Mensuration (2D and 3D)

1The scaling law for lengths, areas and volumes

When every linear dimension of a solid is multiplied by k, lengths scale by k, areas by k^2, and volumes by k^3. The exam trap is treating volume like length: doubling the radius of a sphere multiplies its volume by eight, not two, because volume picks up three factors of k while radius carries only one.

The same rule governs every solid in the syllabus — cube, cylinder, cone, sphere — because each volume formula is a product of three linear measures. Halving every edge of a cube cuts its volume by 2^3 = 8, not by half.

Figure. Two cubes drawn to the same scale: the larger face is exactly twice as wide and twice as tall as the smaller one, so its volume is 2 \times 2 \times 2 = 8 times as much — not twice.

How it works

1. Count the dimensionsA length uses one copy of k, an area two, a volume three — perimeter scales like length, surface area like area.
2. Raise k to that powerTripling every edge multiplies volume by 3^3 = 27 and surface area by 3^2 = 9.
3. Apply to the given changeIf only the radius doubles, read k = 2 and write the volume factor as 2^3 = 8 before touching a formula.

2Plane areas — rectangle, triangle, circle and sector

Every composite 2D figure in CAT decomposes into a handful of plane formulas: rectangle lb, triangle \frac12 bh, circle \pi r^2, equilateral triangle \frac{\sqrt3}{4}a^2, and sector area \frac{\theta}{360^\circ}\pi r^2 with arc length \frac{\theta}{360^\circ}\cdot 2\pi r. The skill is spotting which piece you have, not memorising a longer list.

Sectors appear whenever a wire is bent into an arc or a slice is cut from a circular disc. Read \theta in degrees unless the question states radians, and remember that arc length is a one-dimensional measure while sector area is two-dimensional — mixing them up is the commonest 2D trap after using diameter where radius belongs.

Figure. A rectangle and a triangle side by side: the rectangle's area is length times breadth; the triangle's is half base times height. Circles and sectors have no honest primitive here — their formulas sit in the table.

How it works

1. Decompose the figureSplit a composite shape into rectangles, triangles, semicircles or sectors whose formulas you already know.
2. Match the formulaA triangle wants base and height perpendicular; an equilateral triangle wants side a only; a sector wants central angle \theta and radius r.
3. Add or subtract piecesShaded region = big area minus small area, or sum of non-overlapping parts — never double-count an overlap.

3Prisms and cylinders — volume and lateral surface

A prism and a cylinder share one idea: volume equals base area times height, V = A_{\text{base}} \times h. The lateral (curved) surface of a cylinder unwraps to a rectangle whose width is the base circumference 2\pi r and whose height is h, so lateral surface area is 2\pi r h.

Total surface area of a closed cylinder adds the two circular bases: TSA = 2\pi r h + 2\pi r^2 = 2\pi r(r + h). When a question asks only for the material on the curved wall — a label on a can, paint on the side of a pillar — it wants lateral surface, not TSA.

How it works

1. Find the base areaFor a cylinder, A_{\text{base}} = \pi r^2; for a prism, use the cross-section formula.
2. Multiply by heightV = A_{\text{base}} \times h — the height must be perpendicular to the base.
3. Surface: lateral vs totalLateral = 2\pi r h for a cylinder; add 2\pi r^2 for both bases when the solid is closed.

4Cones and pyramids — the one-third volume rule

A cone or pyramid has volume one-third of the matching cylinder or prism with the same base and height: V = \frac13 \pi r^2 h for a cone. The factor \frac13 is not a separate memorisation — it is the same relationship that makes a cone exactly one-third of the cylinder it sits inside.

Surface area of a cone needs the slant height l = \sqrt{r^2 + h^2}, not the vertical height h. Curved surface is \pi r l; add the circular base \pi r^2 for total surface area. The 3\text{-}4\text{-}5 family appears constantly because l is a right triangle on the axial section.

