Topic 5: Mensuration

Almost every IGCSE Mathematics paper contains mensuration questions, and they reward students who know their formulae cold and apply them carefully. The topic is largely about substitution: identify the shape, pick the correct formula, put the numbers in with consistent units, and round sensibly. The traps are predictable. People confuse radius with diameter, mix up area and circumference, forget that area conversions square the length factor, or lose a half somewhere in a triangle or sector. This article walks through each shape type, explains where the formulae come from so you remember them, and shows how to break compound shapes into pieces. Master the structure here and mensuration becomes one of the most reliable sources of marks on the paper.

Perimeter and Area of Standard Plane Shapes

Perimeter is the total distance around the boundary of a 2D shape, measured in length units such as cm. Area is the amount of surface a shape covers, measured in square units such as $\\text{cm}^{2}$. For a rectangle of length $l$ and width $w$ the perimeter is $2(l+w)$ and the area is $l \\times w$. A square is the special case where $l = w$, giving perimeter $4l$ and area $l^{2}$. For a triangle with base $b$ and perpendicular height $h$ the area is $\\frac{1}{2} b h$; the height must be measured at right angles to the chosen base, not along a slanted side. A parallelogram uses the same idea: area $= b h$ where $h$ is the perpendicular distance between the two parallel sides. A trapezium has two parallel sides of lengths $a$ and $b$ separated by perpendicular height $h$, and its area is $\\frac{1}{2}(a+b)h$, which is simply the average of the parallel sides multiplied by the distance between them. When a question gives you the area and asks for a missing length, rearrange the formula rather than guessing. Always check that every measurement is in the same unit before you multiply.

Circles: Circumference and Area

A circle is defined by its centre and its radius $r$, the distance from the centre to the edge. The diameter $d$ is twice the radius, so $d = 2r$. The circumference (the perimeter of a circle) is $C = \\pi d$ or equivalently $C = 2 \\pi r$. The area is $A = \\pi r^{2}$. The single biggest error in circle questions is using the diameter where the formula calls for the radius, so if a question gives the diameter, halve it first. The constant $\\pi$ is approximately $3.142$; use the $\\pi$ button on your calculator for accuracy and only round at the very end. To find the area of a ring (annulus) between two concentric circles of radii $R$ and $r$, subtract: $A = \\pi R^{2} - \\pi r^{2}$. Reverse problems are common: given a circumference, find the radius from $r = \\frac{C}{2 \\pi}$; given an area, find the radius from $r = \\sqrt{\\frac{A}{\\pi}}$. Keep the difference between circumference (a length) and area (a squared quantity) clear, because mixing them is an easy way to lose marks.

Arcs and Sectors of a Circle

A sector is a 'slice' of a circle bounded by two radii and an arc; the arc is the curved part of the circumference. If the angle at the centre is $\\theta$ degrees, the sector is the fraction $\\frac{\\theta}{360}$ of the whole circle. So the arc length is $\\frac{\\theta}{360} \\times 2 \\pi r$ and the sector area is $\\frac{\\theta}{360} \\times \\pi r^{2}$. A semicircle is the special case $\\theta = 180$ and a quarter circle is $\\theta = 90$. Be careful about the perimeter of a sector: it is the arc length plus the two straight radii, so perimeter $= \\frac{\\theta}{360} \\times 2 \\pi r + 2r$. Forgetting to add the two radii is a frequent mistake. A chord divides a circle into two regions called segments; the area of a segment can be found by taking the area of a sector and subtracting the area of the triangle formed by the two radii and the chord. Whenever you meet a curved-edge shape, ask what fraction of a full circle it represents and scale the relevant circle formula by that fraction.

