Mensuration (2D + 3D with numericals)

Mensuration (2D + 3D with numericals)

🟢 Lite

Key Rule / Formula

2D: Square: area = a², perimeter = 4a. Rectangle: area = l×b, perimeter = 2(l+b). Circle: area = πr², circumference = 2πr. Triangle: area = ½×b×h. 3D: Cube: volume = a³, TSA = 6a². Cylinder: volume = πr²h, CSA = 2πrh. Sphere: volume = (4/3)πr³, TSA = 4πr².

Memory Trick

For cone: slant height l = √(r²+h²). For hemisphere: volume = (2/3)πr³, curved SA = 2πr², total SA = 3πr².

1-Sentence Summary

Mensuration requires memorising area, perimeter, volume, and surface area formulas for 2D and 3D shapes, and knowing when to apply which formula.

Quick Example

Q: Find the volume of a cylinder with radius 7 cm and height 10 cm (π = 22/7). A: Volume = (22/7) × 7² × 10 = 22 × 7 × 10 = 1540 cm³.

🟡 Standard

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 base radius 6 cm. Find the height of the cone. (π = 22/7) Approach: Recasting conserves volume, so the volume of the sphere equals the volume of the cone. Volume of sphere = (4/3)π × 6³ = (4/3)π × 216 = 288π. Volume of cone = (1/3)π × 6² × h = 12πh. Equate the two: 288π = 12πh → h = 288/12 = 24 cm. The technique is always to set the original volume equal to the new volume and solve for the unknown dimension.

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.

🔴 Extended

Full Concept

Advanced 2D Mensuration:

• Sector and Segment: A sector is a “pizza slice” of a circle. Area = (θ/360)πr². A segment is the area between a chord and an arc. Segment area = Sector area − Triangle area (formed by radii and chord).
• Quadrilateral Area: For cyclic quadrilateral with sides a, b, c, d: area = √[(s−a)(s−b)(s−c)(s−d)] where s = semi-perimeter (Brahmagupta’s formula).
• Regular Hexagon: A regular hexagon can be divided into 6 equilateral triangles. Area = (3√3/2)a² where a = side. Each interior angle = 120°.

Advanced 3D Mensuration:

• Hemisphere: Volume = (2/3)πr³. Curved surface area = 2πr². Total surface area = 3πr² (includes base circle).
• Frustum of a Cone: When a cone is sliced parallel to its base, the smaller cut-off top is also a smaller similar cone. Volume of frustum = (1/3)πh(r₁² + r₂² + r₁r₂) where r₁ and r₂ are radii of the two circular ends.
• Inscribed and Circumscribed Shapes:
  • Largest cylinder inside a sphere: for the maximum-volume cylinder inscribed in a sphere of radius R, the cylinder has base radius r = R√(2/3) and height h = 2R/√3, giving maximum volume = (4/3√3)πR³.
  • Largest sphere inside a cylinder: sphere diameter = cylinder’s internal diameter, and the sphere fits only if cylinder height is at least that diameter.

Combination and Conversion Problems:

• When metal sheets are folded into open or closed cylinders, surface area changes but volume (of metal) stays same. The metal volume = area of sheet × thickness (usually negligible).
• When wire is drawn (stretched) to reduce diameter, volume is constant: πr₁²L₁ = πr₂²L₂. The length changes inversely with the square of the radius change.
• For cone + hemisphere + cylinder (composite solid), add individual volumes.

Error in Approximation: SSC uses π = 22/7 or 3.14 in most questions. If not specified, use 22/7 for calculations involving fractions of 7, and 3.14 otherwise.

SSC CGL Deep Analysis

• Frequency: 1–2 questions per paper. Combination solids and percentage increase in area/volume are most common.
• Difficulty: Medium. The hardest are the inscribed shapes (largest cone in a sphere, largest cylinder in a cone).
• Recent trend: Questions combining wire reshaping (constant volume) with the new shape’s dimensions — e.g., a wire bent into a circle, then reshaped into a square.
• Newer patterns: “The radius of a sphere increases by 10%. By what percentage does its volume increase?” Type questions — use (1.1)³ − 1 = 33.1%.
• Total weight in Tier 2: Roughly 3–4% of the quant paper.

High-Scoring Strategy

1. For wire reshaping problems: write volume of original = volume of new, substitute, solve. Volume of cylinder = πr²L.
2. For inscribed shapes, maximise volume by setting the derivative to zero, or simply recall the standard results (largest cone in a sphere, largest cylinder in a cone).
3. When comparing old and new area/volume after percentage change in dimensions, use the square/cube expansion: (1 ± x/100)² or ³.
4. For composite solids, draw a rough sketch, label each component, calculate each volume separately, add.
5. If a shape is painted both inside and outside with some thickness, use surface area × thickness for volume of paint.

SSC-Level Practice

Q1: A sphere of radius 6 cm is dropped into a cylindrical vessel of radius 6 cm containing water. By how much does the water level rise? Answer: 8 cm — Working: Volume of sphere = (4/3)π × 216 = 288π. Cylindrical cross-section area = π × 36 = 36π. Rise in water level = Volume / cross-section area = 288π / 36π = 8 cm.

Q2: A wire of radius 2 mm is melted and recast into a thinner wire of radius 1 mm. If the original wire is 50 cm long, find the length of the new wire. Answer: 200 cm — Working: Volume is conserved, so π × r₁² × L₁ = π × r₂² × L₂. Here r₁ = 2 mm, L₁ = 50 cm, r₂ = 1 mm. So 2² × 50 = 1² × L₂ → 4 × 50 = L₂ → L₂ = 200 cm. The length grows by the square of the radius ratio: halving the radius quadruples the length.

Q3: The radius of a sphere increases from 7 cm to 14 cm. Find the ratio of new volume to original volume. Answer: 8:1 — Working: Volume ∝ r³. New r = 2 × old r. New volume = 8 × old volume. Ratio = 8:1.

Common Traps

• Trap 1: Confusing CSA (Curved Surface Area) and TSA (Total Surface Area). CSA excludes the base/top; TSA includes all surfaces.
• Trap 2: Forgetting that for a hollow cylinder, the inner and outer curved surfaces have different areas. You can’t just use average radius.
• Trap 3: In wire reshaping problems, using wrong radius when the wire is described as “diameter 4 mm” instead of “radius 2 mm.” Always use radius in the volume formula.

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