Mensuration — Areas and Perimeters of 2D Shapes

What Mensuration Covers

Mensuration deals with measuring the perimeter (boundary length) and area (surface enclosed) of 2D shapes. For Classes 6-8, we work with triangles, quadrilaterals, circles, and their combinations. This chapter is formula-heavy, but every formula has a clear geometric reason behind it.

graph TD
    A[2D Shape] --> B{Which shape?}
    B -->|Triangle| C["A = ½bh"]
    B -->|Rectangle| D[A = l × b]
    B -->|Square| E[A = s²]
    B -->|Parallelogram| F[A = b × h]
    B -->|Trapezium| G["A = ½(a+b)h"]
    B -->|Rhombus| H["A = ½d₁d₂"]
    B -->|Circle| I[A = πr²]
    C --> J[No height? Use Heron]
    A --> K{Composite?}
    K -->|Yes| L[Split into known shapes]
    L --> M[Add or subtract areas]

Key Terms & Definitions

Perimeter — Total boundary length of a 2D shape. For polygons, sum of all sides.

Area — Surface enclosed within a shape. Measured in square units.

Base and Height — In triangles and parallelograms, the height is the perpendicular distance from the base to the opposite vertex or side.

Heron’s Formula — Area of a triangle from all three sides: $A = \sqrt{s(s-a)(s-b)(s-c)}$ where $s = \frac{a+b+c}{2}$ .

All Formulas

Formula	Expression
Area (base-height)	$\frac{1}{2} \times b \times h$
Area (Heron’s)	$\sqrt{s(s-a)(s-b)(s-c)}$
Perimeter	$a + b + c$
Equilateral triangle	$\frac{\sqrt{3}}{4} a^2$

Shape	Area	Perimeter
Rectangle	$l \times b$	$2(l + b)$
Square	$s^2$	$4s$
Parallelogram	$b \times h$	$2(a + b)$
Rhombus	$\frac{1}{2} d_1 d_2$	$4s$
Trapezium	$\frac{1}{2}(a + b)h$	Sum of all sides

Formula	Expression
Circumference	$2\pi r$
Area	$\pi r^2$
Sector area	$\frac{\theta}{360} \times \pi r^2$
Arc length	$\frac{\theta}{360} \times 2\pi r$

A parallelogram’s area ( $b \times h$ ) equals a rectangle’s because you can cut and rearrange any parallelogram into a rectangle. The triangle formula is half of that — a triangle is half a parallelogram.

Solved Examples — Easy to Hard

Example 1 (Easy — Class 6)

Rectangle: length 12 cm, breadth 8 cm. Find area and perimeter.

Area $= 12 \times 8 = \mathbf{96 \text{ cm}^2}$ . Perimeter $= 2(12+8) = \mathbf{40 \text{ cm}}$ .

Example 2 (Medium — Class 7)

Triangle with sides 13, 14, 15 cm. Find area using Heron’s formula.

$s = 21$ . Area $= \sqrt{21 \times 8 \times 7 \times 6} = \sqrt{7056} = \mathbf{84 \text{ cm}^2}$ .

Example 3 (Medium — Class 8)

Rhombus with diagonals 24 cm and 10 cm. Find area and side.

Area $= \frac{1}{2}(24)(10) = \mathbf{120 \text{ cm}^2}$

Half-diagonals: 12 and 5. Side $= \sqrt{144 + 25} = \mathbf{13 \text{ cm}}$ .

Example 4 (Hard — Class 8)

Isosceles trapezium: parallel sides 10 m and 16 m, non-parallel sides 5 m each. Find area.

Extra length per side $= \frac{16-10}{2} = 3$ m. Height $= \sqrt{25 - 9} = 4$ m.

Area $= \frac{1}{2}(10+16)(4) = \mathbf{52 \text{ m}^2}$ .

Exam-Specific Tips

CBSE Class 6-7: Direct formula application. Word problems about fencing (perimeter) and painting/tiling (area) are common. Know rectangle, square, triangle, circle formulas cold.

