Mensuration & Trigonometry - Complete SSC CGL Guide
What are Mensuration & Trigonometry? Mensuration deals with measurement of geometric figures (area, volume, perimeter). Trigonometry deals with relationships between angles and sides of triangles, especially right triangles. Both are essential for SSC CGL's geometry and practical math questions.
Pro Tip – Remember Key Formulas!
Mensuration: Area formulas for 2D shapes, Volume for 3D shapes. Trigonometry: SOH-CAH-TOA for ratios, Pythagorean identity sin²θ + cos²θ = 1. Visit SKY Practice for 500+ Mensuration & Trigonometry questions with step-by-step solutions.
Understanding geometric relationships is key to solving mensuration problems
1. Mensuration Basics
What is Mensuration? Branch of mathematics dealing with measurement of geometric figures and their parameters like area, perimeter, volume, surface area.
Basic Concepts & Units
Understanding Measurement Units
All measurements depend on consistent units. Common units: cm, m, km for length; cm², m² for area; cm³, m³ for volume.
Perimeter vs Area
Perimeter: Total distance around a 2D shape (linear measurement)
Area: Space enclosed within a 2D shape (square measurement)
Area = 5² = 25 cm²
Area vs Volume
Area: 2D measurement (cm², m²)
Volume: 3D measurement (cm³, m³)
Volume = 3³ = 27 cm³
Conversion Factors
- 1 m = 100 cm
- 1 m² = 10000 cm²
- 1 m³ = 1000000 cm³
- 1 hectare = 10000 m²
- 1 km² = 100 hectares
- 1 liter = 1000 cm³
SSC Shortcut: Quick Conversions
Area conversions: m² to cm² → multiply by 10000
Volume conversions: m³ to cm³ → multiply by 1000000
Practical: 1 liter water = 1 kg weight (approx)
Circle: Circumference ≈ 3.14 × diameter (πd)
Solved Example: Basic Mensuration
Length (l) = 50 m, Width (w) = 30 m
Step 2: Calculate perimeter
Perimeter of rectangle = 2(l + w)
= 2(50 + 30) = 2 × 80 = 160 m
Step 3: Calculate area
Area of rectangle = l × w
= 50 × 30 = 1500 m²
Step 4: Convert to other units
Area in cm² = 1500 × 10000 = 15,000,000 cm²
Perimeter in cm = 160 × 100 = 16,000 cm
Step 5: Practical interpretation
This field is about 1.5 times a football field
Fencing needed: 160 m
Grass needed: 1500 m² area
Final Answer: Perimeter = 160 m, Area = 1500 m²
2. Plane Figures (2D Shapes)
Important 2D Shapes: Square, rectangle, triangle, circle, parallelogram, trapezium, rhombus.
Area & Perimeter Formulas
Memorization Technique
Group formulas by shape type. Rectangle-based, triangle-based, circle-based. Remember derivations.
SSC Shortcut: Important Values
π ≈ 22/7 (better for exact division), π ≈ 3.14 (decimal)
Right triangle: Area = ½ × product of legs
Equilateral triangle: Area = (√3/4) × a²
Circle: Area = πr², Circumference = πd = 2πr
Diagonal of square: a√2, Diagonal of rectangle: √(l² + b²)
Solved Example: Composite Figure
Rectangle: length = 10 m, breadth = 4 m
Semicircle: diameter = 10 m (on longer side)
Radius = 10/2 = 5 m
Step 2: Calculate rectangle area
Area₁ = l × b = 10 × 4 = 40 m²
Step 3: Calculate semicircle area
Area of full circle = πr² = π × 5² = 25π m²
Area of semicircle = ½ × 25π = 12.5π m²
Using π = 22/7: 12.5 × 22/7 = 275/7 ≈ 39.29 m²
Step 4: Total area
Total = 40 + 39.29 = 79.29 m²
Step 5: Alternative calculation
Using π = 3.14: 12.5 × 3.14 = 39.25 m²
Total = 40 + 39.25 = 79.25 m²
Final Answer: Approximately 79.3 m²
Right triangle showing trigonometric ratios: opposite, adjacent, and hypotenuse
3. Solid Figures (3D Shapes)
Important 3D Shapes: Cube, cuboid, cylinder, cone, sphere, hemisphere.
Volume & Surface Area Formulas
Understanding 3D Measurements
Volume = space occupied (3D). Surface Area = total area of all faces (2D). Lateral Surface Area = area excluding top and bottom.
