📐 Mensuration - Complete Theory

Master area, perimeter, and volume calculations for all shapes!

🎯 What is Mensuration?

Mensuration is the branch of mathematics that deals with measurement of:

• Length (1D)
• Area (2D) - Space covered by flat shapes
• Volume (3D) - Space occupied by solid shapes
• Perimeter - Boundary length of flat shapes
• Surface Area - Total area of surfaces of solid shapes

📐 2D Shapes (Plane Figures)

1. Square

Side = a

Perimeter = 4a Area = a² Diagonal = a√2

Key Formulas:

If diagonal = d: Side = d/√2 Area = d²/2

2. Rectangle

Length = l, Breadth = b

Perimeter = 2(l + b) Area = l × b Diagonal = √(l² + b²)

Example: l = 12 cm, b = 5 cm

Perimeter = 2(12 + 5) = 34 cm Area = 12 × 5 = 60 cm² Diagonal = √(144 + 25) = √169 = 13 cm

3. Triangle

General Triangle:

Base = b, Height = h

Area = (1/2) × base × height Perimeter = Sum of all three sides

Equilateral Triangle (all sides equal = a):

Area = (√3/4) × a² Height = (√3/2) × a Perimeter = 3a

Right-Angled Triangle:

Area = (1/2) × base × perpendicular height Hypotenuse = √(base² + height²) [Pythagoras]

Heron’s Formula (when all sides known):

If sides are a, b, c: Semi-perimeter: s = (a + b + c) / 2 Area = √[s(s-a)(s-b)(s-c)]

4. Circle

Radius = r, Diameter = d = 2r

Circumference = 2πr = πd Area = πr²

Important:

π (pi) ≈ 3.14 or 22/7

If diameter = 14 cm: Radius = 7 cm Circumference = 2 × (22/7) × 7 = 44 cm Area = (22/7) × 7² = 154 cm²

Sector & Arc:

Central angle = θ degrees

Arc length = (θ/360) × 2πr Sector area = (θ/360) × πr²

5. Parallelogram

Base = b, Height = h

Area = base × height Perimeter = 2(a + b) where a, b are adjacent sides

6. Rhombus

Diagonals = d₁, d₂

Area = (1/2) × d₁ × d₂ Perimeter = 4 × side

If side = a:

Area = a × h (where h = height)

7. Trapezium

Parallel sides = a, b Height = h

Area = (1/2) × (a + b) × h

🧊 3D Shapes (Solid Figures)

1. Cube

Side = a

Volume = a³ Total Surface Area = 6a² Lateral Surface Area = 4a² (4 walls) Diagonal (body diagonal) = a√3

Example: Cube with side 5 cm

Volume = 5³ = 125 cm³ TSA = 6 × 5² = 150 cm²

2. Cuboid (Rectangular Box)

Length = l, Breadth = b, Height = h

Volume = l × b × h Total Surface Area = 2(lb + bh + hl) Lateral Surface Area = 2h(l + b) Diagonal = √(l² + b² + h²)

3. Cylinder

Radius = r, Height = h

Volume = πr²h Curved Surface Area (CSA) = 2πrh Total Surface Area (TSA) = 2πr(r + h) = 2πr² + 2πrh

Example: r = 7 cm, h = 10 cm

Volume = (22/7) × 7² × 10 = 1,540 cm³ CSA = 2 × (22/7) × 7 × 10 = 440 cm² TSA = 2 × (22/7) × 7 × (7 + 10) = 44 × 17 = 748 cm²

4. Cone

Radius = r, Height = h, Slant height = l

l = √(r² + h²) [Pythagoras]

Volume = (1/3) × πr²h Curved Surface Area = πrl Total Surface Area = πr(l + r)

Key: Cone volume = (1/3) × Cylinder volume

5. Sphere

Radius = r

Volume = (4/3) × πr³ Surface Area = 4πr²

Hemisphere:

Volume = (2/3) × πr³ Curved Surface Area = 2πr² Total Surface Area = 3πr² (curved + base)

💡 Solved Examples

Example 1: Rectangle Problem

Q: Length of rectangle is 20 cm, breadth is 15 cm. Find area and diagonal.

Solution:

Area = l × b = 20 × 15 = 300 cm²

Diagonal = √(20² + 15²) = √(400 + 225) = √625 = 25 cm

Answer: Area = 300 cm², Diagonal = 25 cm

Example 2: Circle to Square

Q: A wire bent in shape of circle (radius 28 cm) is rebent into a square. Find side of square.

Solution:

Circumference of circle = 2πr = 2 × (22/7) × 28 = 176 cm

This becomes perimeter of square: 4a = 176 a = 44 cm

Answer: 44 cm

Example 3: Triangle Area

Q: Triangle with sides 13 cm, 14 cm, 15 cm. Find area.

Solution:

Using Heron’s formula: s = (13 + 14 + 15) / 2 = 21

Area = √[21(21-13)(21-14)(21-15)] = √[21 × 8 × 7 × 6] = √7,056 = 84 cm²

Answer: 84 cm²

Example 4: Cylinder Volume

Q: Radius = 10 cm, height = 21 cm. Find volume and CSA.

