Geometry is the second-largest topic in SSC CGL — accounting for approximately 15–25% of the quantitative aptitude section — yet it receives far less systematic preparation than arithmetic topics like percentage and ratio. Most students memorize area formulas without understanding the relationships between them, learn angle theorems in isolation, and then struggle to apply them when questions combine two or three concepts in a single problem.
The key to fast geometry is not memorizing more formulas — it is recognizing patterns. Most geometry problems tested in competitive exams are variations of 8–10 core configurations. A student who has internalized these configurations and the shortcuts that apply to them can solve the majority of SSC CGL geometry questions in 30–45 seconds.
This guide covers those configurations systematically — triangles first, then circles, then quadrilaterals and polygons — with the fastest available method for each type and exam-format worked examples throughout.
Part 1: Triangles
Core Area Formulas
Formula 1 — Base and Height:
Area = ½ × b × h
Formula 2 — Heron's Formula (all three sides known):
s = (a + b + c) ÷ 2
Area = √(s × (s−a) × (s−b) × (s−c))
Formula 3 — Two sides and included angle:
Area = ½ × a × b × sin C
Formula 4 — Equilateral triangle (side a):
Area = (√3 ÷ 4) × a²
Shortcut value: √3/4 ≈ 0.433 — memorize this decimal for fast calculation.
Worked Example 1: Equilateral triangle, side = 8 cm
- Area = (√3/4) × 64 = 16√3 ≈ 27.7 cm²
Worked Example 2: Triangle with sides 13, 14, 15
- s = (13+14+15) ÷ 2 = 21
- Area = √(21 × 8 × 7 × 6) = √7056 = 84 cm²
Triangle Angle Properties — The Critical Shortcuts
Property 1: Angle Sum
Sum of all angles in any triangle = 180°
Property 2: Exterior Angle Theorem
Exterior angle = Sum of two non-adjacent interior angles
- If interior angles are 50° and 70°, exterior angle = 120°
- Fastest exam application: find an angle without full calculation
Property 3: Isoceles Triangle
Base angles are equal. If apex angle = θ, each base angle = (180° − θ) ÷ 2
Property 4: In a right triangle
- Hypotenuse² = Base² + Height² (Pythagoras)
- Median to hypotenuse = half the hypotenuse
- Altitude to hypotenuse: h = (a × b) ÷ c, where a, b are legs and c is hypotenuse
Pythagorean Triplets — Memorize These 10
Recognizing Pythagorean triplets eliminates all calculation for right triangle problems.
| Triplet | Multiples |
|---|---|
| 3, 4, 5 | 6-8-10, 9-12-15, 12-16-20 |
| 5, 12, 13 | 10-24-26 |
| 8, 15, 17 | 16-30-34 |
| 7, 24, 25 | 14-48-50 |
| 9, 40, 41 | — |
| 11, 60, 61 | — |
| 20, 21, 29 | — |
| 28, 45, 53 | — |
Exam shortcut: When you see two sides of a right triangle matching a known triplet pattern — even as multiples — write the third side instantly without using Pythagoras.
Worked Example: Right triangle with legs 15 and 20. Hypotenuse?
- Recognize 15-20-? as 3 × (5-4-?): multiply 3 × 5 → 25
Special Triangle Shortcuts
30-60-90 Triangle (sides in ratio 1 : √3 : 2):
- Short leg = x → Long leg = x√3, Hypotenuse = 2x
- If hypotenuse = 10: short leg = 5, long leg = 5√3
45-45-90 Triangle (sides in ratio 1 : 1 : √2):
- Leg = x → Hypotenuse = x√2
- If hypotenuse = 8√2: each leg = 8
Worked Example: Equilateral triangle of side 6 cm. Find height.
- Height divides it into two 30-60-90 triangles
- Height = (√3 ÷ 2) × 6 = 3√3 cm
Centers of a Triangle
| Center | Definition | Key Property |
|---|---|---|
| Centroid (G) | Intersection of medians | Divides each median in 2:1 from vertex |
| Circumcenter (O) | Intersection of perpendicular bisectors | Equidistant from all vertices |
| Incenter (I) | Intersection of angle bisectors | Equidistant from all sides |
| Orthocenter (H) | Intersection of altitudes | In right triangle: at the right angle vertex |
Exam-critical property: Centroid divides each median in ratio 2:1 from vertex.
Worked Example: Median from vertex A = 9 cm. Distance from A to centroid?
