Mensuration

The chapter has two connected halves. The first stays in two dimensions and finds the area of shapes beyond rectangles and triangles: the formula for a trapezium (one pair of parallel sides) and a rhombus (all sides equal, diagonals crossing at right angles), then a general method — any polygon can be cut along diagonals into triangles and trapeziums and its area found by adding the pieces. The second half moves into three dimensions, revisiting the cuboid, the cube (all edges equal), and the right circular cylinder. For each it asks two questions: how much surface area does the solid have, split into lateral (side walls only) and total (every face), and how much volume does it enclose? The chapter closes by clarifying that volume and capacity are the same idea in different units, linked by the standard conversions between cubic centimetres, cubic metres and litres. Throughout, keep two questions in mind: is this a 2-D area question or a 3-D surface-area-or-volume question, and, if 3-D, does it call for the whole surface or only part of it?

Area of a trapezium

A trapezium is a quadrilateral with exactly one pair of parallel sides. Call the two parallel sides $a$ (shorter) and $b$ (longer), and let $h$ be the height — the perpendicular distance between the two parallel sides, not the length of the slanted sides. The area of a trapezium is

where: $a$ and $b$ are the lengths of the two parallel sides and $h$ is the perpendicular distance between them. Notice the formula reduces to the ordinary triangle-area idea when one parallel side shrinks to a point ($a \to 0$), giving $A = \tfrac12 h b$ — a useful mental check. This result is proved from first principles in the first derivation below.

Area of a rhombus

A rhombus is a quadrilateral with all four sides equal in length. Its two diagonals $d_1$ and $d_2$ always bisect each other at right angles, and this special property gives a very clean area formula:

where: $d_1$ and $d_2$ are the lengths of the two diagonals. The formula works because the two diagonals cut the rhombus into four congruent right triangles, and stacking their areas together simplifies to half the product of the diagonals — you do not need the side length at all once you know both diagonals.

Area of a general polygon

There is no single formula for the area of a five-sided, six-sided or irregular polygon. The method is always the same: draw diagonals to break the polygon into triangles and trapeziums whose areas you can compute directly, then add. For a pentagon $ABCDE$, diagonals $AC$ and $AD$ split it into three triangles, $\triangle ABC$, $\triangle ACD$ and $\triangle AED$. Alternatively, one diagonal $AD$ with perpendiculars from $B$ and $C$ onto it splits the pentagon into two right triangles and a trapezium. A regular hexagon of side $5\ \text{cm}$ can be split two genuinely different ways and still give the same answer. One long diagonal gives two congruent trapeziums, each of area $32\ \text{cm}^2$, totalling $64\ \text{cm}^2$. Splitting it instead into two congruent triangles (altitude $3\ \text{cm}$, area $12\ \text{cm}^2$ each) plus a central rectangle ($8 \times 5 = 40\ \text{cm}^2$) gives $12+12+40 = 64\ \text{cm}^2$ — the same total, as it must, however you slice a fixed region.

Surface area of the cuboid, cube and cylinder

The surface area of any solid is the sum of the areas of all its faces. A cuboid has six rectangular faces in three congruent pairs — top and bottom ($l \times b$ each), front and back ($l \times h$ each), and two sides ($b \times h$ each) — so its total surface area is

where: $l$, $b$, $h$ are the length, breadth and height. Excluding the top and bottom and counting only the four side walls gives the lateral surface area, $S_{\text{LSA}} = 2h(l+b)$ — the quantity needed when whitewashing a room's walls but not its ceiling or floor. A cube is a cuboid with $l = b = h$, so its total surface area simplifies to $S_{\text{cube}} = 6l^2$, one square face's area times six identical faces. A right circular cylinder has two flat circular ends and one curved side wall. Unrolled flat, the curved wall becomes a rectangle of length equal to the circumference, $2\pi r$, and width equal to the height $h$, so the curved (lateral) surface area is $2\pi r h$. Adding the two circular ends, each of area $\pi r^2$, gives the total surface area

where: $r$ is the radius of the circular base and $h$ is the height; $\pi$ is taken as $\tfrac{22}{7}$ unless a problem says otherwise. The second derivation below walks through this unrolling argument in full.

Volume of the cuboid, cube and cylinder

Volume is the amount of space a three-dimensional solid occupies, measured in cubic units, because we imagine packing the solid with unit cubes. For all three solids here, the base and top are congruent and parallel and every cross-section parallel to the base is identical, so volume is always base area times height:

where: for the cuboid, $l \times b$ is the base area and $h$ the height; for the cube all three edges equal $l$; for the cylinder $\pi r^2$ is the base area. Remember this as one idea, not three separate formulas — it is why stacking paper into a pile, or cans into a tube, multiplies a flat area by a height to give a volume.

Volume and capacity

Volume describes how much space a solid occupies; capacity describes how much a container can hold. The two are numerically the same once units match: $1\ \text{cm}^3 = 1\ \text{mL}$, and since $1\ \text{L} = 1000\ \text{cm}^3$, it follows that $1\ \text{m}^3 = 1{,}000{,}000\ \text{cm}^3 = 1000\ \text{L}$. Asking "how much can the tank hold" (capacity, litres) or "how much space does it occupy" (volume, $\text{m}^3$) is the same calculation — only the final unit label changes.

Board level - area of a rhombus from one diagonal and the area (NCERT Example 2 pattern)

The area of a rhombus is $240\ \text{cm}^2$ and one of its diagonals is $16\ \text{cm}$. Find the length of the other diagonal.

