Mensuration - Theory & Concepts
📐 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:
- Square with side 12 cm. Find area and perimeter.
- Circle with radius 14 cm. Find circumference and area.
- Cube with edge 6 cm. Find volume and TSA.
Level 2:
- Rectangle: length 15 cm, area 180 cm². Find breadth and perimeter.
- Cylinder: radius 7 cm, CSA 440 cm². Find height.
- Triangle with sides 5 cm, 12 cm, 13 cm. Find area.
Level 3:
- A square and circle have same perimeter 88 cm. Which has larger area?
- Cone: volume 1,232 cm³, radius 7 cm. Find height and slant height.
- 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:
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