- AG = 2/3 × 9 = 6 cm
- GD (centroid to midpoint) = 1/3 × 9 = 3 cm
Similarity and Congruence Shortcuts
Area ratio of similar triangles:
Area₁ ÷ Area₂ = (Side₁ ÷ Side₂)²
Worked Example: Two similar triangles with sides in ratio 3:5. Area of larger = 100 cm². Area of smaller?
- Area ratio = 9:25
- Smaller area = 9/25 × 100 = 36 cm²
Mid-Point Theorem:
Line joining midpoints of two sides of a triangle is parallel to the third side and half its length.
Basic Proportionality Theorem (BPT):
If a line is parallel to one side of a triangle, it divides the other two sides proportionally.
Part 2: Circles
Core Formulas — No Confusion Version
| Measurement | Formula |
|---|---|
| Circumference | 2πr |
| Area | πr² |
| Arc length (angle θ°) | (θ ÷ 360) × 2πr |
| Sector area (angle θ°) | (θ ÷ 360) × πr² |
| Segment area | Sector area − Triangle area |
π values for calculation:
- Use π = 22/7 when radius is a multiple of 7
- Use π = 3.14 for all other cases
- Use π = 3 for quick approximations
Circle Theorems — The 7 Most Tested
Theorem 1: Angle at Centre = 2 × Angle at Circumference
If arc AB subtends ∠AOB at centre and ∠ACB at circumference:
∠AOB = 2 × ∠ACB
Exam application: Central angle = 110° → Inscribed angle on same arc = 55°
Theorem 2: Angles in Same Segment Are Equal
All inscribed angles subtending the same arc are equal.
Exam application: If ∠APB = 40°, then ∠AQB = 40° for any Q on the same arc.
Theorem 3: Angle in Semicircle = 90°
Any angle inscribed in a semicircle (subtending the diameter) = 90°.
Exam shortcut: Whenever you see a triangle inscribed in a circle with one side as diameter → that angle is 90°. Instantly apply Pythagoras.
Theorem 4: Opposite Angles of Cyclic Quadrilateral = 180°
In a quadrilateral inscribed in a circle:
∠A + ∠C = 180°, ∠B + ∠D = 180°
Worked Example: Cyclic quadrilateral with ∠A = 75°, ∠B = 110°. Find ∠C and ∠D.
- ∠C = 180° − 75° = 105°
- ∠D = 180° − 110° = 70°
Theorem 5: Tangent-Radius Perpendicularity
A tangent to a circle is perpendicular to the radius at the point of tangency.
Theorem 6: Equal Tangents from External Point
Tangent lengths from an external point to a circle are equal.
Exam application: From external point P, two tangents PA and PB → PA = PB always.
Worked Example: Two tangents from point P are each 12 cm. Distance from P to centre = 13 cm. Find radius.
- Right triangle: r² + 12² = 13²
- r² = 169 − 144 = 25 → r = 5 cm
Theorem 7: Chord-Tangent Angle (Tangent-Chord Angle)
Angle between tangent and chord = Inscribed angle in alternate segment.
Common Circle Problems in Exams
Problem Type 1: Area of Ring (Annulus)
Two concentric circles, radii R and r.
Area of ring = π(R² − r²) = π(R+r)(R−r)
Worked Example: Outer radius = 7 cm, inner = 3 cm.
- Area = π × (7+3) × (7−3) = π × 10 × 4 = 40π ≈ 125.7 cm²
Problem Type 2: Perimeter of Sector
Perimeter = 2r + Arc length = 2r + (θ ÷ 360) × 2πr
Worked Example: Sector, r = 14 cm, θ = 90°.
- Arc = (90 ÷ 360) × 2π × 14 = 1/4 × 88 = 22 cm
- Perimeter = 28 + 22 = 50 cm
Problem Type 3: Largest Circle in Square / Square in Circle
- Circle inscribed in square of side a: r = a/2
- Square inscribed in circle of radius r: side = r√2
Part 3: Quadrilaterals
Area Formulas — All Types
| Shape | Area Formula | Key Condition |
|---|---|---|
| Rectangle | l × b | — |
| Square | a² or d²/2 | d = diagonal |
| Parallelogram | b × h | h = perpendicular height |
| Rhombus | (d₁ × d₂) ÷ 2 | d₁, d₂ = diagonals |
| Trapezium | ½ × (a+b) × h | a, b = parallel sides |
| Kite | (d₁ × d₂) ÷ 2 | Same as rhombus |
Critical Diagonal Properties
Rectangle: Diagonals are equal and bisect each other
d = √(l² + b²)
Square: Diagonals are equal, bisect each other at 90°, bisect vertex angles
d = a√2
Parallelogram: Diagonals bisect each other (not equal, not perpendicular in general)
Rhombus: Diagonals bisect each other at 90°, bisect vertex angles (not equal)
Side² = (d₁/2)² + (d₂/2)²
Worked Example: Rhombus with diagonals 16 cm and 12 cm. Find side and area.
