Mensuration (2D)

high importance~2 Q in Tier 123 formulas⚡ 12 shortcuts5 subtopics

Overview Learn (5)Formulas (23)Examples (4)Practice (62)

Areas and perimeters of triangles, quadrilaterals, circles, sectors and regular polygons, plus paths, wheels and percentage-change effects. A reliable 1-3 questions per shift; the formulas are few and heavily recycled, so this converts to marks faster than almost any other advanced topic.

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Learn — 5 subtopics

One page per subtopic: detailed notes, every question type, formulas, tricks and practice sets.

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Formula sheet

Every formula on one printable page, grouped by subtopic.

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Solved examples

4 exam-level questions worked step by step.

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Practice & test

62 questions — untimed practice or a timed test with analysis.

Track record in the exam

avg 1.0 Q / shift2024: 1–2 Q2025: 1 Q

Questions per shift in recent SSC CGL papers.

Test difficulty mix (62 questions)

18 easy35 medium9 hard

Question patterns exams keep repeating

Taken from previous-year papers. If a pattern is marked "very common", expect to see it in your exam.

Triangle area: base-height, equilateral, right-triangle

very common

Spot it:

A triangle with base and height given, or 'equilateral triangle of side a', or three sides that form a Pythagorean family.

How to solve: General: $K = \frac{1}{2}bh$ . Equilateral: $\frac{\sqrt{3}}{4}a^2$ , height $\frac{\sqrt{3}}{2}a$ . Triplet sides → right triangle → $\frac{1}{2} \times$ legs, skipping Heron.

Example: Find the area of an equilateral triangle of side $12$ cm.

$K = \frac{\sqrt{3}}{4} \times 144 = 36\sqrt{3}$ sq cm.

Learn this in “Areas of triangles” →

Heron's formula for three sides

very common

Spot it:

Three side lengths given with no height mentioned; the numbers are not an obvious triplet.

How to solve: Compute $s = \frac{a+b+c}{2}$ , then $K = \sqrt{s(s-a)(s-b)(s-c)}$ . Check (13,14,15) → 84 and triplet families first.

Example: Find the area of the triangle with sides $13$ , $14$ , $15$ cm.

$s = 21$ ; $K = \sqrt{21 \times 8 \times 7 \times 6} = 84$ sq cm.

Learn this in “Areas of triangles” →

Isosceles triangle area via half-base

common

Spot it:

Equal sides and base given; the height is not stated but implied by the symmetry.

How to solve: $h = \sqrt{a^2 - (b/2)^2}$ , then $K = \frac{1}{2}bh$ . The half-base triplets (5,12,13), (8,15,17) appear constantly.

Example: An isosceles triangle has equal sides $17$ cm and base $16$ cm. Find its area.

$h = \sqrt{17^2 - 8^2} = 15$ ; $K = \frac{1}{2} \times 16 \times 15 = 120$ sq cm.

Learn this in “Areas of triangles” →

Rectangle from two of {perimeter, diagonal, area}

very common

Spot it:

The perimeter and diagonal of a rectangle given (or area and diagonal) — the missing quantity asked.

How to solve: $l + b = \frac{P}{2}$ ; match with the diagonal to a triplet (5-12-13 etc.). General: $K = \left(\frac{l+b}{2}\right)^2 - \left(\frac{d}{2}\right)^2$ .

Example: A rectangle has perimeter $34$ cm and diagonal $13$ cm. Find its area.

$l+b = 17$ ; 5-12-13 → area $60$ sq cm.

Learn this in “Areas of quadrilaterals” →

Rhombus: diagonals, area, perimeter chain

very common

Spot it:

Diagonals given (or area + one diagonal, or diagonals in a ratio) — side, perimeter or the other diagonal asked.

How to solve: $K = \frac{1}{2}d_1d_2$ ; half-diagonals form a right triangle with the side — use half-size triplets. Ratio form: $d_1 = mk, d_2 = nk$ into the area.

Example: The diagonals of a rhombus are $10$ cm and $24$ cm. Find its perimeter.

Halves $5, 12$ → side $13$ → perimeter $52$ cm.

Learn this in “Areas of quadrilaterals” →

Circle: area ↔ circumference forward and reverse

very common

Spot it:

Radius/diameter given for one of area/circumference — or circumference/area given and the other asked.

How to solve: $C = 2\pi r$ , $K = \pi r^2$ with $\pi = 22/7$ . Reverse: $r = \frac{C}{2\pi}$ or $r = \sqrt{K/\pi}$ . Halve diameters before anything else.

Example: The circumference of a circle is $132$ cm. Find its area.

$r = \frac{132 \times 7}{44} = 21$ ; $K = \frac{22}{7} \times 441 = 1386$ sq cm.

Learn this in “Circles, sectors and rings” →

Sector area and arc length

very common

Spot it:

A central angle and radius given; the sector's area or the arc's length asked (sometimes reversed).

How to solve: Multiply the full circle's area (or circumference) by $\frac{\theta}{360}$ ; reduce the fraction first ( $72^\circ \to \frac15$ ).

Example: Find the area of a sector of angle $90^\circ$ in a circle of radius $14$ cm.

$\frac14 \times \frac{22}{7} \times 196 = 154$ sq cm.

Learn this in “Circles, sectors and rings” →

Ring area and wheel revolutions

common

Spot it:

Two concentric circles forming a path/border — or a wheel of given diameter rolling a stated distance.

How to solve: Ring: $\pi(R^2 - r^2) = \pi(R+r)(R-r)$ . Wheel: one revolution = $\pi D$ ; revolutions $= \frac{\text{distance}}{\pi D}$ with units matched first.

Example: A wheel of diameter $70$ cm rolls $231$ m. How many revolutions does it make?

$C = 2.2$ m; $\frac{231}{2.2} = 105$ revolutions.

Learn this in “Circles, sectors and rings” →

Regular polygon: angles, diagonals, hexagon area

very common

Spot it:

An interior/exterior angle, angle sum or diagonal count given — or a regular hexagon with its side/diagonal.

How to solve: Route through the exterior angle $\frac{360^\circ}{n}$ . Diagonals $\frac{n(n-3)}{2}$ — reverse by factor pairs. Hexagon $= 6$ equilateral triangles: $K = \frac{3\sqrt3}{2}a^2$ , long diagonal $2a$ , short $a\sqrt3$ .

Example: Each interior angle of a regular polygon is $150^\circ$ . Find the number of sides.

ext $= 30^\circ$ ; $n = \frac{360}{30} = 12$ .

Learn this in “Regular polygons and inscribed figures” →

Area scaling: side change, cost, map scale

very common

Spot it:

'Side increased by x%', painting/carpeting costs for a resized figure, map scale 1 : n with a map area given.

How to solve: Areas scale as $k^2$ ; lengths as $k$ . Side $+x\%$ → area $+\left(2x + \frac{x^2}{100}\right)\%$ . Map $1:n$ → actual area $=$ map area $\times n^2$ (convert units).

Example: If each side of a square is increased by $20\%$ , by what percent does the area increase?

$1.2^2 = 1.44$ → $44\%$ .

Learn this in “Percentage change, similarity and re-bent shapes” →

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