PSTET · Mathematics and Science (Paper II — Classes VI-VIII) · Mathematics Content (Class VI-VIII)

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Mensuration

Perimeter, area, surface area and volume of solids.

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Mensuration

Perimeter, Area, Surface Area and Volume of Solids


Overview

Mensuration is the branch of mathematics dealing with measurement of geometric figures — their lengths, areas and volumes. You must be comfortable with two-dimensional figures (perimeter and area) as well as three-dimensional solids (surface area and volume).

The topic tests both formula recall and application in word problems. Examiners often frame questions around real-life contexts — fencing a field, painting a room, filling a tank — so understanding what each formula measures is as important as memorising it. Mastery here also supports the pedagogy section, where you may be asked how to teach these concepts using concrete materials.


Key Concepts

  • Perimeter is the total length of the boundary of a 2-D figure. Units are linear (m, cm).
  • Area is the measure of the surface enclosed by a 2-D figure. Units are square (m², cm²).
  • Surface Area is the total area of all outer faces of a 3-D solid. Distinguish between Curved Surface Area (CSA) and Total Surface Area (TSA).
  • Volume is the space occupied by a 3-D solid. Units are cubic (m³, cm³). 1 litre = 1000 cm³.
  • Unit conversion is critical: 1 m = 100 cm, so 1 m² = 10000 cm² and 1 m³ = 10⁶ cm³.
  • For composite figures, break into standard shapes, compute separately, then add or subtract as needed.
  • Pedagogy tip: Use paper cut-outs for area and unit cubes/water displacement for volume to build conceptual understanding.

Formulas / Key Facts

2-D Figures (Perimeter & Area)

FigurePerimeterArea
Rectangle (l × b)2(l + b)l × b
Square (side a)4aa²
Triangle (sides a, b, c; base b, height h)a + b + c½ × b × h
Equilateral Triangle (side a)3a(√3 / 4) × a²
Right Triangle (legs p, q)p + q + hypotenuse½ × p × q
Parallelogram (base b, height h)2(a + b) where a, b are adjacent sidesb × h
Rhombus (diagonals d₁, d₂)4 × side½ × d₁ × d₂
Trapezium (parallel sides a, b; height h)sum of all sides½ × (a + b) × h
Circle (radius r)Circumference = 2πrπr²
Semicircleπr + 2r½ πr²

Use π = 22/7 or 3.14 as specified in the question.

3-D Solids (Surface Area & Volume)

SolidCurved/Lateral SATotal SAVolume
Cuboid (l × b × h)2h(l + b)2(lb + bh + hl)l × b × h
Cube (side a)4a²6a²a³
Cylinder (radius r, height h)2πrh2πr(r + h)πr²h
Cone (radius r, slant height l, height h)πrlπr(r + l)⅓ πr²h
Sphere (radius r)—4πr²(4/3)πr³
Hemisphere (radius r)2πr²3πr²(2/3)πr³

Slant height of cone: l = √(r² + h²)


Worked Examples

Example 1 — Area of a Trapezium

Problem: A trapezium has parallel sides 12 cm and 8 cm. The perpendicular distance between them is 5 cm. Find the area.

Solution: Area = ½ × (sum of parallel sides) × height Area = ½ × (12 + 8) × 5 Area = ½ × 20 × 5 = 50 cm²


Example 2 — Volume and Surface Area of a Cylinder

Problem: A cylindrical water tank has radius 7 m and height 10 m. Find (a) capacity in litres, (b) cost of painting the curved surface at ₹15 per m².

Solution: (a) Volume = πr²h = (22/7) × 7² × 10 = (22/7) × 49 × 10 = 1540 m³ 1 m³ = 1000 litres → Capacity = 1540 × 1000 = 15,40,000 litres

(b) CSA = 2πrh = 2 × (22/7) × 7 × 10 = 440 m² Cost = 440 × 15 = ₹6600


Example 3 — Composite Solid

Problem: A solid is in the form of a cone mounted on a hemisphere. Both have radius 3 cm. The height of the cone is 4 cm. Find the total surface area.

Solution: Slant height of cone, l = √(r² + h²) = √(9 + 16) = √25 = 5 cm CSA of cone = πrl = (22/7) × 3 × 5 = 330/7 cm² CSA of hemisphere = 2πr² = 2 × (22/7) × 9 = 396/7 cm² TSA = 330/7 + 396/7 = 726/7 ≈ 103.7 cm²

(Note: The base of the cone sits on the hemisphere, so we take only curved surfaces.)


Common Mistakes

  1. Confusing perimeter with area → Perimeter is a length (one dimension); area is a surface (two dimensions). Always check units.
  2. Mixing CSA and TSA → Read the question carefully. "Painting the curved surface" needs CSA; "total outer surface" needs TSA.
  3. Forgetting to square or cube during unit conversion → 2 m² ≠ 200 cm². Correct: 2 m² = 2 × 10000 = 20000 cm².
  4. Using diameter instead of radius → Formulas use radius. If the question gives diameter, halve it first.
  5. Ignoring slant height vs vertical height in cones → Volume uses vertical height h; curved surface area uses slant height l. Calculate l if not given.
  6. Adding volumes when asked for surface area of composite solids → Identify which measure is asked; they are independent.

Quick Reference

  • Rectangle area = l × b; perimeter = 2(l + b).
  • Circle area = πr²; circumference = 2πr.
  • Cylinder volume = πr²h; CSA = 2πrh; TSA = 2πr(r + h).
  • Cone volume = ⅓ πr²h; slant height l = √(r² + h²).
  • Sphere volume = (4/3)πr³; surface area = 4πr².
  • 1 m³ = 1000 litres; 1 litre = 1000 cm³.

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Notes generated on 28 Jun 2026