Master Mensuration for RRB Exams (NTPC, Group D, Technician): Formulas, Tricks & Solved Questions

Introduction: Why Mensuration is Crucial for RRB Exams

Welcome, future Railway professionals! If you're gearing up for the RRB NTPC, RRB Group D, RRB Technician Grade I, or Grade III exams, you know that the Mathematics or Quantitative Aptitude section is a game-changer. Among the many topics in this section, Mensuration stands out as a high-scoring and frequently tested area. Mensuration is the branch of geometry that deals with the measurement of length, area, and volume of various geometric shapes. For RRB exams, questions from this topic are direct, formula-based, and can fetch you easy marks if your concepts are clear.

Understanding Mensuration is not just about memorizing formulas; it's about visualizing shapes and applying the correct logic to solve problems. Whether it's calculating the area of a railway platform, the volume of a water tank, or the length of fencing required for a plot, the applications are vast. This comprehensive guide is designed to be your one-stop solution for mastering Mensuration. We will break down complex concepts into simple, understandable parts, cover all essential formulas for both 2D and 3D shapes, provide time-saving tricks, and walk you through a plethora of solved examples and practice questions tailored for RRB exams. Let's begin building a solid foundation to conquer this important topic!

Part 1: Mastering 2D Mensuration (Plane Figures)

2D Mensuration deals with shapes that can be drawn on a plane, like squares, circles, and triangles. The primary measurements we are concerned with here are Perimeter (the total length of the boundary of a figure) and Area (the space enclosed by the figure).

1. The Square (वर्ग)

A square is a quadrilateral with four equal sides and four right angles (90°).

• Side: a
• Perimeter (परिमाप): Sum of all sides = 4a
• Area (क्षेत्रफल): Side × Side = a²
• Diagonal (विकर्ण): d = a√2

Solved Example:

Question: The diagonal of a square park is 20 meters. Find its area and the cost of fencing it at a rate of ₹15 per meter.

Solution:
Given, Diagonal (d) = 20 m.
We know, d = a√2, where 'a' is the side of the square.
20 = a√2 => a = 20/√2 = 10√2 meters.
Area = a² = (10√2)² = 100 × 2 = 200 sq. meters.
Perimeter = 4a = 4 × 10√2 = 40√2 meters.
Cost of fencing = Perimeter × Rate = 40√2 × 15 = 600√2. (Using √2 ≈ 1.414)
Cost ≈ 600 × 1.414 = ₹848.40.
Answer: The area is 200 sq. m and the approximate cost of fencing is ₹848.40.

2. The Rectangle (आयत)

A rectangle is a quadrilateral with opposite sides equal and four right angles (90°).

• Length: l, Breadth: b
• Perimeter (परिमाप): 2(l + b)
• Area (क्षेत्रफल): l × b
• Diagonal (विकर्ण): d = √(l² + b²)

Solved Example:

Question: The length of a rectangular field is 15 m and its area is 150 sq. m. Find the length of its diagonal.

Solution:
Given, Length (l) = 15 m, Area = 150 sq. m.
Area = l × b => 150 = 15 × b => b = 150/15 = 10 meters.
Now we have length (l) = 15 m and breadth (b) = 10 m.
Diagonal (d) = √(l² + b²) = √(15² + 10²) = √(225 + 100) = √325.
√325 = √(25 × 13) = 5√13 meters.
Answer: The length of the diagonal is 5√13 meters.

3. The Triangle (त्रिभुज)

A triangle is a polygon with three sides. The sum of its angles is always 180°.

• General Formula for Area: ½ × base × height
• Heron's Formula (for any triangle with sides a, b, c):
  Semi-perimeter (s) = (a + b + c) / 2
  Area = √[s(s-a)(s-b)(s-c)]
• Equilateral Triangle (समबाहु त्रिभुज) (all sides 'a' are equal):
  Height (h) = (√3/2)a
  Area = (√3/4)a²

Solved Example:

Question: Find the area of an equilateral triangle whose side is 8 cm.

Solution:
Given, side (a) = 8 cm.
For an equilateral triangle, Area = (√3/4)a².
Area = (√3/4) × 8² = (√3/4) × 64 = 16√3 sq. cm.
Answer: The area is 16√3 sq. cm.