Step 1. Write down what is known: area $A = 240\ \text{cm}^2$, one diagonal $d_1 = 16\ \text{cm}$, and the other diagonal $d_2$ is unknown.

Step 2. Substitute into the rhombus area formula $A = \tfrac12 d_1 d_2$: $240 = \tfrac12 \times 16 \times d_2 = 8\,d_2$.

Step 3. Solve for $d_2$ by dividing both sides by $8$: $d_2 = 240 \div 8 = 30\ \text{cm}$. Answer: the other diagonal is $30\ \text{cm}$ long.

Board level - cost of painting cylindrical pillars using curved surface area (NCERT Example 6 pattern)

A building has 24 cylindrical pillars, each of radius $28\ \text{cm}$ and height $4\ \text{m}$. Find the total cost of painting the curved surface area of all the pillars at the rate of Rs 8 per square metre.

Step 1. Convert to consistent units and identify the correct formula. Radius $r = 28\ \text{cm} = 0.28\ \text{m}$, height $h = 4\ \text{m}$. Painting only the curved (lateral) surface of a pillar — not its flat top or bottom — needs $S_{\text{CSA}} = 2\pi r h$, not the total surface area formula.

Step 2. Curved surface area of one pillar: $S_{\text{CSA}} = 2 \times \tfrac{22}{7} \times 0.28 \times 4 = 7.04\ \text{m}^2$.

Step 3. Curved surface area of all $24$ pillars: $7.04 \times 24 = 168.96\ \text{m}^2$.

Step 4. Total cost at Rs $8$ per $\text{m}^2$: $168.96 \times 8 = \text{Rs } 1351.68$. Answer: the total cost of painting all 24 pillars is Rs 1351.68.

Higher-order thinking (HOTS / olympiad-style) - counting painted faces on a cut-up cube

A wooden cube of side $12\ \text{cm}$ has all six outer faces painted red. It is then cut into $64$ identical smaller cubes. How many of the small cubes have exactly three faces painted, exactly two faces painted, exactly one face painted, and no face painted at all? Verify your count using volume.

Step 1. Since $64 = 4 \times 4 \times 4$, the big cube is cut into a $4 \times 4 \times 4$ grid, so each small cube has side $12 \div 4 = 3\ \text{cm}$. Painted faces on a small cube depend only on where it sits in this grid — corner, edge, face, or interior.

Step 2. Corner cubes always show three painted faces, and every cube has exactly $8$ corners, so $8$ small cubes have three faces painted.

Step 3. Edge cubes (on one edge, excluding corners) show two painted faces. Each of the $12$ edges contributes $4-2=2$ such cubes, giving $12 \times 2 = 24$ small cubes with two faces painted.

Step 4. Face cubes (on one face, away from every edge) show one painted face; each of the $6$ faces contributes a $(4-2)\times(4-2)=4$ inner square, giving $6 \times 4 = 24$ small cubes with one face painted. The fully interior cubes form a $(4-2)^3 = 8$ cube buried inside, so $8$ small cubes have no face painted at all.

Step 5. Check: $8+24+24+8=64$. As a volume cross-check, the big cube's volume $12^3 = 1728\ \text{cm}^3$ equals $64$ small cubes of volume $3^3 = 27\ \text{cm}^3$ each, since $64 \times 27 = 1728\ \text{cm}^3$. Answer: 8 cubes with three faces painted, 24 with two faces painted, 24 with one face painted, and 8 with no face painted. Timing note: aim for under four minutes. The pattern is corners give $8$ always, edges give $12(n-2)$, faces give $6(n-2)^2$, interior gives $(n-2)^3$, where $n$ is the cubes per edge (here $n=4$). Recognise the grid position categories first rather than counting all $64$ cubes one by one, and use the volume total as a fast final check.

Confusing perimeter or circumference with area is the most basic slip. In the trapezium and rhombus formulas, students often use the slant side instead of the perpendicular height $h$, or drop the factor of $\tfrac12$ altogether. When splitting a polygon into triangles, check that the pieces cover it exactly once, with no gaps and no overlaps. In three dimensions, the most common error is mixing up lateral surface area with total surface area — forgetting the two circular ends for a cylinder's total surface, or wrongly including the top and bottom when only the side walls need painting. A second error is using diameter where radius is needed inside $\pi r^2$ or $2\pi r h$. Unit mismatches are a frequent silent killer: mixing centimetres and metres in one formula, or forgetting $1\ \text{m}^3 = 1000\ \text{L}$ rather than $100\ \text{L}$. Finally, students sometimes add only three cuboid face areas instead of doubling them for the three pairs — remember the factor of $2$ in $2(lb+bh+hl)$.

• State the trapezium area formula and correctly identify the height $h$ versus a slant side.
• State the rhombus area formula and use it either way — area from both diagonals, or one diagonal from the area and the other.
• Split an irregular polygon into triangles and trapeziums using diagonals, and add the pieces to find its area.
• Distinguish lateral from total surface area for a cuboid and a cylinder, and choose the correct one for a real-world situation.
• Write and apply the surface area formulas for a cuboid, cube and cylinder, including how $2\pi r(r+h)$ comes from unrolling a net.
• Write and apply the volume formulas for a cuboid, cube and cylinder, recognising all three as base area times height.
• Convert confidently between cubic centimetres, cubic metres and litres.
• Solve multi-step cost problems (rate per square or cubic metre) by first finding the correct area or volume, then multiplying by the rate.

Ask the tutor: Want to work through one more trapezium-field or cylindrical-tank problem together, step by step, before trying the exercises on your own?