Mensuration Notes for Competitive Exams

Mensuration is the branch of mathematics that deals with the measurement of lengths, areas and volumes of geometrical figures. It is a regular chapter of SSC, banking, railways, CSAT and state-level examinations, and it also helps in daily life when we calculate the area of a floor, the cost of painting a wall or the capacity of a water tank. The questions are mostly formula-based, so a clear list of formulas and some practice are enough to score full marks. This note gives the formulas for 2D shapes and 3D solids in tables, explains the main ideas, and solves many examples, in simple language. (We use π = 22/7 wherever convenient.)

Basic Terms

Perimeter is the total length of the boundary of a 2D shape (measured in cm, m). Area is the surface covered by a 2D shape (measured in square units such as cm² or m²). Volume is the space occupied by a 3D solid (measured in cubic units such as cm³ or m³). Curved surface area (CSA) or lateral surface area (LSA) is the area of the curved or side surfaces without the base, and total surface area (TSA) includes all surfaces. Remember: 1 m = 100 cm, 1 m² = 10,000 cm², 1 m³ = 10 lakh cm³ = 1,000 litres, and 1 hectare = 10,000 m².

Two-Dimensional Figures

Solved Examples on 2D Figures

Example 1. A rectangle has length 12 m and breadth 5 m. Its area is 60 m², perimeter is 2 × 17 = 34 m, and diagonal is √(144 + 25) = √169 = 13 m.

Example 2. Find the area of a triangle with sides 13 cm, 14 cm and 15 cm. Here s = 21. The area = √[21 × (21 − 13) × (21 − 14) × (21 − 15)] = √(21 × 8 × 7 × 6) = √7,056 = 84 cm².

Example 3. For a circle of radius 7 cm, the area = (22/7) × 49 = 154 cm² and the circumference = 2 × (22/7) × 7 = 44 cm.

Example 4. A park is 20 m long and 15 m wide. A path of 2 m width runs around it outside. The outer rectangle is 24 m by 19 m, so its area is 456 m². The park area is 300 m². The area of the path = 456 − 300 = 156 m².

Example 5. A wheel has a diameter of 70 cm. Its circumference = (22/7) × 70 = 220 cm. In covering 4.4 km (4,40,000 cm), the number of revolutions = 4,40,000/220 = 2,000.

Three-Dimensional Solids

Solved Examples on 3D Solids

Example 6. A cuboid is 10 cm long, 8 cm wide and 6 cm high. Volume = 480 cm³; TSA = 2(80 + 48 + 60) = 376 cm²; diagonal = √(100 + 64 + 36) = √200 = 10√2 cm.

Example 7. For a cube of side 5 cm, volume = 125 cm³ and TSA = 6 × 25 = 150 cm².

Example 8. A cylinder has radius 7 cm and height 10 cm. Volume = (22/7) × 49 × 10 = 1,540 cm³; CSA = 2 × (22/7) × 7 × 10 = 440 cm².

Example 9. A cone has radius 3 cm and height 4 cm. The slant height = √(9 + 16) = 5 cm. Volume = ⅓ × π × 9 × 4 = 12π cm³ and CSA = π × 3 × 5 = 15π cm².

Example 10. A sphere has a radius of 21 cm. Volume = (4/3) × (22/7) × 21³ = (4/3) × (22/7) × 9,261 = 38,808 cm³. A hemisphere of radius 7 cm has TSA = 3 × (22/7) × 49 = 462 cm².

Changes in Dimensions: Quick Percent Rules

• If the radius of a circle is doubled, the area becomes 4 times; if the side of a square is doubled, the area becomes 4 times and the volume of a cube with doubled side becomes 8 times.
• If the side of a square increases by 10 percent, the area increases by 21 percent (1.1² = 1.21). If the edge of a cube increases by 10 percent, the volume increases by 33.1 percent (1.1³ = 1.331).
• If the length of a rectangle increases by 20 percent and the breadth decreases by 10 percent, the area changes by 20 − 10 − (20 × 10)/100 = +8 percent.
• The ratio of the areas of two similar figures is the square of the ratio of their sides; the ratio of volumes is the cube of the ratio of their sides.