4. The Circle (वृत्त)

A circle is a set of all points in a plane that are at a given distance from a central point.

• Radius: r (Diameter d = 2r)
• Circumference (परिधि) (Perimeter): 2πr
• Area (क्षेत्रफल): πr²
• Area of a Semicircle: ½πr²
• Perimeter of a Semicircle: πr + 2r (arc length + diameter)

Solved Example:

Question: A car wheel has a diameter of 70 cm. How many complete revolutions must it make to cover a distance of 1.1 km?

Solution:
Given, Diameter (d) = 70 cm, so Radius (r) = 35 cm.
Distance covered in one revolution = Circumference of the wheel.
Circumference = 2πr = 2 × (22/7) × 35 = 2 × 22 × 5 = 220 cm.
Total distance to be covered = 1.1 km = 1.1 × 1000 m = 1100 m = 1100 × 100 cm = 110000 cm.
Number of revolutions = Total Distance / Distance in one revolution.
Number of revolutions = 110000 / 220 = 11000 / 22 = 500.
Answer: The wheel must make 500 revolutions.

Quick Formula Reference Table for 2D Shapes

Part 2: Conquering 3D Mensuration (Solid Figures)

3D Mensuration involves solid shapes that have three dimensions: length, breadth, and height. Here, we calculate Volume (the space occupied by the object), Curved Surface Area (CSA) or Lateral Surface Area (LSA) (the area of only the curved surfaces), and Total Surface Area (TSA) (the sum of the areas of all its surfaces).

1. The Cube (घन)

A cube is a 3D shape with six equal square faces.

• Edge (Side): a
• Volume (आयतन): a³
• Lateral Surface Area (LSA): 4a²
• Total Surface Area (TSA): 6a²
• Longest Diagonal: a√3

Solved Example:

Question: The total surface area of a cube is 150 sq. cm. What is its volume?

Solution:
Given, TSA = 150 sq. cm.
We know, TSA of a cube = 6a².
150 = 6a² => a² = 150/6 = 25 => a = 5 cm.
Volume of the cube = a³ = 5³ = 125 cubic cm.
Answer: The volume of the cube is 125 cm³.

2. The Cuboid (घनाभ)

A cuboid is a 3D shape with six rectangular faces.

• Length: l, Breadth: b, Height: h
• Volume (आयतन): l × b × h
• Lateral Surface Area (LSA): 2h(l + b)
• Total Surface Area (TSA): 2(lb + bh + hl)
• Longest Diagonal: √(l² + b² + h²)

Solved Example:

Question: Find the length of the longest pole that can be placed in a room 12 m long, 8 m broad, and 9 m high.

Solution:
The length of the longest pole that can be placed in a room is its longest diagonal.
Given, l = 12 m, b = 8 m, h = 9 m.
Longest Diagonal = √(l² + b² + h²) = √(12² + 8² + 9²) = √(144 + 64 + 81) = √289 = 17 meters.
Answer: The length of the longest pole is 17 meters.

3. The Cylinder (बेलन)

A cylinder is a solid with two parallel circular bases and a curved surface connecting them.

• Radius: r, Height: h
• Volume (आयतन): πr²h
• Curved Surface Area (CSA): 2πrh
• Total Surface Area (TSA): 2πr(r + h)

Solved Example:

Question: The radius of a cylindrical tank is 7 m and its height is 10 m. Find the volume of water it can hold and the area of the sheet required to make it (assuming it's a closed tank).

Solution:
Given, r = 7 m, h = 10 m.
Volume = πr²h = (22/7) × 7² × 10 = (22/7) × 49 × 10 = 22 × 7 × 10 = 1540 cubic meters.
Area of sheet required = Total Surface Area (TSA).
TSA = 2πr(r + h) = 2 × (22/7) × 7 × (7 + 10) = 44 × 17 = 748 sq. meters.
Answer: The volume is 1540 m³ and the sheet area required is 748 m².

4. The Cone (शंकु)

A cone is a 3D shape that tapers smoothly from a flat circular base to a point called the apex.

• Radius: r, Height: h, Slant Height: l
• Slant Height (l): √(r² + h²)
• Volume (आयतन): (1/3)πr²h
• Curved Surface Area (CSA): πrl
• Total Surface Area (TSA): πr(r + l)

Solved Example:

Question: The height of a cone is 24 cm and the radius of its base is 7 cm. Find its slant height and curved surface area.

