Length, area, volume and everyday shapes

Choose between a length and a whole boundary

A length measures an extent or a path. A table’s edge length, the distance along a piece of string and the length of a route are length quantities, reported in units such as metres. A segment length is nonnegative. Name the endpoints or path so the number has a clear meaning: a straight separation and a winding route can have different lengths.

A perimeter is the total length around a closed plane figure’s boundary. For a polygon, add the lengths of all its boundary sides, each once. A garden fence question may ask for perimeter, while grass coverage asks for area. Before calculating, identify the quantity and which boundaries the question includes. A line drawn inside a region merely to divide it into parts is not part of its outer boundary.

Area measures a region with square units

Area measures the extent of a plane region. One square metre, 1 m², is the area of a square with side length 1 m. A rectangle covered by unit squares without gaps or overlaps has area equal to their total count in that unit. Partial squares contribute fractions of a square unit. The unit’s two length factors explain why area is not expressed simply in metres.

Because 1 m = 100 cm, the same unit square has area (100 cm) × (100 cm) = 10 000 cm². Thus 1 m² = 10 000 cm² exactly. Apply a unit conversion to both length directions, or square the complete length conversion factor. An area of 1 m² is not 100 cm². Perimeter and area describe different properties: figures with equal areas can have different boundary lengths.

A rectangle’s right angles justify the product

For a rectangle with adjacent side lengths a and b, area A = ab and perimeter P = 2(a + b). The product uses two perpendicular length directions; the perimeter counts two sides of each length. With stipulated exact sides 4 m and 3 m, A = 12 m² and P = 14 m. Express both lengths in a common unit before calculating a result in that unit.

A square is a rectangle whose sides all have the same length a, so A = a² and P = 4a. A square of side 0.6 m has area 0.36 m² and perimeter 2.4 m. Squaring a multiplies it by itself; multiplying a by two does not calculate its area. Two adjacent side lengths alone do not justify a rectangle formula unless the shape’s right-angle condition is given or established.

A triangle’s height is perpendicular to its chosen base

Choose one side of a triangle as base b. Its corresponding height h is the perpendicular distance from the opposite vertex to the line containing that base. The area is A = bh/2. Two copies of the triangle can be arranged into a parallelogram with the same base and height, which explains the factor one half. The height depends on the chosen base and may meet the extended base line outside an obtuse triangle.

A sloping side is not automatically the height. In the explorer’s right triangle, the 4 m and 3 m legs are perpendicular, so choosing base 4 m gives height 3 m and area 6 m². The stipulated third side is 5 m, giving perimeter 4 m + 3 m + 5 m = 12 m. For a general triangle, base and perpendicular height determine area, but they do not by themselves determine all three side lengths or the perimeter.

Rearrange pieces while preserving their area

A parallelogram with base b and perpendicular height h has area A = bh. By making suitable cuts and moving the pieces, it can be rearranged into a rectangle with the same base and height without changing area. Its sloping side length cannot replace h unless it is perpendicular to the base. A parallelogram with base 8 cm and height 5 cm therefore has area 40 cm², even if a sloping side has another length.

For a trapezium with parallel bases b₁ and b₂, let h be the perpendicular distance between their lines. Its area is A = (b₁ + b₂)h/2. Two congruent copies form a parallelogram with base b₁ + b₂ and height h, so one copy has half that area. Parallel bases 8 cm and 12 cm with height 5 cm give 50 cm². Each rearrangement requires the stated geometry, and each calculation uses compatible length units.

Keep the circle’s boundary separate from its enclosed area

For an ideal circle with radius r > 0, the radius runs from the centre to the boundary. A diameter passes through the centre and has length d = 2r. The boundary length, or circumference, is C = 2πr = πd. The area of the enclosed disc is A = πr². These formulas answer different questions and have different units: circumference is a length, while disc area is an area.