SSC Shortcut: Volume Relationships
Cone volume = ⅓ cylinder volume (same base and height)
Hemisphere volume = ½ sphere volume
Sphere fits exactly in cube: Side of cube = diameter of sphere
Water flow problems: Volume = Area × Speed × Time
Solved Example: Solid Figures
Radius (r) = 7 cm, Height (h) = 10 cm
Use π = 22/7 for exact calculations
Step 2: Calculate volume
Volume of cylinder = πr²h
= (22/7) × 7² × 10
= (22/7) × 49 × 10
= 22 × 7 × 10 = 1540 cm³
Step 3: Calculate total surface area
TSA = 2πr(h + r)
= 2 × (22/7) × 7 × (10 + 7)
= 2 × 22 × 17 = 748 cm²
Step 4: Calculate lateral surface area
LSA = 2πrh = 2 × (22/7) × 7 × 10
= 2 × 22 × 10 = 440 cm²
Step 5: Verify with decimal π
Using π = 3.14:
Volume = 3.14 × 49 × 10 = 1538.6 cm³
TSA = 2 × 3.14 × 7 × 17 = 747.32 cm²
(Close to exact values)
Final Answer: Volume = 1540 cm³, TSA = 748 cm²
4. Trigonometry Basics
What is Trigonometry? Study of relationships between angles and sides of triangles, especially right triangles.
Trigonometric Ratios
The SOH-CAH-TOA Mnemonic
Sine = Opposite/Hypotenuse, Cosine = Adjacent/Hypotenuse, Tangent = Opposite/Adjacent. Remember this for all right triangle problems.
Right Triangle showing sides relative to angle θ
SSC Shortcut: Standard Angle Values
0°: sin=0, cos=1, tan=0
30°: sin=½, cos=√3/2, tan=1/√3
45°: sin=1/√2, cos=1/√2, tan=1
60°: sin=√3/2, cos=½, tan=√3
90°: sin=1, cos=0, tan=∞
Memorize: 0, ½, 1/√2, √3/2, 1 pattern
Solved Example: Trig Ratios
Using Pythagoras: 13² = 5² + other²
169 = 25 + other²
other² = 144 → other = 12 cm
Step 2: Identify smallest angle
Smallest angle opposite smallest side (5 cm)
Opposite to this angle = 5 cm
Adjacent to this angle = 12 cm
Hypotenuse = 13 cm
Step 3: Calculate basic ratios
sin θ = Opposite/Hypotenuse = 5/13
cos θ = Adjacent/Hypotenuse = 12/13
tan θ = Opposite/Adjacent = 5/12
Step 4: Calculate reciprocal ratios
cosec θ = 1/sin θ = 13/5
sec θ = 1/cos θ = 13/12
cot θ = 1/tan θ = 12/5
Step 5: Verify using identity
sin²θ + cos²θ = (25/169) + (144/169) = 169/169 = 1 ✓
1 + tan²θ = 1 + (25/144) = 169/144 = (13/12)² = sec²θ ✓
Final Answer: sin=5/13, cos=12/13, tan=5/12, cosec=13/5, sec=13/12, cot=12/5
5. Trigonometric Identities
Important Identities: Equations that are true for all values of the variables where both sides are defined.
Fundamental Identities
Three Basic Groups of Identities
1. Pythagorean identities (based on right triangle)
2. Reciprocal identities (flip ratios)
3. Quotient identities (division relationships)
Pythagorean Identities
1 + tan²θ = sec²θ
1 + cot²θ = cosec²θ
Derived from a² + b² = c²
Most frequently used in SSC
Reciprocal Identities
cos θ = 1/sec θ
tan θ = 1/cot θ
cosec θ = 1/sin θ
sec θ = 1/cos θ
cot θ = 1/tan θ
Simply flip the ratio
Quotient Identities
cot θ = cos θ/sin θ
Definition of tangent and cotangent
tan θ = sin θ/cos θ = (3/5)/(4/5) = 3/4
SSC Shortcut: Identity Applications
Simplify expressions: Convert everything to sin and cos
Prove identities: Start with complex side, simplify to match other side
Find unknown ratio: Use sin²θ + cos²θ = 1
Remember: 1 = sin²θ + cos²θ = sec²θ - tan²θ = cosec²θ - cot²θ
Solved Example: Using Identities
tan θ = Opposite/Adjacent = 3/4
Let opposite = 3k, adjacent = 4k
Hypotenuse = √(3k)² + (4k)² = √(9k² + 16k²) = √(25k²) = 5k
sin θ = Opposite/Hypotenuse = 3k/5k = 3/5
cos θ = Adjacent/Hypotenuse = 4k/5k = 4/5
Expression = (3/5 + 4/5)/(3/5 - 4/5) = (7/5)/(-1/5) = -7
Method 2: Direct division
(sin θ + cos θ)/(sin θ - cos θ)
Divide numerator and denominator by cos θ:
= (tan θ + 1)/(tan θ - 1)
= (3/4 + 1)/(3/4 - 1)
= (7/4)/(-1/4) = -7
Method 3: Using identity
sec²θ = 1 + tan²θ = 1 + 9/16 = 25/16
sec θ = 5/4 (positive in first quadrant)
cos θ = 4/5
sin θ = tan θ × cos θ = (3/4) × (4/5) = 3/5
Then calculate as in method 1
Final Answer: -7
6. Heights & Distances
Practical Applications: Using trigonometry to find heights of buildings, widths of rivers, distances between objects.