Solution:
Given, h = 24 cm, r = 7 cm.
Slant Height (l) = √(r² + h²) = √(7² + 24²) = √(49 + 576) = √625 = 25 cm.
Curved Surface Area (CSA) = πrl = (22/7) × 7 × 25 = 22 × 25 = 550 sq. cm.
Answer: The slant height is 25 cm and the CSA is 550 cm².

5. The Sphere (गोला)

A sphere is a perfectly round 3D object, like a ball.

• Radius: r
• Volume (आयतन): (4/3)πr³
• Surface Area: 4πr²

6. The Hemisphere (अर्धगोला)

A hemisphere is exactly half of a sphere.

• Radius: r
• Volume (आयतन): (2/3)πr³
• Curved Surface Area (CSA): 2πr²
• Total Surface Area (TSA): 3πr² (CSA + Area of circular base)

Solved Example (Sphere/Hemisphere):

Question: A metallic sphere of radius 10.5 cm is melted and recast into small cones, each of radius 3.5 cm and height 3 cm. Find the number of cones so formed.

Solution:
This is a volume conservation problem. The volume of the sphere will be equal to the total volume of all the small cones.
Let 'n' be the number of cones.
Volume of Sphere = n × Volume of one Cone
(4/3)π(R)³ = n × (1/3)π(r)²h
Given, R = 10.5 cm, r = 3.5 cm, h = 3 cm.
(4/3)π(10.5)³ = n × (1/3)π(3.5)²(3)
Canceling (1/3)π from both sides:
4 × (10.5)³ = n × (3.5)² × 3
4 × 10.5 × 10.5 × 10.5 = n × 3.5 × 3.5 × 3
Notice that 10.5 is 3 times 3.5.
4 × (3 × 3.5) × (3 × 3.5) × (3 × 3.5) = n × 3.5 × 3.5 × 3
4 × 27 × (3.5)³ = n × 3 × (3.5)²
(4 × 27 × 3.5) / 3 = n
4 × 9 × 3.5 = n
36 × 3.5 = n
n = 126.
Answer: 126 small cones can be formed.

Smart Tricks & Tips for RRB Mensuration Questions

1. Memorize Formulas: Create a chart of all the 2D and 3D formulas and revise it daily. This is non-negotiable.
2. Know Your Pythagorean Triplets: Triplets like (3, 4, 5), (5, 12, 13), (7, 24, 25), (8, 15, 17) will help you calculate diagonals and slant heights much faster. In the cone example above, (7, 24, 25) was used.
3. Unit Consistency: Always ensure all measurements are in the same unit before you start calculating. If length is in meters and breadth is in centimeters, convert one of them.
4. Use π = 22/7 Judiciously: If the radius or diameter is a multiple of 7, always use π = 22/7. It simplifies calculations significantly. Otherwise, use 3.14 or keep π as it is if the options have π in them.
5. Percentage Change Shortcut: For a 2D shape, if the sides are increased by x%, the percentage increase in area is approximately (2x + x²/100)%. For a 3D shape, if sides increase by x%, the volume increase is approximately (3x + 3x²/100 + x³/10000)%.
6. Visualization: Try to draw a rough diagram of the shape described in the problem. It helps in understanding the question better and applying the correct formula.

Practice Questions with Solutions (RRB Exam Level)

Question 1:

The perimeter of a rhombus is 52 cm and the length of one of its diagonals is 24 cm. What is the area of the rhombus?

Solution:
Perimeter of a rhombus = 4a = 52 cm => Side (a) = 52/4 = 13 cm.
The diagonals of a rhombus bisect each other at right angles. Let the diagonals be d1 and d2.
Given, d1 = 24 cm. So, half of d1 = 12 cm.
The side of the rhombus, half of d1, and half of d2 form a right-angled triangle, with the side as the hypotenuse.
Using Pythagoras theorem: a² = (d1/2)² + (d2/2)²
13² = 12² + (d2/2)²
169 = 144 + (d2/2)²
(d2/2)² = 169 - 144 = 25
d2/2 = 5 => d2 = 10 cm.
Area of rhombus = ½ × d1 × d2 = ½ × 24 × 10 = 120 sq. cm.
Answer: 120 sq. cm.