CBSE Class 8: Heron’s formula, trapezium, and composite figures. Questions often ask you to decompose an irregular plot into triangles and quadrilaterals — practise this splitting skill.

Common Mistakes to Avoid

Mistake 1 — Using slant side as height. In parallelograms and trapeziums, “height” is the perpendicular distance, not the slant side.

Mistake 2 — Forgetting $\frac{1}{2}$ in the triangle formula. Area is $\frac{1}{2}bh$ , not $bh$ .

Mistake 3 — Using full perimeter instead of semi-perimeter in Heron’s. The formula needs $s = \frac{a+b+c}{2}$ .

Mistake 4 — Mixing circumference and area for circles. Circumference ( $2\pi r$ ) is a length. Area ( $\pi r^2$ ) is a surface measure.

Mistake 5 — Not converting units before calculation. If length is in metres and breadth in centimetres, convert first.

Practice Questions

Q1. Parallelogram: base 15 cm, height 8 cm. Find area.

$15 \times 8 = 120$ cm $^2$ .

Q2. Circle circumference 44 cm. Find area.

$r = 7$ cm. Area $= \frac{22}{7} \times 49 = 154$ cm $^2$ .

Q3. Trapezium: parallel sides 8, 14 cm; height 6 cm. Find area.

$\frac{1}{2}(8+14)(6) = 66$ cm $^2$ .

Q4. Triangle: sides 120, 160, 200 m. Find area.

Right triangle ( $120^2 + 160^2 = 200^2$ ). Area $= \frac{1}{2}(120)(160) = 9600$ m $^2$ .

Q5. Circular ring: outer radius 10 cm, inner 6 cm. Find area.

$\pi(100 - 36) = 64\pi \approx 201.1$ cm $^2$ .

Q6. Park $60 \times 40$ m with 2 m path around. Find path area.

Outer $= 64 \times 44 = 2816$ . Path $= 2816 - 2400 = 416$ m $^2$ .

Q7. Equilateral triangle, side 10 cm. Find area.

$\frac{\sqrt{3}}{4}(100) = 25\sqrt{3} \approx 43.3$ cm $^2$ .

Q8. Sector: radius 14 cm, angle 90°. Find area and arc length.

Area $= \frac{1}{4}\pi(196) = 154$ cm $^2$ . Arc $= \frac{1}{4}(2\pi)(14) = 22$ cm.

FAQs

What’s the difference between perimeter and area?

Perimeter is boundary length (cm, m). Area is enclosed surface (cm $^2$ , m $^2$ ). Perimeter = fencing needed. Area = paint needed.

When to use Heron’s formula vs. base-height?

Use $\frac{1}{2}bh$ when height is known. Use Heron’s when all three sides are given but height isn’t.

How to find area of irregular shapes?

Split into triangles, rectangles, trapeziums. Find each area and add. For shapes with holes, subtract the removed part.

Why is circle area $\pi r^2$ ?

Cut a circle into thin sectors, rearrange into a near-rectangle. Length $= \pi r$ , width $= r$ . Area $= \pi r \times r = \pi r^2$ .

What units for area and perimeter?

Perimeter in cm, m, km. Area in cm $^2$ , m $^2$ , km $^2$ . Always include the correct unit — examiners deduct marks for missing units.

3D Mensuration (Class 8+)

Once you master 2D, the jump to 3D is natural. We add one more measurement — surface area (paint needed for the outside) and volume (water the shape can hold).

Shape	Lateral/Curved SA	Total SA	Volume
Cuboid	$2h(l+b)$	$2(lb + bh + hl)$	$lbh$
Cube	$4s^2$	$6s^2$	$s^3$
Cylinder	$2\pi rh$	$2\pi r(r+h)$	$\pi r^2 h$
Cone	$\pi r l$	$\pi r(r+l)$	$\frac{1}{3}\pi r^2 h$
Sphere	$4\pi r^2$	$4\pi r^2$	$\frac{4}{3}\pi r^3$
Hemisphere	$2\pi r^2$	$3\pi r^2$	$\frac{2}{3}\pi r^3$

Where $l$ = slant height for cone: $l = \sqrt{r^2 + h^2}$ .