The constant π is the exact ratio of circumference to diameter for a circle. Its decimal expansion does not terminate, so 3.14 or 3.14159 is an approximation, not an exact replacement. With exact radius 2 m, d = 4 m, C = 4π m and A = 4π m². The repeated coefficient 4 does not make the quantities interchangeable. Retain π during calculation and unit conversion; for a requested report, use a calculator and mark the rounded result with ≈.

Count each part of a composite region once

Divide a composite region into familiar pieces with non-overlapping interiors. Their areas add, because each portion of the region is counted once. A cut-out region can instead be found by subtracting the missing area from an enclosing figure. For example, a 6 m by 4 m rectangle with a 2 m by 1.5 m corner removed has remaining area 24 m² − 3 m² = 21 m². The corner’s position and dimensions must be specified so the construction is clear.

If component regions overlap, simply adding their areas counts the overlap twice. For two regions, subtract the overlap area once from the sum to obtain the area of their union. For perimeter, trace the boundary actually requested; do not add every component perimeter or count internal decomposition lines. A hole has its own boundary, so specify whether a task concerns only the outside edge or all boundary components. Area and boundary calculations need their own geometric information.

Volume counts space in three length directions

Volume measures the space occupied by a solid or a specified three-dimensional region. One cubic metre, 1 m³, is the volume of a cube with edge length 1 m. Unit cubes can build layers; their total volume adds when their interiors do not overlap. Length, area and volume have dimensions L, L² and L³ respectively, so identify the quantity before choosing a unit or formula.

A cuboid, or rectangular prism, has mutually perpendicular edge lengths a, b and c. Its volume is V = abc: one layer has area ab and the perpendicular depth supplies c. The exact 4 m by 3 m by 2 m cuboid has volume 24 m³. Since 1 m = 100 cm, 1 m³ = (100 cm)³ = 1 000 000 cm³. The complete length factor is cubed; a single factor 100 cannot convert a cubic unit.

Use base area times perpendicular height under the right conditions

A prism has parallel congruent bases and cross-sections parallel to them that preserve the same area. If the base area is B and H is the perpendicular distance between the base planes, its volume is V = BH. In a right prism, the joining edges are perpendicular to the bases and their lengths equal H. For an oblique prism, a sloping joining edge is not H. Knowing an arbitrary solid’s base area and height does not establish this volume rule without the required cross-section condition.

For a right circular cylinder, the parallel bases are congruent discs, the radius is r, and H is the height perpendicular to the base planes. Each base has area B = πr², so V = πr²H. The exact explorer cylinder has r = 2 m and H = 3 m: B = 4π m² and V = 12π m³. The cylinder’s height is distinct from a diameter or its circumference. A base diagram alone does not provide height; the three-dimensional dimensions must be stated.

Surface area and usable capacity answer separate questions

Surface area measures a solid’s surface with square units, rather than the space it occupies with cubic units. A cuboid’s total surface area is S = 2(ab + ac + bc), because its six rectangular faces form three equal pairs. For a 4 m by 3 m by 2 m cuboid, S = 2(12 + 8 + 6) m² = 52 m². Count every exposed face required by the question once; a material estimate may concern a different set of faces from the total closed surface.

A right circular cylinder’s curved surface can be unrolled into a rectangle of width 2πr and height H. The curved area is 2πrH. Adding both circular ends gives the total closed surface area S = 2πr² + 2πrH. With r = 2 m and H = 3 m, S = 8π m² + 12π m² = 20π m². An open container without a top has a different surface-area calculation, so the closed formula requires both ends to be included.

Capacity concerns the usable internal volume that a vessel can hold. The litre relation is exact: 1 L = 1 dm³ = 0.001 m³ = 1000 cm³, so 1 m³ = 1000 L. External dimensions describe an outside shape and do not alone give capacity when walls or other features occupy space. If an ideal vessel’s internal region is a cuboid of dimensions 0.4 m, 0.3 m and 0.2 m, its capacity is 0.024 m³ = 24 L. The formula and unit conversion do not establish an actual vessel’s internal dimensions.