Question 2:

A rectangular grassy plot 112 m by 78 m has a gravel path 2.5 m wide all around it on the inside. Find the area of the path.

Solution:
Outer dimensions of the plot (Outer Rectangle): Length (L) = 112 m, Breadth (B) = 78 m.
Area of Outer Rectangle = L × B = 112 × 78 = 8736 sq. m.
The path is 2.5 m wide on the inside. So, the dimensions of the inner grassy area will be reduced by 2.5 m on both sides.
Inner Length (l) = 112 - (2.5 + 2.5) = 112 - 5 = 107 m.
Inner Breadth (b) = 78 - (2.5 + 2.5) = 78 - 5 = 73 m.
Area of Inner Rectangle = l × b = 107 × 73 = 7811 sq. m.
Area of the path = Area of Outer Rectangle - Area of Inner Rectangle.
Area of the path = 8736 - 7811 = 925 sq. m.
Answer: 925 sq. m.

Question 3:

How many bricks, each measuring 25 cm × 11.25 cm × 6 cm, will be needed to build a wall 8 m × 6 m × 22.5 cm?

Solution:
First, make all units consistent. Let's convert everything to cm.
Wall dimensions: Length = 8 m = 800 cm, Height = 6 m = 600 cm, Width = 22.5 cm.
Brick dimensions: Length = 25 cm, Width = 11.25 cm, Height = 6 cm.
Volume of the wall = 800 × 600 × 22.5 cm³.
Volume of one brick = 25 × 11.25 × 6 cm³.
Number of bricks = Volume of wall / Volume of one brick
Number of bricks = (800 × 600 × 22.5) / (25 × 11.25 × 6)
Number of bricks = (800/25) × (600/6) × (22.5/11.25)
Number of bricks = 32 × 100 × 2 = 6400.
Answer: 6400 bricks will be needed.

Question 4:

A hollow iron pipe is 21 cm long and its external diameter is 8 cm. If the thickness of the pipe is 1 cm and iron weighs 8 g/cm³, then find the weight of the pipe.

Solution:
This is a problem of a hollow cylinder. We need to find the volume of the material (iron).
Length of pipe (h) = 21 cm.
External diameter = 8 cm, so External Radius (R) = 4 cm.
Thickness = 1 cm.
Internal Radius (r) = External Radius - Thickness = 4 - 1 = 3 cm.
Volume of iron = Volume of external cylinder - Volume of internal cylinder
Volume = πR²h - πr²h = πh(R² - r²)
Volume = (22/7) × 21 × (4² - 3²)
Volume = 22 × 3 × (16 - 9) = 66 × 7 = 462 cm³.
Weight of the pipe = Volume × Density = 462 cm³ × 8 g/cm³ = 3696 grams.
In kilograms, Weight = 3696 / 1000 = 3.696 kg.
Answer: The weight of the pipe is 3.696 kg.

Question 5:

The circumference of the base of a 9 m high wooden cone is 44 m. Find its volume.

Solution:
Given, Height (h) = 9 m, Circumference = 44 m.
Circumference of circular base = 2πr = 44.
2 × (22/7) × r = 44
(44/7) × r = 44 => r = 7 meters.
Now we have radius (r) = 7 m and height (h) = 9 m.
Volume of cone = (1/3)πr²h
Volume = (1/3) × (22/7) × 7² × 9
Volume = (1/3) × (22/7) × 49 × 9 = 22 × 7 × 3 = 462 cubic meters.
Answer: The volume of the cone is 462 m³.

Conclusion: Your Path to Success

Mensuration, at its core, is a test of your understanding of formulas and your ability to apply them correctly. As we have seen, the questions in RRB exams are designed to be solvable within a minute if you have a strong command of the basics. The key to success is consistent practice. Solve as many different types of problems as you can. Use the formula tables provided in this guide for quick revisions. Work on your calculation speed and accuracy. By dedicating regular time to this topic, you can turn Mensuration into one of your strongest scoring areas, bringing you one step closer to securing your dream job in the Indian Railways. Keep practicing, stay focused, and success will be yours!