Mensuration

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Mensuration — Quick Facts for SSC CGL

Mensuration is the branch of mathematics dealing with measurement of geometric figures — their area, perimeter, volume, and surface area.

2D Figures:

Triangle:

• Area = ½ × base × height
• Heron’s Formula (for any triangle, given 3 sides): Area = √[s(s−a)(s−b)(s−c)] where s = (a+b+c)/2 (semi-perimeter)
• Equilateral triangle (all sides equal, area = (√3/4) × a²; height = (√3/2)a
• Right-angled triangle: Area = ½ × product of perpendicular sides
• Perimeter: a + b + c

Quadrilateral:

• Rectangle: Area = l × b; Perimeter = 2(l + b); Diagonal = √(l² + b²)
• Square: Area = a²; Perimeter = 4a; Diagonal = a√2
• Parallelogram: Area = base × height
• Rhombus: Area = ½ × d₁ × d₂ (product of diagonals); Perimeter = 4a
• Trapezium: Area = ½ × (a + b) × h (sum of parallel sides × height)

Circle:

• Area = πr²; Circumference = 2πr; Arc length = (θ/360) × 2πr
• Area of sector = (θ/360) × πr²
• Chord length = 2√(r² − d²) where d = perpendicular distance from centre to chord

Key Formulas to Memorise:

• π ≈ 22/7 or 3.1416
• Area of ring (annulus) = π(R² − r²)

⚡ Exam tip: When a circle’s radius doubles, area quadruples (π(2r)² = 4πr²). This is a common trick in SSC CGL.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Mensuration — SSC CGL Study Guide

3D Figures — Surface Areas and Volumes:

Solid	TSA	Volume	Key Formulas
Cuboid	2(lb + bh + hl)	l × b × h	Diagonal = √(l² + b² + h²)
Cube	6a²	a³	Diagonal = a√3
Right Circular Cylinder	2πr(h + r)	πr²h	Curved SA = 2πrh
Cone	πr(l + r) where l = √(r² + h²)	(1/3)πr²h	Slant height l = √(r² + h²)
Sphere	4πr²	(4/3)πr³	—
Hemisphere	3πr² (including base)	(2/3)πr³	—

Important Results:

• Volume of cone = ⅓ × volume of cylinder of same base and height
• Volume of sphere = ⅔ × volume of circumscribing cylinder
• Volume of hemisphere = ½ × volume of sphere
• Surface area of sphere = 4 × area of circle of same radius

Combination of Solids: When two solids are joined:

• Total surface area = Sum of individual TSAs − area of contact (since internal surface is not exposed)
• Total volume = Sum of volumes

Hemisphere on a cylinder (solid figure in many SSC questions): A solid metal piece consisting of a hemisphere mounted on a cylinder:

• Volume = (2/3)πr³ + πr²h
• Total surface area (if base of hemisphere is included) = 3πr² + 2πrh

Frustum of a Cone: Volume = (1/3)πh(r₁² + r₂² + r₁r₂) where r₁ = larger radius, r₂ = smaller radius, h = height

⚡ Common mistake: Students forget to subtract the base area when combining solids, or forget that the curved surface area of a cone excludes the base. Always check whether the problem asks for Total Surface Area (TSA) or Curved/Lateral Surface Area (CSA/LSA).

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Mensuration — Comprehensive Notes

Solved Examples (SSC CGL Pattern):

Example 1: A room is 12 m long, 8 m wide, and 5 m high. Find the cost of plastering its four walls at ₹25 per sq. metre.

• Area of four walls = 2(l + b) × h = 2(12 + 8) × 5 = 200 m²
• Cost = 200 × 25 = ₹5,000

Example 2: The radius of a sphere is doubled. By what ratio does the volume increase?

• Original: V₁ = (4/3)πr³
• New: V₂ = (4/3)π(2r)³ = (4/3)π × 8r³ = 8 × V₁
• Ratio = 8:1

Example 3: The height of a right circular cone is 12 cm and the radius of its base is 5 cm. Find its slant height, total surface area, and volume.

• Slant height l = √(r² + h²) = √(25 + 144) = √169 = 13 cm
• Volume = (1/3)πr²h = (1/3) × π × 25 × 12 = 100π ≈ 314 cm³
• TSA = πr(l + r) = π × 5 × (13 + 5) = 90π ≈ 282.6 cm²

Example 4: A cylindrical tank 7 m in diameter is filled with water to a height of 3 m. A sphere of radius 3.5 m is immersed in it. How much does the water level rise?

• Volume of sphere = (4/3)π × (3.5)³ = (4/3)π × 42.875 = 57.17π m³
• Tank radius = 3.5 m (since diameter = 7 m)
• Rise in water level = Volume displaced / Area of base = (57.17π) / (π × 3.5²) = 57.17/12.25 ≈ 4.67 m

Example 5: A field is in the shape of a trapezium with parallel sides of 30 m and 20 m, and the distance between them is 15 m. The cost of fencing it at ₹12 per metre is:

• Perimeter = sum of all sides (need all sides to calculate cost). But for just the parallel sides: 30 + 20 = 50 m. However, for a trapezium we need the non-parallel sides too — assume they’re 13 m each (from Pythagorean theorem if it’s a right trapezium).

Important Derivations:

Area of equilateral triangle: For equilateral triangle of side a: Height = √[a² − (a/2)²] = √(a² − a²/4) = √(3a²/4) = a√3/2 Area = ½ × base × height = ½ × a × (a√3/2) = (a²√3)/4 ✓

Surface area to volume ratio (for SSC advanced): For a sphere: SA/V = 3/r. Smaller spheres have larger SA/V ratio → heat up/cool down faster (why small animals lose heat faster). For a cube: SA/V = 6/a. Larger cubes have smaller SA/V ratio.

Ratio of volumes of a cone, hemisphere, and cylinder of same base and height: Cone: Hemisphere: Cylinder = (1/3)πr²h : (2/3)πr³ : πr²h Since h = r for hemisphere: = (1/3)πr³ : (2/3)πr³ : πr³ = 1 : 2 : 3

This is a frequently tested relationship!

NEET/SSC Pattern Analysis: Mensuration is one of the most heavily tested quantitative aptitude topics in SSC CGL Tier-II. Key areas: volume and surface area of 3D solids, combination figures, and ratio problems. Questions often involve finding the number of small objects that can be made from a bigger object (e.g., how many spheres can be made from a cylinder).

⚡ SSC CGL 2022 Qn: If the diameter of a sphere is doubled, its surface area becomes how many times the original? Answer: 4 times. Since area ∝ r² and r doubles → new area = 4 × original. Note: Volume increases by 8 times.