Geometry — SSC CGL Quantitative Aptitude

SSC geometry is not school geometry. Nobody asks you to prove the angle-bisector theorem — they ask you to use it, numerically, in under a minute. The exam draws from a closed set of theorems (roughly twenty), dresses them in different triangles each year, and rewards the aspirant who can see which theorem the figure is hiding. This chapter is that theorem set, each with its exam-ready form.

1. What SSC actually asks

Across recent Tier 1 papers geometry contributes 2–4 questions, and in Tier 2's Mathematical Abilities module it swells to 4–6 questions — the largest single block in the paper. The distribution inside the topic is stable:

2. Triangle basics you must have on tap

• Angle sum $=180°$; exterior angle = sum of the two opposite interior angles (SSC's favourite one-liner).
• Triangle inequality: each side lies strictly between the difference and the sum of the other two. Questions like "which of these can be sides of a triangle?" are free marks.
• Sides ↔ angles: the largest side faces the largest angle.

The four centres — the most-tested table in SSC geometry

Memorise the three angle formulas as a family: $90+\frac{A}{2}$ (incentre), $2A$ (circumcentre), $180-A$ (orthocentre). One of these appears in almost every SSC cycle.

Equilateral special case (constant SSC filler): all four centres coincide; $R=2r$; height $=\frac{\sqrt{3}}{2}a$; area $=\frac{\sqrt{3}}{4}{a}^2$; $r=\frac{a}{2\sqrt{3}}$, $R=\frac{a}{\sqrt{3}}$.

Two length workhorses

• Apollonius (median length): $A{B}^2+A{C}^2=2(A{M}^2+B{M}^2)$ where M is the midpoint of BC.
• Angle-bisector theorem: the bisector from A splits BC in the ratio $AB:AC$.

3. Similarity — the ratio machine

Triangles are similar by AA / SAS / SSS. Once similar with side ratio $k$:

$\text{perimeter ratio}=k,\text{area ratio}=k^2$

Midpoint theorem: the segment joining midpoints of two sides is parallel to the third and half of it — and cuts off a triangle of ¼ the area.

Basic Proportionality (Thales): a line parallel to one side divides the other two sides in equal ratio. SSC phrasing: "DE ∥ BC, AD = 4, DB = 6, DE = 8 → BC?" Answer: $BC=DE\times \frac{AB}{AD}=8\times \frac{10}{4}=20$.

Right-triangle altitude configuration (huge in Tier 2): altitude $h$ from the right angle to the hypotenuse creates three similar triangles, giving

$h^2=pq,a^2=pc,b^2=qc$

where $p,q$ are the hypotenuse segments and $c=p+q$.

4. Circle theorems — where the marks live

1. Angle at the centre = 2 × angle at the circumference (same arc). Angle in a semicircle = 90°.
2. Same segment: angles subtended by the same chord on the same side are equal.
3. Cyclic quadrilateral: opposite angles sum to $180°$; exterior angle = opposite interior angle.
4. Chord bisection: the perpendicular from the centre bisects the chord; equal chords are equidistant from the centre. Half-chord: ${\left(\frac{\ell }{2}\right)}^2+d^2=r^2$.
5. Tangent ⟂ radius at the point of contact; the two tangents from an external point are equal and subtend equal angles at the centre.
6. Alternate segment theorem: angle between tangent and chord = angle in the alternate segment.

The three length theorems (interchangeable via "power of a point")

Two circles

• Direct common tangent length: $\sqrt{d^2-(r_1-r_2{)}^2}$
• Transverse common tangent length: $\sqrt{d^2-(r_1+r_2{)}^2}$
• Number of common tangents: separate → 4, externally touching → 3, intersecting → 2, internally touching → 1, one inside another → 0.

5. Polygons — thirty seconds of theory, one guaranteed question

For an $n$-sided polygon: interior-angle sum $=(n-2)\times 180°$; exterior angles always sum to $360°$.

Regular polygon: each exterior angle $=\frac{360°}{n}$, each interior $=180°-\frac{360°}{n}$.

SSC's two standard asks: "each interior angle is 150° — how many sides?" (exterior = 30° → n = 12) and "the ratio of interior to exterior angle is 7 : 2 — find n" (ext = 40° → n = 9). Both are 15-second questions via the exterior angle.

Quadrilateral rapid-fire: parallelogram diagonals bisect each other; rhombus diagonals bisect at 90° (${\text{side}}^2=(\frac{d_1}{2}{)}^2+(\frac{d_2}{2}{)}^2$); rectangle diagonals are equal; a parallelogram with equal diagonals is a rectangle; with perpendicular diagonals, a rhombus.

6. Solved PYQ-style examples

Q1. In △ABC, ∠A = 70°. Find ∠BIC (I = incentre), ∠BOC (O = circumcentre) and ∠BHC (H = orthocentre). Solution. $\angle BIC=90+35=125°$; $\angle BOC=140°$; $\angle BHC=180-70=110°$. (Straight from the centres table — this exact trio cycles through SSC papers.)

Q2. The centroid divides median AD of △ABC at G with AG = 8 cm. Find AD. Solution. $AG:GD=2:1$ → $AD=\frac{3}{2}\times 8=12$ cm.

Q3. Two chords AB and CD intersect at P inside a circle. AP = 6, PB = 8, CP = 4. Find PD. Solution. $6\times 8=4\times PD\Rightarrow PD=12$.

Q4. From a point P, 17 cm from the centre of a circle of radius 8 cm, the tangent length is? Solution. $PT=\sqrt{17^2-8^2}=\sqrt{225}=15$ cm. (8–15–17 triple; SSC lives on Pythagorean triples.)

Q5. △ABC ~ △DEF with area ratio 16 : 25. If BC = 12 cm, find EF. Solution. Side ratio $=\sqrt{16:25}=4:5$ → $EF=12\times \frac{5}{4}=15$ cm.

Q6 (Tier-2 pattern). In a right triangle, the altitude to the hypotenuse divides it into segments of 4 cm and 9 cm. The altitude is? Solution. $h=\sqrt{4\times 9}=6$ cm — geometric mean, no drawing needed.

7. How to train this topic

Geometry rewards recognition, not derivation. Build a one-page theorem sheet (the tables above), then run PYQ sets topic-wise: 20 questions on centres, 20 on circles, 20 on similarity. On every question, name the theorem before calculating. Within two weeks the figures start "announcing" their theorem — that recognition is exactly the skill the exam prices at 12–18 marks.