Learn what area, volume, surface area and capacity measure, how to convert their units, and how to find them for plane shapes, circles, sectors, prisms, cylinders, cones, pyramids and spheres, including composite solids, scale factors and bounds.
Area and volume are two of the most useful ideas in measurement. Area is the amount of flat surface a shape covers. Volume is the amount of space a solid takes up. They appear in building, farming, packaging, science and everyday decisions such as how much paint, how many tiles or how many litres of water you need. The real skill is not memorising formulas. It is recognising the shape, finding the right dimensions, choosing the right formula, keeping units consistent, and checking that the answer makes sense. This guide covers units and conversions, every common plane shape, circles and sectors, surface area, volume, capacity, composite solids, scale factors and bounds. It has 26 worked examples checked by computer, five diagrams, three calculators, a practice drill, exam questions with answers, and a 20-question quiz.
Area measures a two-dimensional region and is written in square units such as cm², m² or km². Volume measures the space inside a three-dimensional solid and is written in cubic units such as cm³ or m³. Capacity is the volume of liquid or gas a container can hold, usually in millilitres (mL) or litres (L).
| Quantity | Dimensions | Typical units | Meaning |
|---|---|---|---|
| Length, perimeter, circumference | 1D | mm, cm, m, km | Distance along a line or round a boundary |
| Area | 2D | cm², m², ha, km² | Flat surface covered |
| Surface area | 2D (on a 3D solid) | cm², m² | Total area of the outside of a solid |
| Volume | 3D | cm³, m³ | Space occupied by a solid |
| Capacity | 3D | mL, L | How much a container holds |
A classroom floor measures 8 m by 6 m. The floor is flat, so we want area: 8 × 6 = 48 m².
A storage room measures 8 m by 6 m by 2 m. The space inside is three-dimensional, so we want volume: 8 × 6 × 2 = 96 m³.
A number without its unit is incomplete. Length conversions use a scale factor. Because area is length × length, the factor for area is squared. Because volume is length × length × length, the factor for volume is cubed.
| Quantity | Conversion |
|---|---|
| Length | 1 m = 100 cm = 1000 mm; 1 km = 1000 m |
| Area | 1 m² = 10 000 cm²; 1 cm² = 100 mm²; 1 km² = 1 000 000 m² |
| Land area | 1 hectare (ha) = 10 000 m²; 1 km² = 100 ha |
| Volume | 1 m³ = 1 000 000 cm³; 1 cm³ = 1000 mm³ |
| Capacity | 1 L = 1000 mL = 1000 cm³; 1 mL = 1 cm³; 1 m³ = 1000 L |
Since 1 m = 100 cm, a 1 m by 1 m square is 100 cm by 100 cm. Its area is 100 × 100 = 10 000 cm², so 1 m² = 10 000 cm², not 100 cm².
A 1 m cube is 100 cm × 100 cm × 100 cm, so 1 m³ = 100³ = 1 000 000 cm³.
(a) 2.5 m² = 2.5 × 10 000 = 25 000 cm².
(b) 3500 cm² = 3500 ÷ 10 000 = 0.35 m².
(c) 0.4 m³ = 0.4 × 1 000 000 = 400 000 cm³ = 400 000 ÷ 1000 = 400 L.
(d) 2.5 km² = 2.5 × 100 = 250 hectares.
Does the factor 100 apply to areas?
The fix: 100 is the factor for lengths. For areas the factor is 100² = 10 000. So 3 m² = 3 × 10 000 = 30 000 cm². For volumes it would be 100³ = 1 000 000.
The area of a rectangle is A = lw. The area of a square of side s is A = s². The area of a parallelogram is A = bh, where b is the base and h is the perpendicular height. A parallelogram is a rectangle that has been slanted without changing its base or height, so the area stays the same. A rhombus (or any kite) with diagonals d₁ and d₂ has area ½d₁d₂.
A garden is 12 m long and 7.5 m wide. Area = 12 × 7.5 = 90 m². Fencing round the edge needs the perimeter, not the area: 2(12 + 7.5) = 39 m.
(a) A parallelogram has base 14 cm and perpendicular height 9 cm. A = 14 × 9 = 126 cm².