Change the shape and unit while preserving the stated geometry

Begin with the explorer’s stipulated exact rectangle 4 m by 3 m in metre units. It displays perimeter 14 m and area 12 m². Choose the right triangle with base 4 m, perpendicular height 3 m and third side 5 m, or the circle with radius 2 m. The triangle’s area is 6 m² and perimeter 12 m; the circle’s diameter is 4 m, circumference 4π m and disc area 4π m². Every dimension is supplied rather than estimated from the screen.

Choose the cuboid with edge lengths 4 m, 3 m and 2 m or the closed right circular cylinder with radius 2 m and perpendicular height 3 m. The solid diagrams show their bases; use the separately stated height and other three-dimensional dimensions for volume and total closed surface area. The cuboid gives 24 m³ and 52 m²; the cylinder gives 12π m³ and 20π m². A plane figure has no solid volume or total solid surface-area output in this explorer.

Switch the common unit between metres and centimetres. The five presets and two units give 5 × 2 = 10 exact states, all preserving the selected geometry. Length values receive factor 100, area values 10 000 and volume values 1 000 000. For example, the rectangle becomes 400 cm by 300 cm, with perimeter 1400 cm and area 120 000 cm². The diagrams keep a fixed drawing scale per represented metre; changing unit labels does not resize the object. Expressions involving π remain exact symbols here, and the explorer does not assign uncertainty to its stipulated dimensions.

Verify the shape, the information and the requested measure

Name whether you need an edge or boundary length, area, volume, surface area or capacity. Check the model’s conditions: right angles for a rectangle or cuboid, a perpendicular height for a chosen triangle base, parallel bases for a trapezium, and suitable constant cross-sections for a prism or cylinder. Check which surfaces are included and whether dimensions are internal or external. Decompose carefully, keep units compatible, and verify the resulting length, square or cubic unit.

A stipulated exact model and measurements of an object carry different information. For an object modeled as a rectangle, reported side readings 4.00 m and 3.00 m give a nominal product 12 m². If each side has stated measurement uncertainty 0.01 m, retain that information: the exact formula does not make the object’s area known exactly, and the result’s digits alone do not determine its area uncertainty. Shape approximation can also matter. The explorer’s exact dimensions are model inputs and do not represent such measurements.

Retain exact fractions and π until the requested final reporting step, then identify any rounded result with ≈. Convert before rounding when a different unit is requested; converting an already rounded cubic-metre value can give a different litre report. Check a result against its meaning and size as well as its dimension. The next lesson, Approximation, precision and measurement uncertainty, examines how rounding and stated measurement limitations affect a reported result.

Worked examples

Try it yourself

1. A task asks for a table edge, a garden’s complete fence line, grass coverage, a box’s occupied space, wrapping over all its closed faces and a vessel’s usable capacity. Name the quantity and a suitable unit for each.

   Show the answer and reasoning

   Edge: length in m. Fence line: perimeter in m. Grass coverage: area in m². Occupied space: volume in m³. Wrapping: surface area in m². Usable capacity: internal volume reported, for example, in L.

   Length measures an extent or boundary, area a region or surface, and volume a three-dimensional region. Capacity is a use of internal volume, not an outside dimension or a surface area. Naming the question first chooses the type of unit and information needed.

2. Find perimeter and area of an exact rectangle with sides 3 m and 80 cm. Separately find perimeter and area of an exact square with side 0.6 m. State why the rectangle formula needs its shape condition.

   Show the answer and reasoning

   Rectangle: 80 cm = 0.8 m, P = 7.6 m and A = 2.4 m². Square: P = 2.4 m and A = 0.36 m². The rectangle’s adjacent sides must be perpendicular.

   Use common length units: P = 2(3 + 0.8) m and A = 3 × 0.8 m². For the square use 4a and a². Adjacent side lengths alone do not make a general quadrilateral a rectangle, and the different dimensions of perimeter and area remain visible in their units.