Surface Area and Volume of Prisms and Cylinders

A prism is a solid with the same cross-section all along its length. Its volume is always area of cross-section multiplied by length, $V = A \\times l$. For a cuboid with dimensions $l$, $w$ and $h$ this gives $V = l w h$, and its surface area is the sum of the six rectangular faces, $2(lw + lh + wh)$. A cylinder is a prism with a circular cross-section of radius $r$ and height $h$, so its volume is $V = \\pi r^{2} h$. The curved (lateral) surface of a cylinder unrolls into a rectangle whose width is the height $h$ and whose length is the circumference $2 \\pi r$, giving curved surface area $= 2 \\pi r h$. A closed cylinder also has two circular ends, so its total surface area is $2 \\pi r h + 2 \\pi r^{2}$. For a triangular prism, find the triangular cross-section area with $\\frac{1}{2} b h$ first, then multiply by the length. Volume is always measured in cubic units such as $\\text{cm}^{3}$, while surface area stays in square units, so check your units match the quantity you are reporting.

Cones, Spheres and Pyramids

These curved and pointed solids need formulae that are given on the IGCSE formula list, but you must know how to use them. A pyramid has volume $V = \\frac{1}{3} \\times \\text{base area} \\times h$, where $h$ is the perpendicular height from base to apex. A cone is a pyramid with a circular base, so its volume is $V = \\frac{1}{3} \\pi r^{2} h$. The curved surface area of a cone is $\\pi r l$, where $l$ is the slant height, not the perpendicular height; the two are linked by Pythagoras as $l = \\sqrt{r^{2} + h^{2}}$. A solid cone's total surface area adds the base, giving $\\pi r l + \\pi r^{2}$. A sphere of radius $r$ has volume $V = \\frac{4}{3} \\pi r^{3}$ and surface area $A = 4 \\pi r^{2}$; note the volume uses $r^{3}$ while the surface area uses $r^{2}$. A hemisphere is half a sphere: its volume is $\\frac{2}{3} \\pi r^{3}$, and a solid hemisphere's total surface area is the curved half $2 \\pi r^{2}$ plus the flat circular face $\\pi r^{2}$, giving $3 \\pi r^{2}$. The classic trap with cones is using $h$ where the formula needs the slant height $l$, so read carefully which length is given.

Compound Shapes and Composite Solids

Many exam questions combine basic shapes, and the strategy is always the same: split the figure into standard parts, work out each part separately, then add or subtract. A running track shape might be a rectangle with a semicircle at each end, so its area is the rectangle plus one full circle (two semicircles). A washer is a large disc with a small disc removed, so subtract the areas. For solids the same logic applies. An ice-cream shape is a cone topped by a hemisphere; its volume is the sum of the two volumes, but its surface area is the curved cone surface plus the curved hemisphere surface only, because the joining circular faces are hidden inside and must not be counted. This hidden-face point is where marks are commonly lost. When subtracting, make sure both pieces are in the same units. A neat method is to label each component, write its formula, compute it, then combine with a clear plus or minus. Sketching the decomposition helps you avoid double counting or omitting a face.

Units and Conversions of Length, Area and Volume

Length conversions are straightforward multiples of ten: $1 \\text{ cm} = 10 \\text{ mm}$, $1 \\text{ m} = 100 \\text{ cm}$, and $1 \\text{ km} = 1000 \\text{ m}$. Area and volume are where students slip, because the conversion factor is raised to a power. Since area is length squared, the factor is squared too: $1 \\text{ cm}^{2} = 10^{2} = 100 \\text{ mm}^{2}$, and $1 \\text{ m}^{2} = 100^{2} = 10000 \\text{ cm}^{2}$. Since volume is length cubed, the factor is cubed: $1 \\text{ cm}^{3} = 10^{3} = 1000 \\text{ mm}^{3}$, and $1 \\text{ m}^{3} = 100^{3} = 1000000 \\text{ cm}^{3}$. Capacity links to volume through $1 \\text{ ml} = 1 \\text{ cm}^{3}$ and $1 \\text{ litre} = 1000 \\text{ cm}^{3}$. The rule to remember: for area multiply by the length factor squared, and for volume multiply by the length factor cubed. A common error is converting $\\text{m}^{2}$ to $\\text{cm}^{2}$ using $100$ instead of $10000$. Before any calculation, convert all measurements to a single consistent unit, and decide early whether your final answer should be a length, an area or a volume so you can sanity-check the units.