(b) A rhombus has diagonals 10 cm and 24 cm. A = ½ × 10 × 24 = 120 cm². Its side is √(5² + 12²) = 13 cm, because the diagonals cross at right angles and halve each other.
The area of a triangle is A = ½bh. The height must be perpendicular to the chosen base. The sloping side of a triangle is usually not the height.
(a) Base 18 cm and perpendicular height 11 cm: A = ½ × 18 × 11 = 99 cm².
(b) An isosceles triangle has base 10 cm and equal sides 13 cm. The height splits the base into two halves of 5 cm, so h = √(13² − 5²) = √144 = 12 cm. A = ½ × 10 × 12 = 60 cm².
Is the sloping side the perpendicular height?
The fix: the height must be at right angles to the base. The sloping side is longer than the height. Using Pythagoras, h = √(13² − 5²) = 12, so A = ½ × 10 × 12 = 60 cm².
If you know two sides and the angle between them, use A = ½ab sin C (see our sine and cosine rule article). If you know all three sides, use Heron’s formula: let s be half the perimeter, s = (a + b + c) ÷ 2, then A = √[s(s − a)(s − b)(s − c)].
(a) Two sides 8 cm and 10 cm enclose an angle of 30°. A = ½ × 8 × 10 × sin 30° = ½ × 80 × 0.5 = 20 cm².
(b) A triangle has sides 13 cm, 14 cm and 15 cm. s = 21. A = √(21 × 8 × 7 × 6) = √7056 = 84 cm².
A trapezium has one pair of parallel sides, a and b. Its area is A = ½(a + b)h, where h is the perpendicular distance between the parallel sides. It is the mean of the two parallel sides multiplied by the height.
Parallel sides 10 cm and 16 cm, height 7 cm. A = ½(10 + 16) × 7 = 13 × 7 = 91 cm².
For a compound shape, split it into familiar shapes, find each area, then add (or subtract for a hole or a missing corner).
(a) A 12 cm × 8 cm rectangle has a triangle attached on one 12 cm side, with perpendicular height 5 cm. Rectangle = 96 cm². Triangle = ½ × 12 × 5 = 30 cm². Total = 126 cm².
(b) A 10 cm × 8 cm rectangle has a 4 cm × 3 cm corner removed. Area = 80 − 12 = 68 cm².
A circle with radius r has diameter d = 2r. The circumference is C = 2πr = πd and the area is A = πr². Circumference is a length (a linear unit). Area is a surface (a square unit). If the question gives the diameter, halve it first.
(a) Radius 6 cm: area = π × 6² = 36π ≈ 113.1 cm². Circumference = 2π × 6 = 12π ≈ 37.7 cm.
(b) Diameter 10 cm: the radius is 5 cm, so area = π × 5² = 25π cm² (not 100π).
What should be squared?
The fix: the formula uses the radius. The radius is 10 ÷ 2 = 5 cm, so A = π × 5² = 25π cm² ≈ 78.5 cm². The wrong answer is four times too large.
Leave answers in terms of π when the question asks for an exact answer. When rounding is needed, keep full calculator precision until the last step.
A sector is a slice of a circle. If the angle at the centre is θ degrees, the sector is the fraction θ ÷ 360 of the whole circle.
(a) Radius 8 cm and angle 90°: area = ¼ × π × 64 = 16π cm². Arc = ¼ × 2π × 8 = 4π cm. The perimeter of the sector is 8 + 8 + 4π ≈ 28.6 cm, because the two straight radii are part of the boundary.
(b) Radius 12 cm and angle 150°: the fraction is 150 ÷ 360 = 5/12. Area = 5/12 × π × 144 = 60π ≈ 188.5 cm². Arc = 5/12 × 2π × 12 = 10π ≈ 31.4 cm.
A window is a rectangle 80 cm wide and 120 cm high with a semicircle of radius 40 cm on top. Area = 80 × 120 + ½ × π × 40² = 9600 + 800π ≈ 12 113.3 cm². Its perimeter goes round the two sides, the bottom and the curved top: 120 + 120 + 80 + π × 40 ≈ 445.7 cm. The straight line where the semicircle joins the rectangle is not part of the perimeter.