- Side = √(8² + 6²) = √(64 + 36) = √100 = 10 cm
- Area = (16 × 12) ÷ 2 = 96 cm²
Parallelogram Shortcuts
Key properties:
- Opposite sides equal and parallel
- Opposite angles equal
- Consecutive angles supplementary (sum = 180°)
- Diagonals bisect each other
Area shortcut using diagonals and angle:
Area = ½ × d₁ × d₂ × sin θ
Where θ = angle between diagonals
Worked Example: Parallelogram with diagonals 10 cm and 8 cm intersecting at 30°.
- Area = ½ × 10 × 8 × sin 30° = ½ × 10 × 8 × ½ = 20 cm²
Trapezium Shortcuts
Area = ½ × (sum of parallel sides) × height
Worked Example: Trapezium with parallel sides 13 cm and 7 cm, height 8 cm.
- Area = ½ × (13+7) × 8 = ½ × 20 × 8 = 80 cm²
Midsegment of trapezium = average of parallel sides = (a+b) ÷ 2
Isoceles trapezium: Non-parallel sides equal, base angles equal, diagonals equal.
Part 4: Polygons
Interior and Exterior Angles
Sum of interior angles of n-sided polygon:
S = (n − 2) × 180°
Each interior angle of regular polygon:
Each interior angle = (n − 2) × 180° ÷ n
Each exterior angle of regular polygon:
Each exterior angle = 360° ÷ n
Note: Interior + Exterior = 180° always.
Quick Reference Table
| Polygon | Sides | Sum of Interior Angles | Each Interior Angle |
|---|---|---|---|
| Triangle | 3 | 180° | 60° |
| Quadrilateral | 4 | 360° | 90° |
| Pentagon | 5 | 540° | 108° |
| Hexagon | 6 | 720° | 120° |
| Octagon | 8 | 1,080° | 135° |
| Decagon | 10 | 1,440° | 144° |
Number of diagonals in n-sided polygon:
Diagonals = n × (n − 3) ÷ 2
- Hexagon: 6 × 3 ÷ 2 = 9 diagonals
- Octagon: 8 × 5 ÷ 2 = 20 diagonals
Area of Regular Polygon
Area of regular polygon with n sides, side length a:
Area = (n × a²) ÷ 4 × cot(180° ÷ n)
For common polygons (memorize these):
- Regular hexagon, side a: Area = (3√3 ÷ 2) × a²
- Regular hexagon = 6 equilateral triangles → Area = 6 × (√3/4) × a²
Worked Example: Regular hexagon with side 6 cm.
- Area = 6 × (√3/4) × 36 = 6 × 9√3 = 54√3 cm²
Part 5: Mensuration Shortcuts
Volume and Surface Area — Quick Reference
| Solid | Volume | Total Surface Area |
|---|---|---|
| Cube (side a) | a³ | 6a² |
| Cuboid (l, b, h) | l × b × h | 2(lb + bh + hl) |
| Cylinder (r, h) | πr²h | 2πr(r + h) |
| Cone (r, h, l) | (1/3)πr²h | πr(r + l) |
| Sphere (r) | (4/3)πr³ | 4πr² |
| Hemisphere (r) | (2/3)πr³ | 3πr² |
Most-Tested Mensuration Shortcuts
Shortcut 1: Cube side doubles → Volume increases by 8×, Surface area by 4×
Shortcut 2: Cylinder radius doubles, height halves → Volume doubles
- πr²h → π(2r)²(h/2) = 2πr²h ✓
Shortcut 3: Cone and Cylinder with same base and height:
- Volume of cone = (1/3) × Volume of cylinder
Shortcut 4: Sphere melted into smaller shapes — equate volumes:
- (4/3)πR³ = n × πr²h → solve for n
Worked Example: A sphere of radius 6 cm melted into cones of radius 2 cm and height 3 cm. How many cones?
- Sphere volume = (4/3) × π × 216 = 288π
- Cone volume = (1/3) × π × 4 × 3 = 4π
- Number = 288π ÷ 4π = 72 cones