Geometry

high importance~4 Q in Tier 136 formulas⚡ 19 shortcuts6 subtopics

Overview Learn (6)Formulas (36)Examples (4)Practice (73)

Lines and angles, triangles and their four centres, similarity and BPT, Pythagoras, quadrilaterals, polygons and circles. The single largest advanced-maths block in CGL (3–5 questions per shift), and nearly every question runs off a memorised angle result, a triplet, or a ratio rule.

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Learn — 6 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

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

Track record in the exam

avg 3.5 Q / shift2024: 3–4 Q2025: 3–4 Q

Questions per shift in recent SSC CGL papers.

Test difficulty mix (73 questions)

26 easy36 medium11 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.

Angles between parallel lines (expressions in x)

very common

Spot it:

Two angles formed by a transversal cutting parallel lines are given as (ax ± b) degrees; x or one of the angles is asked.

How to solve: Classify the pair first: corresponding/alternate → equal; co-interior → sum $180^\circ$ . That gives one linear equation in $x$ . Solve, substitute back, and answer the angle actually asked (not $x$ ).

Example: Two co-interior angles between parallel lines are $(5x-40)^\circ$ and $(3x+20)^\circ$ . The smaller angle is:

$5x-40+3x+20=180 \Rightarrow 8x=200 \Rightarrow x=25$ ; the angles are $85^\circ$ and $95^\circ$ , so the smaller is $85^\circ$ .

Learn this in “Lines and angles” →

Complement and supplement of an angle

very common

Spot it:

'The supplement is k times the complement', 'find the complement of x degrees', or the angle stated through its relation to its complement/supplement.

How to solve: Write complement $= 90^\circ - x$ and supplement $= 180^\circ - x$ , translate the sentence into one linear equation, solve for $x$ . Instant check: supplement minus complement is always $90^\circ$ .

Example: The supplement of an angle is three times its complement. Find the angle.

$180-x = 3(90-x) \Rightarrow 180-x = 270-3x \Rightarrow 2x = 90 \Rightarrow x = 45^\circ$ .

Learn this in “Lines and angles” →

Angle at incentre / circumcentre / orthocentre

very common

Spot it:

The centre is named or defined (angle bisectors, perpendicular bisectors, altitudes) and ∠A is given — or the angle at the centre is given and ∠A asked.

How to solve: Identify the centre, then apply: incentre $\angle BIC = 90^\circ + \frac{A}{2}$ ; circumcentre $\angle BOC = 2A$ ; orthocentre $\angle BHC = 180^\circ - A$ . Invert the same formulas when the centre angle is given.

Example: In $\triangle ABC$ , $\angle A = 50^\circ$ . If $O$ is the circumcentre, find $\angle BOC$ .

$\angle BOC = 2\angle A = 2 \times 50^\circ = 100^\circ$ .

Learn this in “Triangles and their centres” →

Centroid divides the median 2 : 1

common

Spot it:

A median and the centroid G appear; AG, GD or AD is asked, sometimes through a sum or difference condition.

How to solve: On median $AD$ : $AG = \frac{2}{3}AD$ , $GD = \frac{1}{3}AD$ , and $AG - GD = \frac{1}{3}AD$ . Set the smaller piece as $k$ when a condition is given.

Example: The median $AD$ of a triangle is $15$ cm and $G$ is the centroid. Find $AG$ .

$AG = \frac{2}{3} \times 15 = 10$ cm (and $GD = 5$ cm).

Learn this in “Triangles and their centres” →

Similar triangles: perimeter ratio vs area ratio

very common

Spot it:

Perimeters (or a pair of corresponding sides/altitudes) of two similar triangles are given and an area is asked — or areas are given and a perimeter asked.

How to solve: Perimeter/side ratio is linear ( $k$ ); areas scale as $k^2$ . Going from areas back to lengths needs the square root. Then scale the known quantity.

Example: The areas of two similar triangles are $25$ and $81$ sq cm; the smaller perimeter is $40$ cm. Find the larger perimeter.