3. A triangle has base 8 cm and perpendicular height 5 cm. A parallelogram has the same base and perpendicular height. A trapezium has parallel bases 8 cm and 12 cm with perpendicular height 5 cm. Find their areas. Could a sloping side labelled 13 cm replace a stated height without further justification?

   Show the answer and reasoning

   Triangle: 20 cm². Parallelogram: 40 cm². Trapezium: 50 cm². A sloping side cannot replace the perpendicular height merely because its length is known.

   Use bh/2, bh and (b₁ + b₂)h/2 respectively. Each height belongs to the selected base or parallel base lines and is perpendicular to them. A side could serve as a height only when the necessary perpendicular relationship is established.

4. An ideal disc has exact radius 3 cm. Find its diameter, boundary circumference and area exactly. Report circumference and area to two decimal places, keeping their units distinct.

   Show the answer and reasoning

   d = 6 cm, C = 6π cm and A = 9π cm². The rounded reports are C ≈ 18.85 cm and A ≈ 28.27 cm².

   Use d = 2r, C = 2πr and A = πr². Retain π through the exact formulas, then evaluate for the final requested report. The circumference measures the boundary and the area measures the enclosed region, so one cannot replace the other.

5. Two exact rectangular regions each have dimensions 3 m by 2 m. Their overlap has area exactly 1 m². Find their union area. Can those area data alone determine the union’s perimeter? What changes if a line is only an internal decomposition line?

   Show the answer and reasoning

   Union area = 6 m² + 6 m² − 1 m² = 11 m². The stated area data alone do not determine its perimeter. A purely internal decomposition line is not an outside boundary.

   Subtract the shared area once because simple addition would count it twice. Boundary length requires the actual placement and exposed edge lengths, which an overlap-area number alone does not supply. Adding the two rectangle perimeters would generally count edges that are not on the union’s outer boundary.

6. Find volume and total closed surface area of an exact cuboid 2 m by 3 m by 4 m. Find volume of a right prism with base area 7 cm² and perpendicular height 8 cm. Finally find volume and total closed surface area of a right circular cylinder with radius 1 m and perpendicular height 4 m.

   Show the answer and reasoning

   Cuboid: V = 24 m³ and S = 52 m². Prism: V = 56 cm³. Cylinder: V = 4π m³ and S = 10π m².

   For the cuboid use abc and 2(ab + ac + bc). For the prism use BH, with area times perpendicular height. For the cylinder use πr²H and 2πr² + 2πrH; its total surface includes both ends. Surface area and volume keep square and cubic units respectively.

7. An ideal vessel’s usable internal region is a cuboid 0.4 m by 0.3 m by 0.2 m. Find its exact capacity in litres. Would the same three external dimensions alone establish actual capacity? If dimensions instead come from rounded measurements with stated uncertainty, does an exact formula make their volume result exact?

   Show the answer and reasoning

   The stipulated internal capacity is 0.024 m³ = 24 L. External dimensions alone do not establish actual capacity. An exact formula does not make a result from uncertain measured dimensions exact.

   Multiply the internal dimensions and use the exact conversion 1 m³ = 1000 L. Wall thickness and the usable internal shape need their own information. Retain stated measurement uncertainty and shape assumptions; the digits of a computed product alone do not establish its uncertainty or improve the measurements.

What to carry forward

• Name the requested quantity: length or perimeter, area, volume, surface area or usable capacity, and choose its corresponding unit.
• Use formulas only with the stated shape conditions, including perpendicular heights and the required parallel or congruent bases.
• Square and cubic units contain two and three length factors; retain exact π in circle and cylinder expressions until a rounded report is requested.
• Add non-overlapping areas or subtract a known overlap once, and trace the requested boundary rather than counting internal lines.
• Capacity requires usable internal volume; preserve stated measurement information and distinguish an exact model result from an approximate report.