Surface area is the total area of the outside faces of a solid. It is measured in square units. A net is the solid unfolded flat, so its surface area is the area of the net.
(a) A 10 cm × 6 cm × 4 cm cuboid: 2(60 + 40 + 24) = 248 cm².
(b) A cube of side 5 cm: 6 × 25 = 150 cm².
The volume of a cuboid is V = lwh. A prism is a solid with the same cross-section all the way along its length, so V = cross-sectional area × length. If you can find the cross-section and the length, you can find the volume of any prism, including a cylinder.
(a) A tank 2.4 m × 1.5 m × 1.2 m: V = 4.32 m³.
(b) A prism has cross-section 35 cm² and length 18 cm: V = 35 × 18 = 630 cm³.
(c) A triangular prism has a right-angled triangle cross-section with sides 3, 4 and 5 cm, and length 10 cm. The cross-section area is ½ × 3 × 4 = 6 cm², so V = 6 × 10 = 60 cm³. Its surface area is two triangles, 2 × 6 = 12 cm², plus three rectangles with total width (3 + 4 + 5) = 12 and length 10, which is 120 cm². Total = 132 cm².
A cylinder is a prism with a circle as its cross-section. V = πr²h. Its curved surface unrolls into a rectangle of width 2πr and height h, so the curved surface area is 2πrh. The total surface area adds the two circular ends: 2πr² + 2πrh.
Radius 5 cm and height 12 cm. V = π × 25 × 12 = 300π ≈ 942.5 cm³. Curved surface area = 2π × 5 × 12 = 120π cm². Total surface area = 2π × 25 + 120π = 170π cm². Always check whether the question asks for the curved or the total surface area.
| Solid | Volume | Surface area |
|---|---|---|
| Cuboid | lwh | 2(lw + lh + wh) |
| Cylinder | πr²h | 2πr² + 2πrh |
| Pyramid | ⅓ × base area × height | base + all the triangular faces |
| Cone | ⅓πr²h | πr² + πrl (l = slant height) |
| Sphere | 4/3 πr³ | 4πr² |
| Hemisphere | 2/3 πr³ | 3πr² (curved 2πr² + flat πr²) |
The perpendicular height h is used for volume. The slant height l is used for the sloping surface. For a cone, l² = r² + h².
(a) A square-based pyramid has base side 6 cm and perpendicular height 4 cm. V = ⅓ × 36 × 4 = 48 cm³. The slant height of each triangular face is √(3² + 4²) = 5 cm, so each face has area ½ × 6 × 5 = 15 cm². Total surface area = 36 + 4 × 15 = 96 cm².
(b) A cone has radius 3 cm and height 4 cm. The slant height is √(3² + 4²) = 5 cm. V = ⅓ × π × 9 × 4 = 12π cm³. Curved surface area = π × 3 × 5 = 15π cm². Total surface area = 9π + 15π = 24π cm².
(a) A sphere of radius 3 cm: V = 4/3 × π × 27 = 36π cm³ and surface area = 4π × 9 = 36π cm².
(b) A hemisphere of radius 6 cm: V = 2/3 × π × 216 = 144π cm³. Its total surface area, including the flat circle, is 3π × 36 = 108π cm².
1 cm³ holds 1 mL, so 1000 cm³ = 1 L, and 1 m³ = 1000 L. Use internal dimensions for capacity, because the thickness of the walls takes up room.
(a) A tank measures 80 cm × 50 cm × 40 cm inside. V = 160 000 cm³ = 160 L. A tap that delivers 20 L per minute fills it in 160 ÷ 20 = 8 minutes.
(b) A cylindrical tank has radius 0.5 m and height 1.4 m. V = π × 0.25 × 1.4 ≈ 1.0996 m³, which is about 1100 litres.
For a composite solid, split it into familiar solids, find each volume, then add or subtract. For surface area, count only the faces that are actually on the outside. A face where two solids are joined is no longer exposed.
Both have radius 3 cm and the cylinder is 8 cm high. Volume = π × 9 × 8 + 2/3 × π × 27 = 72π + 18π = 90π ≈ 282.7 cm³.
The outside surface, standing on its flat base, is the base circle π × 9 = 9π, the curved side of the cylinder 2π × 3 × 8 = 48π, and the curved hemisphere 2π × 9 = 18π. Total = 9π + 48π + 18π = 75π cm². The circle where they join is inside, so it is not counted.