Side ratio $= \sqrt{\frac{81}{25}} = \frac{9}{5}$ ; larger perimeter $= 40 \times \frac{9}{5} = 72$ cm.

Learn this in “Congruence, similarity and BPT” →

Parallel line inside a triangle (BPT / midpoint theorem)

common

Spot it:

DE parallel to BC is stated inside a triangle — or D, E are described as midpoints — and a segment length is asked.

How to solve: BPT: $\frac{AD}{DB} = \frac{AE}{EC}$ . Midpoint theorem: $DE = \frac{BC}{2}$ (the midpoint triangle has half the perimeter and quarter of the area). Substitute the three known segments and solve.

Example: In $\triangle ABC$ , $DE \parallel BC$ with $AD = 3$ , $DB = 5$ , $AE = 6$ . Find $EC$ .

$\frac{AD}{DB} = \frac{AE}{EC} \Rightarrow \frac{3}{5} = \frac{6}{EC} \Rightarrow EC = 10$ cm.

Learn this in “Congruence, similarity and BPT” →

Pythagoras in a story (ladder, poles, displacement, diagonal)

very common

Spot it:

A ladder against a wall, two poles with a rope between tops, a north-then-east walk, or a rectangle diagonal — a right triangle hiding in a sentence.

How to solve: Draw the right triangle; the ladder/wire/diagonal/displacement is the hypotenuse. Then $a^2+b^2=c^2$ , ideally via a triplet family (3-4-5, 5-12-13, 8-15-17, 7-24-25).

Example: Two poles $6$ m and $11$ m high stand $12$ m apart on level ground. Find the distance between their tops.

Legs: gap $12$ m, height difference $5$ m → $\sqrt{144+25} = 13$ m.

Learn this in “Pythagoras theorem and triplets” →

Regular polygon: angle vs number of sides

common

Spot it:

An interior/exterior angle, the angle sum, or the diagonal count of a regular polygon is given; n or another of these quantities is asked.

How to solve: Convert to the exterior angle: ext $= 180^\circ -$ int, then $n = \frac{360^\circ}{\text{ext}}$ . From the sum: $n = \frac{\text{sum}}{180^\circ} + 2$ . From diagonals: solve $\frac{n(n-3)}{2} = d$ .

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

ext $= 180^\circ - 150^\circ = 30^\circ$ ; $n = \frac{360}{30} = 12$ .

Learn this in “Quadrilaterals and polygons” →

Rhombus from diagonals / cyclic quad opposite angles

very common

Spot it:

Either the diagonals of a rhombus (or its area plus one diagonal) with side/perimeter asked — or a cyclic quadrilateral with one angle (or a ratio) given.

How to solve: Rhombus: half-diagonals and side form a right triangle (a triplet in halves); area $= \frac{1}{2}d_1 d_2$ . Cyclic quad: opposite angles sum to $180^\circ$ ; in ratio questions each opposite pair's parts must total the same count.

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

Halves $5, 12$ → side $13$ (5-12-13) → perimeter $4 \times 13 = 52$ cm.

Learn this in “Quadrilaterals and polygons” →

Circle theorems: chord-distance, power of a point, tangent count, centre angle

very common

Spot it:

A chord with its distance from the centre, a tangent-secant configuration, two circles and their common tangents, or centre-vs-circumference angles on the same arc.

How to solve: Chord: $\left(\frac{c}{2}\right)^2 + d^2 = r^2$ . Power of a point: tangent $^2$ = external × whole secant; crossing chords $ab = cd$ . Common tangents: 4/3/2/1/0 by configuration. Centre angle $= 2 \times$ circumference angle; angle in a semicircle is $90^\circ$ .

Example: A chord $16$ cm long lies in a circle of radius $10$ cm. Find its distance from the centre.

Half-chord $8$ : $d = \sqrt{100-64} = 6$ cm.

Learn this in “Circles: chords, tangents, secants and cyclic angles” →

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