Sec 1 Mensuration: Area, Volume & Surface Area

Secondary 1 Mathematics · G3 focus · Study guide

Choose perpendicular heights, separate perimeter from area, and work with prisms, cylinders and composite solids.

Concept → worked examples → practice22 min read · practice linked below

Decide what is being measured

Perimeter is the total length around a plane figure, measured in units such as cm. Area measures a flat region, in square units such as cm². Volume measures space inside a solid, in cubic units such as cm³. Surface area is the area of a solid's exposed faces, in square units.

Sketch the shape, label the dimensions, and make the units consistent before using a formula. A height in an area formula is the perpendicular distance from the base to the opposite vertex or parallel side. It may lie outside the figure; a sloping side is not automatically a height.

Plane figures and their formulas
Shape	Area	What the letters mean
Rectangle	$lw$	Length $l$ , width $w$
Triangle	$\frac12 bh$	Base $b$ , perpendicular height $h$
Parallelogram	$bh$	Base $b$ , perpendicular height $h$
Trapezium	$\frac12(a+b)h$	Parallel sides $a,b$ , perpendicular distance $h$
Circle	$\pi r^2$	Radius $r$ ; circumference $2\pi r=\pi d$

Keep $\pi$ in exact answers when possible; use the calculator's $\pi$ key and round only at the end when a decimal is requested. A radius is half a diameter.

Perpendicular height gives the area

Cut a triangular piece from one end of a parallelogram and move it to the other end. You obtain a rectangle with the same base and perpendicular height, explaining $A=bh$ . Two identical trapeziums fit together to form a parallelogram with base $a+b$ and height $h$ ; each trapezium therefore has half that area.

Use the marked perpendicular height for area. Use boundary side lengths for perimeter.

Worked example 1: the sloping side has a different job

A parallelogram has base 10 cm, perpendicular height 4 cm and sloping side 5 cm. Find its area and perimeter.

Its area uses the base and perpendicular height: $A=10\times4=\boxed{40\text{ cm}^2}.$
Its perimeter follows the four sides. Opposite sides are equal: $P=10+5+10+5=\boxed{30\text{ cm}}.$

Check: $10\times5$ would use a sloping side as the height. The perimeter answer has length units; the area answer has square units.

Worked example 2: the average of two parallel sides

A trapezium has parallel sides 8 cm and 14 cm, separated by a perpendicular height of 5 cm. Find its area.

Identify the parallel pair: $a=8$ , $b=14$ , $h=5$ .
Add the parallel lengths before multiplying: $A=\frac12(8+14)(5)=\frac12(22)(5)=\boxed{55\text{ cm}^2}.$

Check: the average parallel length is $(8+14)\div2=11$ cm; a rectangle of width 11 cm and height 5 cm also has area 55 cm². The area lies between $8\times5=40$ and $14\times5=70$ cm². The parallel sides alone do not tell you its perimeter.

Split the area; trace the boundary

For a composite figure, add areas of non-overlapping pieces, or subtract a missing piece from a larger familiar shape. For perimeter, trace only the actual outside boundary. Internal joining lines do not count. A cut-out corner introduces new boundary edges, so subtracting the cut-out's perimeter is usually wrong.

Worked example 3: a rectangular corner is removed

Start with a 12 cm by 8 cm rectangle. Remove a 4 cm by 3 cm rectangle from its top-right corner. Find the remaining area and perimeter.

The dashed edges belonged to the removed corner. The solid outline is the perimeter to add.

Subtract the missing area: $A=12(8)-4(3)=96-12=\boxed{84\text{ cm}^2}.$
Find the shortened outside edges: top $=12-4=8$ cm; right $=8-3=5$ cm.
Trace all six boundary sides: $P=8+3+4+5+12+8=\boxed{40\text{ cm}}.$

Check: area also equals $8\times8+4\times5=84$ cm². The removed top 4 cm and right 3 cm are replaced by boundary lengths 4 cm and 3 cm inside the corner, so the perimeter stays $2(12+8)=40$ cm in this particular cut.

A curved boundary uses circumference. For a semicircle of radius $r$ , the curved part is $\pi r$ , while its whole perimeter is $\pi r+2r$ . If that diameter joins a rectangle, the diameter becomes internal and is not counted in the combined outline. Its area remains $\frac12\pi r^2$ .

Prisms: repeat the same cross-section

A prism has a constant cross-section along its length. Its volume is the area of one cross-section multiplied by the perpendicular length between the two ends:

V=A_{\text{cross-section}}L.

For a right prism, the side faces are rectangles. If the cross-section has area $A$ and perimeter $P$ , its total surface area is

\text{surface area}=2A+PL.

The two ends contribute $2A$ ; the side rectangles together form a strip of width $P$ and length $L$ . A cuboid is a rectangular prism, so $V=lwh$ and total surface area is $2(lw+lh+wh)$ .

Worked example 4: a right triangular prism

A right triangular prism has a right-angled cross-section with perpendicular sides 3 cm and 4 cm and third side 5 cm. The prism is 10 cm long. Find its volume and total surface area.

The 3 cm height belongs to the triangular cross-section; the 10 cm length runs along the prism.

Find the area of one triangular end: $A=\tfrac12(3)(4)=6\text{ cm}^2.$
Multiply by the prism length: $V=6(10)=\boxed{60\text{ cm}^3}.$
There are two triangular ends and three rectangles: $\text{surface area}=2(6)+3(10)+4(10)+5(10)=\boxed{132\text{ cm}^2}.$

Check: the end perimeter is $3+4+5=12$ cm, so $2A+PL=12+120=132$ cm². Multiplying all four lengths would not represent a volume.