(a) A pipe has outer radius 5 cm, inner radius 3 cm and length 20 cm. Material = outer − inner = π × 25 × 20 − π × 9 × 20 = 500π − 180π = 320π cm³.
(b) A block 20 cm × 10 cm × 5 cm has a cylindrical hole of radius 2 cm drilled through its 5 cm thickness. Volume = 1000 − π × 4 × 5 = 1000 − 20π ≈ 937.2 cm³.
A cone of radius 6 cm and height 9 cm has its top cut off by a plane parallel to the base. The piece removed is a smaller cone of height 3 cm, which is similar to the large cone with scale factor 3 ÷ 9 = 1/3 (so its radius is 2 cm). Large cone = ⅓ × π × 36 × 9 = 108π. Small cone = ⅓ × π × 4 × 3 = 4π. The remaining frustum has volume 108π − 4π = 104π cm³. As a check, the small cone is (1/3)³ = 1/27 of the large one, and 108π ÷ 27 = 4π.
Substitute the known values into the formula and solve the equation, as in our algebraic equations guide.
(a) A rectangle has area 96 cm² and length 12 cm. 96 = 12w, so w = 8 cm.
(b) A circle has area 49π cm². πr² = 49π, so r² = 49 and r = 7 cm (a radius cannot be negative).
(c) A sphere has volume 36π cm³. 4/3 πr³ = 36π, so r³ = 36 ÷ (4/3) = 27 and r = 3 cm.
(d) A cylinder has volume 200π cm³ and height 8 cm. π × r² × 8 = 200π, so r² = 25 and r = 5 cm.
When a shape is enlarged by a linear scale factor k, lengths are multiplied by k, areas by k² and volumes by k³. This holds for any two similar shapes.
| Quantity | Scale factor | If k = 3 |
|---|---|---|
| Length | k | × 3 |
| Area | k² | × 9 |
| Volume | k³ | × 27 |
(a) A model is built to scale 1 : 10 (k = 10 from model to real). The model has volume 200 cm³ and surface area 150 cm². The real object has volume 10³ × 200 = 200 000 cm³ = 200 L and surface area 10² × 150 = 15 000 cm².
(b) Two similar shapes have areas 20 cm² and 45 cm². The scale factor of lengths is √(45 ÷ 20) = √2.25 = 1.5. If the smaller solid has volume 80 cm³, the larger has volume 80 × 1.5³ = 80 × 27/8 = 270 cm³.
How many small boxes fit in the big one?
The fix: doubling every length multiplies the volume by 2³ = 8, not 2. The surface area is multiplied by 2² = 4. A box twice as tall, wide and deep holds eight times as much.
A measurement rounded to the nearest centimetre could be up to half a centimetre either side of the stated value. 8 cm to the nearest cm means 7.5 ≤ length < 8.5. For the largest area or volume, use the upper bounds of the dimensions. For the smallest, use the lower bounds.
(a) A rectangle is 12 cm by 8 cm, each to the nearest cm. Lower bound of area = 11.5 × 7.5 = 86.25 cm². Upper bound = 12.5 × 8.5 = 106.25 cm².
(b) The side of a cube is 5.0 cm to 1 decimal place, so 4.95 ≤ s < 5.05. The volume lies between 4.95³ ≈ 121.3 cm³ and 5.05³ ≈ 128.8 cm³.
(a) Tiles. A floor is 4.2 m by 3.6 m. Tiles are 30 cm by 30 cm. The floor needs 4.2 ÷ 0.3 = 14 tiles one way and 3.6 ÷ 0.3 = 12 tiles the other, so 14 × 12 = 168 tiles. In practice, buy extra for cutting and breakage.
(b) Paint. A wall is 5 m by 2.8 m, with a door 0.9 m by 2.1 m and a window 1.2 m by 1.0 m. Paintable area = 14 − 1.89 − 1.2 = 10.91 m². If one litre covers 8 m², you need 10.91 ÷ 8 ≈ 1.36 L, so buy 2 L.
