Mensuration (2D + 3D with numericals)

Mensuration (2D + 3D with numericals)

Concept

Mensuration is pure formula recall. The key challenge in SSC Tier 2 is not just remembering formulas but knowing which one applies and when dimensions are given in mixed units. Always convert to consistent units before calculating.

2D Figures:

• Triangle: area = ½ × base × height. Equilateral triangle area = (√3/4)a².
• Parallelogram: area = base × height. Rhombus: area = ½ × d₁ × d₂ (diagonals).
• Trapezium: area = ½ × (sum of parallel sides) × height.
• Circle: area = πr², circumference = 2πr. Arc length = (θ/360) × 2πr. Sector area = (θ/360) × πr².

3D Figures:

• Cuboid: volume = l×b×h, TSA = 2(lb+bh+hl).
• Right circular cone: volume = (1/3)πr²h, slant height = √(r²+h²), CSA = πrl.
• Frustum of a cone: volume = (1/3)πh(r₁² + r₂² + r₁r₂).
• Hollow cylinder: volume = πh(R² − r²). Outer SA + inner SA + 2π(R²−r²) for ends.

Combination of Solids: When shapes are combined (e.g., a cone on top of a cylinder), add volumes of individual shapes. For surface area, add areas of exposed surfaces and subtract any hidden faces.

Key Points

• When a wire is reshaped from one shape to another, volume remains constant (no wire is lost). Set volumes equal.
• If radius doubles, area increases by 4× (square relationship), volume increases by 8× (cube relationship).
• For a sphere inscribed in a cube (touching all faces): sphere diameter = cube side. For a cube inscribed in a sphere (corners touching): cube diagonal = sphere diameter.
• Percentage increase in area when side increases by x%: new area = (1 + x/100)² times old area → increase = (2x + x²/100)%.
• Percentage increase in volume when side increases by x%: new volume = (1 + x/100)³ times old volume.

Worked Example

Q: A solid sphere of radius 6 cm is melted and recast into a cone of radius 6 cm and height 12 cm. Find the height of the cone. (π = 22/7) Approach: Volume of sphere = (4/3)π × 6³ = (4/3)π × 216 = 288π. Volume of cone = (1/3)π × 6² × 12 = (1/3)π × 36 × 12 = 144π. Since volume is the same: 288π = 144π? That doesn’t work — the sphere volume is larger. Actually let me check: sphere volume = 4/3 × π × 216 = 904.78. Cone volume = 1/3 × π × 36 × 12 = 452.39. These are unequal. Let me reconsider the problem: Actually the problem states the sphere is melted into a cone — the numbers must work out. Perhaps the height is what we need to find: Volume of sphere = (4/3)π × 216 = 288π. For the cone: 288π = (1/3)π × 36 × h → 288 = 12h → h = 24 cm. Yes that’s the correct approach: find h from equating volumes.

SSC Pattern / Tips

• Wire reshaping problems: volume of wire (cylinder of very small radius) = πr² × length. When reshaped, equate volumes.
• For a sphere in a cube: diagonal of cube = diameter of sphere = a√3.
• For percentage increase problems, use the square/cube of the change factor, not the percentage change.