In this guide
Mensuration questions in SSC CHSL are some of the most practical in the paper: fencing a field, tiling a floor, filling a tank, melting a metal ball into smaller ones. They depend on two things. You need a small set of formulas you can recall instantly, and you need to handle units without slipping. Most wrong answers here come from a missed conversion, not a wrong formula.
The good news is that the formulas are few and they connect. A cylinder is a circle stretched upwards; a cone is one-third of that cylinder. If you understand why each formula looks the way it does, you will not mix them up under pressure.
2D formulas
| Figure | Perimeter | Area |
|---|---|---|
| Square (side a) | 4a | a²; also diagonal²/2 |
| Rectangle (l × b) | 2(l + b) | l × b; diagonal = √(l² + b²) |
| Triangle | sum of sides | ½ × base × height |
| Equilateral triangle (side a) | 3a | (√3/4) × a² |
| Circle (radius r) | 2πr | πr² |
| Sector (angle θ) | arc = θ/360 × 2πr | θ/360 × πr² |
| Trapezium | sum of sides | ½ × (sum of parallel sides) × height |
Why ½ × base × height: any triangle is half of a parallelogram with the same base and height. Why the sector formula: a sector with angle θ is just θ/360 of the full circle.
When only the three sides of a triangle are given, use Heron's formula: area = √(s(s − a)(s − b)(s − c)), where s is half the perimeter. For sides 13, 14 and 15, s = 21 and area = √(21 × 8 × 7 × 6) = √7,056 = 84.
Paths and roads
Path questions are common, and a formula saves drawing time.
| Path | Area |
|---|---|
| Outside a rectangle l × b, width w | 2w(l + b + 2w) |
| Inside a rectangle, width w | 2w(l + b − 2w) |
| Two crossroads through the middle, width w | w(l + b − w) |
Why the crossroads subtract w²: the two roads overlap in a w × w square at the centre. Adding the two strips counts that square twice, so you remove it once.
Percentage change in area
If every length changes, area changes by the product of the factors. A 10% rise in each side gives 1.1 × 1.1 = 1.21, so area rises 21%. A rectangle with length +20% and breadth −10% gives 1.2 × 0.9 = 1.08, a rise of 8%. This is the successive-percentage rule a + b + ab/100 applied to two dimensions.
3D formulas
| Solid | Volume | Surface area |
|---|---|---|
| Cube (a) | a³ | 6a²; diagonal a√3 |
| Cuboid (l, b, h) | lbh | 2(lb + bh + hl); diagonal √(l² + b² + h²) |
| Cylinder (r, h) | πr²h | curved 2πrh; total 2πr(r + h) |
| Cone (r, h, slant l) | ⅓πr²h | curved πrl; l = √(r² + h²) |
| Sphere (r) | (4/3)πr³ | 4πr² |
| Hemisphere (r) | (2/3)πr³ | curved 2πr²; total 3πr² |
Why a cylinder is πr²h: it is a stack of circles, base area times height. Why a cone is one-third of that: a cone fits inside a cylinder of the same base and height exactly three times by volume. Why curved area of a cylinder is 2πrh: unroll it and you get a rectangle whose length is the circumference.
Units and recasting
- 1 m³ = 1,000 litres; 1 litre = 1,000 cm³; 1 hectare = 10,000 m².
- Convert everything to one unit before multiplying. Converting after multiplying is where most errors happen, because areas scale by 100² and volumes by 100³ when you switch between metres and centimetres.
- Melting and recasting: volume is conserved; surface area is not. Number of small solids = big volume ÷ small volume. For spheres, the count is (R/r)³.
Six worked questions
Q1. A rectangular field has area 2,400 m² and length 60 m. Find the cost of fencing it at ₹15 per metre.
Breadth = 2,400 ÷ 60 = 40 m. Perimeter = 2(60 + 40) = 200 m. Cost = 200 × 15 = ₹3,000.
Q2. A garden 20 m × 15 m has a 2 m wide path around it on the outside. Find the cost of gravelling the path at ₹12 per m².
Path area = 2 × 2 × (20 + 15 + 4) = 156 m². Check: outer 24 × 19 = 456, minus 300. Cost = 156 × 12 = ₹1,872.
Q3. Two roads, each 3 m wide, run through the middle of a park 50 m × 40 m, parallel to its sides. Find the area of the roads.
3 × (50 + 40 − 3) = 3 × 87 = 261 m².
Q4. A wheel has radius 21 cm. How many revolutions does it make in covering 1.32 km? (π = 22/7)
Circumference = 2 × 22/7 × 21 = 132 cm. Distance = 1,32,000 cm. Revolutions = 1,32,000 ÷ 132 = 1,000.
Q5. A metal sphere of radius 6 cm is melted and recast into spheres of radius 2 cm. How many are made?
Count = (6/2)³ = 27.
Q6. A cylindrical drum has radius 70 cm and height 1 m. Find its capacity in litres. (π = 22/7)
In centimetres: 22/7 × 70 × 70 × 100 = 15,40,000 cm³. Divide by 1,000: 1,540 litres.
Common mistakes
| Mistake | Fix |
|---|---|
| Mixing metres and centimetres | Convert all lengths first |
| Using total surface area when curved is asked | Read "curved", "lateral" or "total" carefully |
| Using height instead of slant height for a cone's curved area | l = √(r² + h²) |
| Adding path widths once instead of twice | Outer length = l + 2w |
| Assuming surface area is conserved in recasting | Only volume is conserved |
Practice
- Find the area of a square whose perimeter is 48 cm.
- Find the cost of fencing a circular park of radius 35 m at ₹20 per metre. (π = 22/7)
- The radius of a circle increases by 10%. By what percentage does its area increase?
- How many litres can a cuboidal tank 3 m × 2 m × 1.5 m hold?
- Find the curved surface area of a cylinder with radius 7 cm and height 10 cm. (π = 22/7)
- Find the area of an equilateral triangle of side 12 cm.
- Find the volume of a cone with radius 3.5 cm and height 12 cm. (π = 22/7)
- How many bricks of 20 cm × 10 cm × 5 cm are needed for a wall 4 m × 3 m × 0.2 m, ignoring mortar?
Answers:
- 144 cm². Side = 12 cm.
- ₹4,400. Circumference = 2 × 22/7 × 35 = 220 m.
- 21%. 1.1 × 1.1 = 1.21.
- 9,000 litres. 9 m³ × 1,000.
- 440 cm². 2 × 22/7 × 7 × 10.
- 36√3 cm², about 62.4 cm². √3/4 × 144.
- 154 cm³. ⅓ × 22/7 × 12.25 × 12 = 462 ÷ 3.
- 2,400. Wall = 400 × 300 × 20 = 24,00,000 cm³; each brick is 1,000 cm³.
What to do next
- Rewrite both formula tables from memory, including the reason for each formula.
- Solve 10 path and wheel questions, then 10 recasting questions.
- Before each answer, check that every length is in the same unit.
- Revise geometry for the triangle and circle properties, and keep squares, cubes and roots sharp for the arithmetic.
A note on dates and numbers. Exam patterns, vacancies and schedules change from year to year. Always confirm the current details in the latest notification on the Staff Selection Commission website .
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