(c) Land. In Pakistan, plots are often measured in marla and kanal. A marla is commonly taken as 272.25 square feet (about 25.3 m²), and 1 kanal is 20 marla (5445 square feet, about 506 m²). The size of a marla can vary by region and by authority, so always check the definition used in your own area before calculating.
A cylinder has radius 4 cm and height 10 cm. It is filled to 75% of its capacity. Full volume = π × 16 × 10 = 160π cm³. Liquid volume = 0.75 × 160π = 120π ≈ 377.0 cm³. The percentage applies to the volume, not to the radius or the height.
People measured land long before they had formulas. Ancient Egyptian and Babylonian scribes calculated areas of fields and volumes of granaries. Greek mathematicians gave proofs. Archimedes (3rd century BCE) is credited with showing that a sphere has two-thirds of the volume of the cylinder that just encloses it. With radius r, that cylinder has height 2r, so its volume is πr² × 2r = 2πr³, and two-thirds of that is 4/3 πr³, the formula we use today. Archimedes also found that π lies between 3 10/71 and 3 1/7, a range that makes 22/7 a good approximation. Dates and details from the ancient world are approximate.
Choose a shape, type its measurements (all in the same unit) and see the formula, the working and the result. Try a triangle with sides 13, 14 and 15 using Heron’s formula, or a sector with radius 8 and angle 90.
Find the volume, surface area and capacity of a solid. Use the perpendicular height for the cone and pyramid. The tool works out the slant height for you.
Convert between units of area or volume. Notice how the factor is squared for area and cubed for volume.
Tap the correct answer for each question. Your best score is saved on this device.
1. 0.35 × 1000 = 350 L. 7500 ÷ 10 000 = 0.75 m².
2. Half of π × 10² = 50π ≈ 157.1 cm².
3. The side is the cube root of 125, which is 5 cm. The surface area is 6 × 5² = 150 cm².
1. (a) The height is √(13² − 5²) = 12 cm, so A = ½ × 10 × 12 = 60 cm². (b) A = ½(9 + 15) × 6 = 72 cm².
2. The fraction is 150 ÷ 360 = 5/12. (a) Area = 5/12 × π × 144 = 60π ≈ 188.5 cm². (b) Arc = 5/12 × 2π × 12 = 10π ≈ 31.4 cm. (c) Perimeter = 12 + 12 + 10π ≈ 55.4 cm.
3. The radius is 7 cm. (a) V = π × 49 × 20 = 980π ≈ 3078.8 cm³. (b) 3078.8 ÷ 1000 ≈ 3.08 L. (c) Curved surface area = 2π × 7 × 20 = 280π ≈ 879.6 cm².
4. (a) l = √(6² + 8²) = 10 cm. (b) V = ⅓ × π × 36 × 8 = 96π cm³. (c) Total surface area = π × 36 + π × 6 × 10 = 36π + 60π = 96π cm².
5. The scale factor is 25 ÷ 10 = 2.5. (a) The ratio of areas is 2.5² = 6.25, which is 25 : 4 (large to small). (b) The volume is 400 × 2.5³ = 400 × 15.625 = 6250 mL = 6.25 L.
6. V = 20 × 10 × 5 − π × 2² × 5 = 1000 − 20π ≈ 937.2 cm³.
7. The side lies between 7.35 and 7.45. Lower bound of area = 7.35² = 54.0225 cm². Upper bound = 7.45² = 55.5025 cm².
Let learners build understanding with real objects. Cover a rectangle with squared paper to see that area is a count of squares, then show that a parallelogram can be cut and rearranged into a rectangle. Build a 1 m square from newspaper and fill it with 10 cm squares to see why 1 m² is 10 000 cm², not 100. Use a measuring jug and a small box to show that 1 cm³ holds 1 mL. For the sphere, fill a cone and a hemisphere with rice and pour them into a cylinder to compare. The Shape Area Calculator, Solid Calculator and Unit Converter on this page let learners check their own working, but they should always write the formula and substitute before using a calculator. Ask learners to say the unit aloud at the end of every answer.
This article is original EDUSAMBAM writing. Examination requirements differ between boards, so compare your practice with the specification of your own board.
20 questions covering area, surface area, volume, capacity, circles, scale and problem-solving. Answer every question, then submit to see your score instantly.