Solution:

Volume = πr²h = (22/7) × 10² × 21 = (22/7) × 100 × 21 = 6,600 cm³

CSA = 2πrh = 2 × (22/7) × 10 × 21 = 1,320 cm²

Answer: Volume = 6,600 cm³, CSA = 1,320 cm²

Example 5: Cube Diagonal

Q: Surface area of cube is 150 cm². Find edge and diagonal.

Solution:

TSA = 6a² = 150 a² = 25 a = 5 cm

Diagonal = a√3 = 5√3 ≈ 8.66 cm

Answer: Edge = 5 cm, Diagonal = 8.66 cm

Example 6: Cone Slant Height

Q: Cone with radius 6 cm and height 8 cm. Find slant height and volume.

Solution:

Slant height: l = √(r² + h²) = √(36 + 64) = √100 = 10 cm

Volume = (1/3) × πr²h = (1/3) × (22/7) × 36 × 8 = 301.71 cm³

Answer: Slant height = 10 cm, Volume ≈ 302 cm³

Example 7: Path Around Rectangle

Q: Rectangular field 40m × 30m has 2m wide path around it. Find area of path.

Solution:

Outer dimensions: Length = 40 + 2(2) = 44 m Breadth = 30 + 2(2) = 34 m

Outer area = 44 × 34 = 1,496 m² Inner area = 40 × 30 = 1,200 m²

Path area = 1,496 - 1,200 = 296 m²

Answer: 296 m²

Example 8: Sphere Surface to Volume

Q: Surface area of sphere is 616 cm². Find volume.

Solution:

Surface area = 4πr² = 616

4 × (22/7) × r² = 616 r² = 616 × 7 / (4 × 22) = 49 r = 7 cm

Volume = (4/3) × πr³ = (4/3) × (22/7) × 343 = 1,437.33 cm³

Answer: ≈ 1,437 cm³

📊 Important Patterns

Pattern 1: Doubling Dimensions

If side/radius doubles:

• Perimeter/Circumference doubles (2×)
• Area becomes 4× original
• Volume becomes 8× original

Pattern 2: Shape Conversions

When wire/ribbon converted from one shape to another: Perimeter remains SAME

Circle → Square: 2πr = 4a Square → Rectangle: 4a = 2(l + b)

Pattern 3: Scaling

If all dimensions scaled by factor k:

• Perimeter → k times
• Area → k² times
• Volume → k³ times

⚡ Quick Shortcuts

Shortcut 1: Right Triangle Recognition

Pythagorean triplets (memorize!): 3-4-5 (and multiples: 6-8-10, 9-12-15) 5-12-13 8-15-17 7-24-25

Shortcut 2: Common π Calculations

π ≈ 22/7 (use when radius/diameter is multiple of 7) π ≈ 3.14 (use otherwise)

Quick: If r = 7, Area = (22/7) × 49 = 154

Shortcut 3: Square Diagonal

Diagonal = Side × 1.414 ≈ Side × (7/5)

Quick: Side = 10, Diagonal ≈ 14

Shortcut 4: Equilateral Triangle

Area ≈ 0.433 × side² Height ≈ 0.866 × side

🔢 Common Values (Memorize!)

Areas of Common Dimensions

Square (10 cm) = 100 cm² Rectangle (12×5) = 60 cm² Circle (r=7) = 154 cm² Circle (r=14) = 616 cm² Equilateral triangle (side 10) ≈ 43.3 cm²

Volumes

Cube (5 cm) = 125 cm³ Cuboid (10×5×2) = 100 cm³ Cylinder (r=7, h=10) = 1,540 cm³ Sphere (r=7) ≈ 1,437 cm³

⚠️ Common Mistakes

❌ Mistake 1: Confusing Area and Perimeter

Wrong: Area of square = 4a ✗ Right: Perimeter = 4a, Area = a² ✓

❌ Mistake 2: Surface Area vs Volume

Wrong: Using volume formula for area ✗ Right: Check units - area in cm², volume in cm³ ✓

❌ Mistake 3: Circle Formulas

Wrong: Circumference = πr² ✗ Right: Area = πr², Circumference = 2πr ✓

❌ Mistake 4: Cone Volume

Wrong: Volume = πr²h ✗ Right: Volume = (1/3)πr²h (one-third of cylinder!) ✓

❌ Mistake 5: Diagonal Formula

Wrong: Using a² + b² = c for square diagonal ✗ Right: Square diagonal = a√2, not a√(1²+1²) ✓

📝 Practice Problems

Level 1:

1. Square with side 12 cm. Find area and perimeter.
2. Circle with radius 14 cm. Find circumference and area.
3. Cube with edge 6 cm. Find volume and TSA.

Level 2:

4. Rectangle: length 15 cm, area 180 cm². Find breadth and perimeter.
5. Cylinder: radius 7 cm, CSA 440 cm². Find height.
6. Triangle with sides 5 cm, 12 cm, 13 cm. Find area.

Level 3:

7. A square and circle have same perimeter 88 cm. Which has larger area?
8. Cone: volume 1,232 cm³, radius 7 cm. Find height and slant height.
9. Cost to fence rectangular field 80m × 60m at ₹20/m?

🔗 Related Topics

Prerequisites:

• Number System - For calculations
• Percentage - For area increase/decrease

Related:

• Profit & Loss - Cost of painting, fencing
• Ratio & Proportion - Scaling shapes

Practice:

• Mensuration Questions
• Formula Sheet

Master Mensuration - Visualize shapes and formulas will stick! 📐