How it works

1. Compute slant height firstl = \sqrt{r^2 + h^2} from the right triangle with legs r and h.
2. Volume with vertical heightV = \frac13 \pi r^2 h — here h is the perpendicular height, not l.
3. Surface with slant heightCSA = \pi r l; TSA = \pi r l + \pi r^2.

5Spheres and hemispheres — volume and surface parts

A sphere has volume \frac43\pi r^3 and surface area 4\pi r^2. A hemisphere is half the sphere in volume — \frac23\pi r^3 — but its surface is not half of 4\pi r^2: the curved part is 2\pi r^2 and the flat circular base adds \pi r^2, giving total surface 3\pi r^2.

Questions that say "surface area of a hemisphere" without the word "curved" usually mean the total including the flat base. When only the dome is painted or polished, they want curved surface alone — not the flat base as well.

How it works

1. Sphere volume and surfaceV = \frac43\pi r^3, SA = 4\pi r^2.
2. Hemisphere volumeHalf the sphere: V = \frac23\pi r^3.
3. Hemisphere surface — read the wordingCurved = 2\pi r^2; flat base = \pi r^2; total = 3\pi r^2.

6Melt and recast — equate volumes

When a solid is melted and recast into another shape, total volume is conserved — no material is added or lost. Set the two volume formulas equal and solve for the single unknown dimension; every common factor cancels before you touch arithmetic.

The same principle covers wire bent into a circle, a sphere hammered into a cylinder, or metal from one mould poured into another. Identify which dimension is free, write both volumes, equate, and cancel shared terms first — that is faster than substituting numbers into both formulas separately.

Figure. Melting destroys shape but keeps metal volume. Equate old volume to new volume, then solve for the unknown dimension — surface areas are not conserved.

How it works

1. Write both volumesUse the correct formula for each shape — sphere, cylinder, cone, cuboid.
2. Equate and cancelV_1 = V_2; strike \pi and any shared powers of the radius before solving.
3. Solve for the unknownOne linear dimension (height, side, radius) is usually the only unknown left.

7Standard volume ratios — cone, cylinder and hemisphere

When a cone, a cylinder and a hemisphere share the same radius r and the cone's vertical height equals r while the cylinder's height also equals r, their volumes are in the ratio 1 : 3 : 2. The cone contributes \frac13\pi r^3, the cylinder \pi r^3, and the hemisphere \frac23\pi r^3.

Memorising the ratio 1:3:2 is faster than recomputing three volumes on every question, but you should be able to rebuild it from the formulas when the height is not r or the radius differs between solids.

Figure. With the same radius and height equal to that radius, volumes sit in the ratio cone : hemisphere : cylinder =1:2:3. The stacked heights show that ratio only — not a 3D drawing of the solids.

How it works

1. Fix the common radiusWrite each volume with the same r before comparing.
2. Set heights as givenFor the standard ratio, cone height = r and cylinder height = r.
3. Divide by the smallestCone volume \frac13\pi r^3 is the unit; cylinder is 3 times that; hemisphere is 2 times.

8Cuboids, cubes and the largest inscribed sphere

A cuboid has volume V = lbh and space diagonal \sqrt{l^2 + b^2 + h^2}. For a cube of side a, the diagonal is a\sqrt3 and the largest sphere that fits inside has diameter a — the sphere touches the centre of each face, so its radius is \frac{a}{2}.

Inscribed-sphere questions often pair with melt-and-recast or scaling: the sphere's diameter equals the shortest edge of the box it sits in. For a cube, all three edges match, so diameter = side.

How it works

1. Cuboid volume and diagonalV = lbh; diagonal = \sqrt{l^2 + b^2 + h^2}.
2. Cube special caseSet l = b = h = a: volume = a^3, diagonal = a\sqrt3.
3. Largest inscribed sphere in a cubeDiameter = a, radius = \frac{a}{2}, volume = \frac43\pi\left(\frac{a}{2}\right)^3 = \frac{\pi a^3}{6}.