📦 Area and Volume Formulae

GCSE Maths · Geometry and Measures (G16)

Ages 15–16 · Foundation & Higher

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1 The Area Formulae
ShapeFormulaNote
Rectangle$A = bh$base $\times$ height
Square$A = s^2$a special rectangle
Triangle$A = \tfrac{1}{2}bh$$h$ is the perpendicular height
Parallelogram$A = bh$$h$ is the perpendicular height, not the slant side
Trapezium$A = \tfrac{1}{2}(a+b)h$$a$ and $b$ are the parallel sides
Kite / rhombus$A = \tfrac{1}{2}d_1 d_2$half the product of the diagonals
Circle$A = \pi r^2$see the circles page
h b ½bh h b bh h a b ½(a+b)h
Always use the perpendicular height. In a triangle or parallelogram the sloping side is longer than the perpendicular height, so using it gives an answer that is too big. Look for the little right-angle square in the diagram.
Why $\tfrac{1}{2}bh$ for a triangle? Two identical triangles fit together to make a parallelogram of area $bh$. So one triangle is half of that.

Why $\tfrac{1}{2}(a+b)h$ for a trapezium? Two identical trapezia fit together to make a parallelogram whose base is $a + b$ and whose height is $h$. So one trapezium is half of $(a+b)h$.
Worked Example 1 — Triangle

A triangle has base $14$ cm and perpendicular height $9$ cm. Find its area.

①$A = \tfrac{1}{2} \times 14 \times 9$
②$= \tfrac{1}{2} \times 126 = 63\text{ cm}^2$
Worked Example 2 — Trapezium

A trapezium has parallel sides $9$ cm and $15$ cm, and a perpendicular height of $6$ cm. Find its area.

①$A = \tfrac{1}{2}(9 + 15) \times 6$
②$= \tfrac{1}{2} \times 24 \times 6 = 12 \times 6 = 72\text{ cm}^2$
Worked Example 3 — Working backwards

A parallelogram has area $84\text{ cm}^2$ and base $12$ cm. Find its perpendicular height.

①$A = bh$, so $84 = 12 \times h$.
②$h = 84 \div 12 = 7$ cm
2 Compound Shapes

A compound shape is made from simpler ones. There are two strategies, and it is worth choosing the easier one.

Split it up — divide the shape into rectangles and triangles, find each area and add.
Cut it out — find the area of a big surrounding rectangle and subtract the missing pieces.
Worked Example 4 — An L-shape

An L-shaped room is formed from a $10$ m by $8$ m rectangle with a $4$ m by $3$ m rectangle removed from one corner. Find its area.

①Big rectangle: $10 \times 8 = 80\text{ m}^2$
②Missing corner: $4 \times 3 = 12\text{ m}^2$
③Area $= 80 - 12 = 68\text{ m}^2$
Worked Example 5 — A house shape

A shape consists of a rectangle $12$ cm wide and $9$ cm tall with a triangle on top. The triangle has the same $12$ cm base and a height of $5$ cm. Find the total area.

①Rectangle: $12 \times 9 = 108\text{ cm}^2$
②Triangle: $\tfrac{1}{2} \times 12 \times 5 = 30\text{ cm}^2$
③Total $= 108 + 30 = 138\text{ cm}^2$
Mark each piece on the diagram and label the missing lengths before you start calculating.
3 Volume of Prisms
The one formula for every prism
$V = $ area of cross-section $\times$ length
PrismVolume
Cuboid$V = lwh$
Cube$V = s^3$
Triangular prism$V = \tfrac{1}{2}bh \times L$
Trapezoidal prism$V = \tfrac{1}{2}(a+b)h \times L$
Cylinder$V = \pi r^2 h$
Worked Example 6 — Triangular prism

A triangular prism has a cross-section with base $10$ cm and height $7$ cm. The prism is $20$ cm long. Find its volume.

①Cross-section area $= \tfrac{1}{2} \times 10 \times 7 = 35\text{ cm}^2$
②$V = 35 \times 20 = 700\text{ cm}^3$
Worked Example 7 — Cylinder

A cylinder has radius $6$ cm and height $11$ cm. Find its volume, to 1 decimal place.

①Cross-section is a circle: $A = \pi r^2 = \pi \times 36 = 36\pi$
②$V = 36\pi \times 11 = 396\pi$
③$= 1244.07\ldots = 1244.1\text{ cm}^3$
Keep the answer as $396\pi$ until the very end, then convert to a decimal. Rounding early loses accuracy.
Worked Example 8 — Working backwards

A cylinder has volume $500\text{ cm}^3$ and height $10$ cm. Find its radius, to 2 decimal places.

①$V = \pi r^2 h$, so $500 = \pi \times r^2 \times 10$
②$r^2 = \dfrac{500}{10\pi} = \dfrac{50}{\pi} = 15.9155$
③$r = \sqrt{15.9155} = 3.99$ cm (2 d.p.)
4 Surface Area of Prisms

Surface area is the total area of all the faces. The safest method is to imagine the net and add up every face.

SolidSurface area
Cuboid$2(lw + lh + wh)$
Cube$6s^2$
Cylinder (closed)$2\pi r^2 + 2\pi r h$
Cylinder (open tube)$2\pi r h$ only
Any prism$2 \times$ cross-section $+$ perimeter of cross-section $\times$ length
Read whether the solid is open or closed. An open-topped tank has no lid, so you count five faces, not six. An open pipe has no circular ends at all.
Worked Example 9 — Cuboid surface area

Find the surface area of a cuboid measuring $9$ cm by $5$ cm by $4$ cm.

