5 cm, in terms of π.
    3. A sector of radius 10 cm and angle 144° is bent into a cone. Find the base radius.
    Answers

    1. l = √(9 + 16) = 5 cm. SA = π × 9 + π × 3 × 5 = 24π cm² · 2. ⅓ × π × 36 × 5 = 60π cm³ · 3. Arc = × 20π = 8π, so 2πr = 8π and r = 4 cm

    Next: summary and quiz →

    Chapter summary

    • Triangle ½bh · parallelogram bh · trapezium ½h(a + b). Height is at right angles to the base.
    • Circle: C = 2πr, A = πr². Arc = φ/360 × 2πr. Sector = φ/360 × πr².
    • Prism: V = cross-section × length. Cylinder: V = πr²h, SA = 2πr(r + h). Cuboid: SA = 2(lw + lh + wh).
    • Pyramid: V = ⅓ × base × h. Cone: V = ⅓πr²h, SA = πr² + πrl, l² = r² + h².
    • Sphere: V = πr³, SA = 4πr². Solid hemisphere: SA = 3πr².
    • Same unit throughout. Leave π in the answer for "exact form". Cube root for a missing radius from a volume.

    📄 Practice questions

    The book's Exercises 27.1 to 27.23 and Student assessments 1, 2 and 3 are in the printable assignment for this chapter.

    Mini-quiz

    1. A triangle has base 10 cm and perpendicular height 12 cm. What is its area?
    2. A trapezium has parallel sides 13 cm and 8 cm, 8 cm apart. What is its area?
    3. What is the circumference of a circle of radius 3.5 cm, to 3 significant figures?
    4. A circle has diameter 10 cm. What is its area in exact form?
    5. What is the length of an arc of angle 90° in a circle of radius 8 cm?
    6. What is the area of a sector of angle 60° and radius 6 cm?
    7. A cuboid measures 4 cm by 6 cm by 8 cm. What is its total surface area?
    8. A cylinder has radius 3 cm and height 7 cm. What is its volume?
    9. What is the volume of a sphere of radius 3 cm?
    10. A square-based pyramid has base side 6 cm and perpendicular height 5 cm. Its volume is:
    11. A cone has base radius 3 cm and vertical height 4 cm. What is its volume?
    12. A cone and a cylinder have the same radius and height. The cone holds what fraction?

    Source: Cambridge IGCSE Mathematics Core and Extended (4th edition, Pimentel and Wall), Chapter 27 (Perimeter, area and volume). Explanations, the Try it answers, the tool and the quiz are Guru Jazzy's own. This page is not made or endorsed by Cambridge or Hodder Education.