Cylinders: circles and an unrolled rectangle

A right cylinder has a circular cross-section of area $\pi r^2$ . Its volume is $\pi r^2h$ . Unroll its curved surface: the rectangle has height $h$ and width equal to the circumference $2\pi r$ . Therefore:

\text{curved area}=2\pi rh,\qquad\text{closed surface area}=2\pi rh+2\pi r^2.

An open container with one circular base has surface area $2\pi rh+\pi r^2$ , ignoring material thickness. A tube open at both ends has only its curved area in this thin-wall model. Count the actual circular faces required by the question.

Worked example 5: closed cylinder or open container?

A cylinder has radius 3 cm and height 8 cm. Find its volume, its surface area when closed, and the area of material for the same shape with its top removed. Ignore thickness.

The volume is $V=\pi(3)^2(8)=\boxed{72\pi\text{ cm}^3}\approx226.2\text{ cm}^3.$
The curved area is $2\pi(3)(8)=48\pi$ cm². Two circular ends add $2\pi(3)^2=18\pi$ cm². The closed area is $48\pi+18\pi=\boxed{66\pi\text{ cm}^2}\approx207.3\text{ cm}^2.$
Remove just one circle: $66\pi-9\pi=\boxed{57\pi\text{ cm}^2}\approx179.1\text{ cm}^2.$

Check: opening the top changes the amount of surface material, not the internal capacity. The removed area is $9\pi$ cm², not its circumference $6\pi$ cm.

Build a cylinder from its net

For radius 3 cm and height 8 cm, the curved rectangle is 6π cm wide and 8 cm high. Two circles add 18π cm², giving a closed surface area of 66π cm². Its volume is 72π cm³.

The rectangle's width is a circumference, not a diameter. Choose how many circular ends belong to the required surface.

Volume = 72π cm³ ≈ 226.19 cm³. Surface area = 48π + 18π = 66π cm² ≈ 207.35 cm².

Composite solids: joins disappear from the outside

Add the volumes of separate parts if they do not overlap; subtract a removed cavity. For total surface area, count only the exposed surfaces. If you start by adding the complete surface areas of two joined solids, subtract the contact area twice, because one face from each solid is hidden.

Worked example 6: a cylinder fixed to a cuboid

A solid cylinder of radius 3 cm and height 6 cm stands on the 12 cm by 10 cm top of a cuboid of height 4 cm. Its entire circular base is joined to the cuboid without overlap in volume. Find the total volume and exposed surface area, including the bottom of the cuboid.

The dashed contact circle hides the cylinder's base and an equal patch of the cuboid's top. Its area must be removed twice if complete surface areas were added first.

The cuboid volume is $12(10)(4)=480$ cm³. The cylinder volume is $\pi(3)^2(6)=54\pi$ cm³. Add them: $V=\boxed{480+54\pi\text{ cm}^3}\approx649.6\text{ cm}^3.$
The complete cuboid surface area is $2(12\times10+12\times4+10\times4)=416$ cm².
The complete cylinder surface area is $2\pi(3)(6)+2\pi(3)^2=54\pi$ cm². Their shared circular contact has area $9\pi$ cm² and hides a face on each solid.
Subtract both hidden areas: $S=416+54\pi-2(9\pi)=\boxed{416+36\pi\text{ cm}^2}\approx529.1\text{ cm}^2.$

Check by exposed parts: exposed cuboid $=416-9\pi$ ; cylinder's curved side and top $=36\pi+9\pi$ . Adding gives $416+36\pi$ . The cylinder's diameter is 6 cm, so its base fits within the 12 cm by 10 cm top.

Square and cube the conversion factor

Since $1\text{ m}=100\text{ cm}$ , a square metre contains $100\times100=10\,000$ square centimetres; a cubic metre contains $100\times100\times100=1\,000\,000$ cubic centimetres.

1\text{ m}^2=10\,000\text{ cm}^2,\qquad 1\text{ m}^3=1\,000\,000\text{ cm}^3.

For capacity, $1\text{ cm}^3=1\text{ mL}$ , $1000\text{ cm}^3=1\text{ L}$ , and $1\text{ m}^3=1000\text{ L}$ . Converting to a smaller unit increases the numerical value; converting to a larger unit decreases it.

Worked example 7: area, volume and capacity conversions

Convert $0.024\text{ m}^2$ to cm², and $3500\text{ cm}^3$ to m³ and litres.

For area, multiply by $100^2$ : $0.024\times10\,000=\boxed{240\text{ cm}^2}.$
For volume in a larger unit, divide by $100^3$ : $3500\div1\,000\,000=\boxed{0.0035\text{ m}^3}.$
Each litre holds 1000 cm³, so $3500\div1000=\boxed{3.5\text{ L}}.$

Check: $0.0035\text{ m}^3\times1000\text{ L per m}^3=3.5\text{ L}$ . A factor of 100 alone applies to lengths, not areas or volumes.

Common mistakes and checks

Sloping side used as a height: look for the perpendicular distance and a right-angle mark.
Adding an internal join to the perimeter: trace the final outer outline once.
Using the diameter for $r$ : halve it before calculating $\pi r^2$ .
Counting every cylinder as closed: decide whether there are zero, one or two circular faces.
Leaving a joined face exposed: subtract the contact area from both surfaces when adding complete surface areas.
Mixing cm and m: convert lengths consistently before the formula, or convert the final square or cubic unit with the correct factor.

Practise choosing the right measurement

Continue with the Sec 1 perimeter, area and volume question bank. Before calculating, write “boundary”, “flat region”, “space” or “exposed faces” beside the quantity requested. This helps you choose both the formula and the unit.