①Top and bottom: $2(9 \times 5) = 90$
②Front and back: $2(9 \times 4) = 72$
③Two ends: $2(5 \times 4) = 40$
④Total $= 90 + 72 + 40 = 202\text{ cm}^2$
Worked Example 10 — Cylinder surface area

A closed cylinder has radius $5$ cm and height $12$ cm. Find its total surface area, to 1 d.p.

①Two circular ends: $2\pi r^2 = 2\pi \times 25 = 50\pi$
②Curved surface: $2\pi r h = 2\pi \times 5 \times 12 = 120\pi$
③Total $= 50\pi + 120\pi = 170\pi$
④$= 534.07\ldots = 534.1\text{ cm}^2$
5 Quick Reference

Triangle

$A = \tfrac{1}{2}bh$ with the perpendicular height.

Parallelogram

$A = bh$, again perpendicular height.

Trapezium

$A = \tfrac{1}{2}(a+b)h$; $a$ and $b$ are the parallel sides.

Kite / rhombus

$A = \tfrac{1}{2}d_1 d_2$.

Compound shapes

Split and add, or surround and subtract.

Any prism

$V = $ cross-section area $\times$ length.

Cylinder

$V = \pi r^2 h$; curved surface $= 2\pi r h$.

Cuboid surface

$2(lw + lh + wh)$ — three pairs of faces.

Units

Area in units², volume in units³. Convert before you calculate.

6 Practice Questions
Question 1

Find the area of a triangle with base $18$ cm and perpendicular height $11$ cm.

▶ Show solution

$A = \tfrac{1}{2} \times 18 \times 11 = 99\text{ cm}^2$

Question 2

Find the area of a trapezium with parallel sides $6$ cm and $10$ cm and height $7$ cm.

▶ Show solution

$A = \tfrac{1}{2}(6 + 10) \times 7 = \tfrac{1}{2} \times 16 \times 7 = 56\text{ cm}^2$

Question 3

A parallelogram has area $150\text{ cm}^2$ and perpendicular height $10$ cm. Find its base.

▶ Show solution

$150 = b \times 10$

$b = 15$ cm

Question 4

An L-shape is made from a $12$ cm by $9$ cm rectangle with a $5$ cm by $4$ cm rectangle cut from one corner. Find the area.

▶ Show solution

Big rectangle: $12 \times 9 = 108\text{ cm}^2$

Cut-out: $5 \times 4 = 20\text{ cm}^2$

Area $= 108 - 20 = 88\text{ cm}^2$

Question 5

Find the volume of a cuboid measuring $7$ cm by $4$ cm by $2.5$ cm.

▶ Show solution

$V = 7 \times 4 \times 2.5 = 70\text{ cm}^3$

Question 6

A cylinder has diameter $14$ cm and height $20$ cm. Find its volume, to the nearest cm³.

▶ Show solution

Radius $= 14 \div 2 = 7$ cm.

$V = \pi r^2 h = \pi \times 49 \times 20 = 980\pi$

$= 3078.76\ldots \approx 3079\text{ cm}^3$

Question 7

A prism has a trapezium cross-section with parallel sides $5$ cm and $9$ cm and height $4$ cm. The prism is $25$ cm long. Find its volume.

▶ Show solution

Cross-section $= \tfrac{1}{2}(5 + 9) \times 4 = \tfrac{1}{2} \times 14 \times 4 = 28\text{ cm}^2$

$V = 28 \times 25 = 700\text{ cm}^3$

Question 8

Find the surface area of a cube of side $6$ cm, and the surface area of an open-topped box of the same size.

▶ Show solution

Closed cube: $6s^2 = 6 \times 36 = 216\text{ cm}^2$

Open-topped: only $5$ faces, so $5 \times 36 = 180\text{ cm}^2$

Question 9

A swimming pool is $25$ m long and $10$ m wide. It slopes evenly from $1$ m deep at one end to $3$ m at the other.

(a) Sketch the cross-section and name it.   (b) Find the volume of water when full.   (c) How many litres is this?

▶ Show solution

(a) Looking along the length, the cross-section is a trapezium with parallel sides $1$ m and $3$ m (the two depths) and "height" $25$ m (the length of the pool).

(b) Cross-section area $= \tfrac{1}{2}(1 + 3) \times 25 = \tfrac{1}{2} \times 4 \times 25 = 50\text{ m}^2$

$V = 50 \times 10 = 500\text{ m}^3$

(c) $1\text{ m}^3 = 1000$ litres, so $500 \times 1000 = 500\,000$ litres.

Question 10

A cylindrical water butt has radius $30$ cm and height $80$ cm. It is open at the top.

(a) Find its capacity in litres, to the nearest litre.   (b) Find the area of metal needed to make it, to the nearest cm².   (c) Water flows in at $4$ litres per minute. How long, to the nearest minute, does it take to fill?

▶ Show solution

(a) $V = \pi r^2 h = \pi \times 900 \times 80 = 72\,000\pi = 226\,194.7\text{ cm}^3$

$226\,194.7 \div 1000 = \mathbf{226}$ litres (nearest litre).

(b) Open top, so one circular base plus the curved surface.

Base: $\pi r^2 = 900\pi$

Curved: $2\pi r h = 2\pi \times 30 \times 80 = 4800\pi$

Total $= 5700\pi = \mathbf{17\,907}\text{ cm}^2$ (nearest cm²).

(c) $226.19 \div 4 = 56.5$ minutes, so about $\mathbf{57}$ minutes.

Area & Volume Formulae (G16) · GCSE Maths Revision · Created with MathJax