Application Problems
Understanding Angle Terminology
Angle of elevation: Looking upward from horizontal
Angle of depression: Looking downward from horizontal
Both angles are equal if lines are parallel (alternate interior angles).
Angle of Elevation
tan θ = Height/Distance
Angle of Depression
tan θ = Depth/Distance
Two Observer Problems
When same object observed from two points:
Where:
h = height of object
d = distance between observers
θ₁, θ₂ = angles of elevation
SSC Shortcut: Heights & Distances
tan 45° = 1: Height = Distance when angle is 45°
tan 30° = 1/√3 ≈ 0.577: Height ≈ 0.577 × Distance
tan 60° = √3 ≈ 1.732: Height ≈ 1.732 × Distance
Shadow problems: Sun's angle same for all objects at same time
Solved Example: Heights & Distances
Right triangle with:
Base (distance) = 100 m
Angle at observer = 45°
Height of tower = h (opposite side)
Step 2: Apply trigonometric ratio
tan 45° = Height/Distance
1 = h/100
h = 100 m
Step 3: Verify with other ratios
Using sin 45° = 1/√2 = Height/Hypotenuse
Hypotenuse = h/(1/√2) = 100√2 ≈ 141.4 m
Using cos 45° = 1/√2 = Distance/Hypotenuse
Hypotenuse = 100/(1/√2) = 100√2 ≈ 141.4 m ✓
Step 4: Practical interpretation
When angle is 45°, height equals distance
This is a useful relationship to remember
The line of sight forms isosceles right triangle
Final Answer: Height of tower = 100 m
Ready to Master Mensuration & Trigonometry?
Access 500+ Mensuration & Trigonometry questions with detailed solutions, diagrams, and shortcut techniques
Start Mensuration & Trigonometry PracticeIncludes all types: area-volume problems, trigonometric ratios, identities, heights & distances
Frequently Asked Questions
Q1: How many Mensuration & Trigonometry questions in SSC CGL?
Answer: Typically 6-8 questions in Tier I and 10-12 questions in Tier II. These include direct formula applications, composite figures, and practical trigonometry problems.
Q2: What are the must-know formulas for SSC CGL?
Answer: Area of triangle, rectangle, circle; Volume of cube, cuboid, cylinder; sin²θ+cos²θ=1; tanθ=sinθ/cosθ; SOH-CAH-TOA; values for 0°, 30°, 45°, 60°, 90°.
Q3: How to quickly solve composite figure problems?
Answer: Break into basic shapes (rectangle, triangle, circle). Calculate area/volume of each part separately. Add or subtract as needed. Watch for overlapping regions.
Q4: What's the trick to remember trigonometric values?
Answer: Use the pattern: sin 0°, 30°, 45°, 60°, 90° = √0/2, √1/2, √2/2, √3/2, √4/2 = 0, ½, 1/√2, √3/2, 1. cos values are reversed.
Q5: How to approach "heights and distances" problems?
Answer: Always draw a diagram. Mark known angles and distances. Identify right triangles. Use tan for height/distance, sin for height/hypotenuse, cos for distance/hypotenuse.
Q6: What's the difference between total surface area and lateral surface area?
Answer: Total Surface Area (TSA) includes all faces/surfaces. Lateral Surface Area (LSA) excludes top and bottom surfaces. For cylinder: TSA = 2πr(h+r), LSA = 2πrh.
Final Exam Strategy for Mensuration & Trigonometry
Time Allocation: Direct formula: 30-45 seconds, Composite figures: 60-90 seconds, Trig proofs: 90-120 seconds.
Priority Order: 1) Basic area/volume, 2) Trigonometric ratios, 3) Heights & distances, 4) Composite figures, 5) Identity proofs.
Accuracy Check: Area/volume should be positive. Trig ratios between -1 and 1 (sin, cos). tan 45° = 1. In right triangle, hypotenuse is longest side.
👉 For complete mastery with 1000+ questions, visit SKY Practice!