Geometry

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Geometry is one of the most scoring topics in SSC CGL Quantitative Aptitude. It tests your understanding of shapes, their properties, areas, volumes, and relationships between angles. The key to mastering geometry is memorising formulas and understanding theorems.

Essential Formulas:

Triangle:

• Area = $\frac{1}{2} \times base \times height$
• Perimeter = a + b + c (sum of all three sides)
• Heron’s Formula: $A = \sqrt{s(s-a)(s-b)(s-c)}$ where $s = \frac{a+b+c}{2}$
• Pythagorean Theorem (Right-angled): $a^2 + b^2 = c^2$

Quadrilateral:

• Parallelogram: Area = base × height; Perimeter = 2(a + b)
• Rectangle: Area = l × b; Perimeter = 2(l + b); Diagonal = $\sqrt{l^2 + b^2}$
• Square: Area = side²; Perimeter = 4 × side; Diagonal = side × √2

Circle:

• Area = πr²
• Circumference = 2πr
• Arc length = $\frac{\theta}{360} \times 2\pi r$ (where θ is in degrees)

⚡ SSC CGL Exam Tips:

• Always check if triangle is right-angled before applying Pythagorean theorem
• π = 22/7 for rough calculations, 3.1416 for precise
• Similar triangles have equal corresponding angles and proportional sides
• Congruent triangles (SSS, SAS, ASA, RHS, AAS rules)

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Detailed Properties and Worked Examples

Example 1: Triangle Area The sides of a triangle are 5 cm, 12 cm, and 13 cm. Find its area.

Solution: Check if right-angled: $5^2 + 12^2 = 25 + 144 = 169 = 13^2$ ✓ Area = $\frac{1}{2} \times 5 \times 12 = 30$ sq cm

Example 2: Using Heron’s Formula Find area of triangle with sides 7 cm, 8 cm, 9 cm.

Solution: $s = \frac{7+8+9}{2} = 12$ Area = $\sqrt{12(12-7)(12-8)(12-9)} = \sqrt{12 \times 5 \times 4 \times 3} = \sqrt{720} = 26.83$ sq cm

Example 3: Circle Problem Radius of a circle is 7 cm. Find area and circumference.

Area = π × 7² = 22/7 × 49 = 154 sq cm Circumference = 2 × 22/7 × 7 = 44 cm

Angle Relationships:

• Sum of angles in triangle = 180°
• Sum of angles in quadrilateral = 360°
• Vertically opposite angles are equal
• Exterior angle of triangle = sum of two opposite interior angles

Similar vs Congruent Triangles:

Criteria	Similar Triangles	Congruent Triangles
Sides	Proportional	Equal
Angles	Equal	Equal
Area ratio	Square of side ratio	Same area

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage with theorems, proofs, and previous year SSC CGL patterns.

Important Theorems and Proofs:

Theorem 1: Basic Proportionality Theorem (Thales) In a triangle, a line drawn parallel to one side divides the other two sides proportionally. If DE || BC in ΔABC, then $\frac{AD}{DB} = \frac{AE}{EC}$

Theorem 2: Angle Bisector Theorem The angle bisector of an angle of a triangle divides the opposite side in the ratio of the other two sides. In ΔABC, if AD bisects ∠A, then $\frac{BD}{DC} = \frac{AB}{AC}$

Theorem 3: Apollonius’ Theorem In a triangle with sides a, b, c and median to side a being m: $$b^2 + c^2 = 2\left(m^2 + \frac{a^2}{4}\right)$$

Circumcircle and Incircle:

• Circumradius (R): $R = \frac{abc}{4\Delta}$ where Δ is area
• Inradius (r): $r = \frac{\Delta}{s}$ where s is semi-perimeter
• Inradius and Circumradius relation: $R \geq 2r$ (Euler’s inequality)

Polygon Properties:

Polygon	Number of Sides	Sum of Interior Angles
Triangle	3	180°
Quadrilateral	4	360°
Pentagon	5	540°
Hexagon	6	720°
n-gon	n	(n-2) × 180°

Previous Year SSC CGL Patterns:

SSC CGL 2022 Question: The ratio of areas of two similar triangles is 9:16. Find the ratio of their corresponding medians.

Solution: Ratio of areas = (ratio of sides)² $$\frac{9}{16} = \left(\frac{a}{b}\right)^2 \Rightarrow \frac{a}{b} = \frac{3}{4}$$ Since medians correspond to sides, ratio of medians = 3:4

SSC CGL 2023 Question: If the diagonals of a rhombus are 24 cm and 10 cm, find its perimeter.

Solution: In rhombus, diagonals bisect at 90° Half diagonals: 12 cm and 5 cm Side = $\sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13$ cm Perimeter = 4 × 13 = 52 cm

Advanced Topics for Higher Scores:

• Ceva’s Theorem: For concurrent cevians in a triangle
• Menelaus’ Theorem: For transversal intersecting sides
• Stewart’s Theorem: For lengths when a cevian is drawn
• Pedal triangles and their properties
• Euler’s formula for any polygon: $V - E + F = 2$

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