Worked Examples — 3D

A cylindrical tank has radius 7 m and height 10 m. Find the volume of water it can hold.

$V = \pi r^2 h = \frac{22}{7} \times 49 \times 10 = 1540$ m $^3$

Since 1 m $^3$ = 1000 litres, the tank holds $15,40,000$ litres.

A cone has the same base radius and height as a cylinder. What fraction of the cylinder’s volume does the cone occupy?

Cone volume = $\frac{1}{3}\pi r^2 h$ . Cylinder volume = $\pi r^2 h$ .

Fraction = $\frac{1}{3}$ . The cone occupies exactly one-third of the cylinder. This is why the formula has the $\frac{1}{3}$ factor.

A solid sphere of radius 6 cm is melted and recast into a cylinder of radius 2 cm. Find the cylinder’s height.

Volume of sphere = $\frac{4}{3}\pi(6)^3 = 288\pi$ cm $^3$

Volume of cylinder = $\pi(2)^2 h = 4\pi h$

Setting equal: $4\pi h = 288\pi$ , so $h = 72$ cm.

Composite figures — the strategy

Real exam problems combine shapes. A typical problem: “A solid is a cylinder with hemispheres on both ends.” The surface area is the curved surface of the cylinder plus the curved surfaces of two hemispheres. The volume is the cylinder volume plus two hemisphere volumes.

The key rule: never double-count surfaces. Where two shapes join, neither surface is exposed — subtract those areas from the total surface area.

Additional Practice Questions

Q9. A cone has radius 3 cm and height 4 cm. Find slant height, curved SA, and volume.

$l = \sqrt{9 + 16} = 5$ cm. CSA $= \pi r l = \pi \times 3 \times 5 = 15\pi \approx 47.1$ cm $^2$ . Volume $= \frac{1}{3}\pi(9)(4) = 12\pi \approx 37.7$ cm $^3$ .

Q10. A sphere of radius 14 cm. Find surface area and volume.

SA $= 4\pi r^2 = 4 \times \frac{22}{7} \times 196 = 2464$ cm $^2$ . Volume $= \frac{4}{3}\pi r^3 = \frac{4}{3} \times \frac{22}{7} \times 2744 = 11498.67$ cm $^3$ .

For sphere problems, remember that surface area is $4\pi r^2$ (4 times the area of a great circle), and volume is $\frac{4}{3}\pi r^3$ . The 4/3 is hard to derive but easy to memorise: “four-thirds pi r-cubed.”

Unit Conversions That Trip Students

From	To	Factor
1 m $^2$	cm $^2$	$\times 10^4$
1 m $^3$	cm $^3$	$\times 10^6$
1 m $^3$	litres	$\times 1000$
1 hectare	m $^2$	$= 10,000$ m $^2$
1 km $^2$	hectares	$= 100$
1 litre	cm $^3$	$= 1000$

The most common unit error: converting m $^2$ to cm $^2$ by multiplying by 100 instead of 10,000. Area units square the linear conversion: $1\text{ m}^2 = (100\text{ cm})^2 = 10000\text{ cm}^2$ . Volume units cube it: $1\text{ m}^3 = 10^6\text{ cm}^3$ .

Real-world applications

Fencing a garden: perimeter tells you how much wire to buy
Painting a wall: area tells you how much paint to buy (1 litre covers roughly 10–12 m $^2$ )
Filling a tank: volume tells you how much water it holds
Laying tiles: floor area divided by tile area gives number of tiles
Cost of land: area in hectares or square metres multiplied by rate per unit area

These word problem types repeat in every CBSE exam from Class 6 to 8. Translate the story into the correct formula, plug in numbers, and check units.