Melting, Recasting and Flow Problems

When a solid is melted and recast into other solids, the volume remains constant. Example: a solid metal sphere of radius 6 cm is melted into small spheres of radius 2 cm. The number of small spheres = (6³)/(2³) = 216/8 = 27. In problems on water flowing through a pipe, the volume of water = area of the cross-section × length of the flow. For example, a pipe of cross-section 7 cm² and water speed 100 cm per second delivers 700 cm³ per second.

Cost-based Questions

The questions often ask for the cost of painting, tiling or fencing: cost = area (or length) × rate. For a room, the area of four walls = 2h(l + b), the ceiling = lb and the floor = lb. Remember to subtract the doors and windows if they are given. For fencing, use the perimeter. For a circular park with a path, use the area of the ring: π(R² − r²).

Tips and Common Mistakes

Write the formula first and then substitute the values. Keep units consistent: convert cm to m or the other way before calculating. For the circle, use 22/7 when the radius is a multiple of 7. Use the Pythagorean triples (3-4-5, 5-12-13, 8-15-17) to find diagonals fast. Do not mix up the CSA and TSA, or the height and the slant height of a cone. Draw a rough diagram for each problem, because it helps you to see the dimensions. Many students forget to square or cube the units when converting.

Practice Questions With Answers

Try these questions on your own. Question 1: The side of a square field is 40 m. What is the cost of fencing it at 15 rupees per metre? The perimeter is 160 m, so the cost is 2,400 rupees. Question 2: The area of a rectangle is 150 m² and its length is 15 m. What is its perimeter? The breadth is 10 m, so the perimeter is 2 × 25 = 50 m. Question 3: What is the area of an equilateral triangle of side 8 cm? It is (√3/4) × 64 = 16√3 cm². Question 4: A cylindrical tank has a radius of 7 m and a height of 5 m. How many litres can it hold? Volume = (22/7) × 49 × 5 = 770 m³, which is 7,70,000 litres. Question 5: A cube of side 6 cm is cut into small cubes of side 2 cm. How many cubes are formed? (6/2)³ = 27. Check each answer by estimating the size of the result, and note which formula you used, so that you can quickly revise your mistakes.

Quick Revision

• Rectangle: area lb; circle: πr² and 2πr; triangle: ½bh; Heron: √[s(s − a)(s − b)(s − c)].
• Cuboid TSA: 2(lb + bh + hl); cylinder CSA: 2πrh; cone volume: ⅓πr²h; sphere volume: ⁴⁄₃πr³.
• Melting and recasting keep the volume the same; similar figures: areas in the square, volumes in the cube of the ratio.

Mensuration is a formula-driven chapter. Keep the tables on one page, practise the examples and revise them regularly, and you will solve any geometry question with ease.

Related Notes to Read Next

Continue your preparation with these related notes from the same subject:

• Time, Speed and Distance Notes for Competitive Exams
• Simplification and Approximation Notes
• Algebra Notes for Competitive Exams
• All Quantitative Aptitude notes

Frequently Asked Questions (FAQ)

What is the difference between area and perimeter?

Area is the surface enclosed by a 2D shape, measured in square units, while perimeter is the total length of its boundary, measured in linear units.

What is the area of a circle?

The area of a circle is πr², where r is the radius.

What is Heron’s formula?

Heron’s formula gives the area of a triangle from its three sides: √[s(s − a)(s − b)(s − c)], where s is half the perimeter.

What is the volume of a cylinder?

The volume of a cylinder is πr²h, where r is the radius and h is the height.

What is the volume of a cone?

The volume of a cone is one-third of πr²h.

What is the surface area of a sphere?

The surface area of a sphere is 4πr².

What is the total surface area of a cuboid?

The total surface area of a cuboid is 2(lb + bh + hl).

What is the diagonal of a cube?

The diagonal of a cube with side a is a√3.

What is the slant height of a cone?

The slant height is the square root of (r² + h²), where r is the base radius and h is the vertical height.

What happens to the area of a circle if the radius is doubled?

The area becomes four times when the radius is doubled.

What remains constant when a solid is melted and recast?

The volume remains constant when a solid is melted and recast into another shape.

How many litres are there in 1 cubic metre?

One cubic metre equals 1,000 litres.