In this guide
Mensuration is one of the most formula-driven areas in SSC maths. Most questions give you some dimensions and ask for a perimeter or area. What makes them tricky is the setup: a path around a garden versus inside it, a wheel's distance in revolutions, or the change in area when both sides change. Learn the formulas thoroughly, then practise the common setups.
Core formulas
| Figure | Perimeter | Area |
|---|---|---|
| Square (side a) | 4a | a²; also d²/2 from the diagonal |
| Rectangle (l, b) | 2(l + b) | l × b; diagonal = √(l² + b²) |
| Triangle | a + b + c | ½ × base × height; Heron's formula |
| Equilateral triangle (a) | 3a | (√3/4)a² |
| Parallelogram | 2(a + b) | base × height |
| Rhombus | 4a | ½ × d₁ × d₂ |
| Trapezium | sum of sides | ½ × (a + b) × h |
| Circle (r) | 2πr | πr² |
| Semicircle | πr + 2r | ½πr² |
Circles, sectors and arcs
For a sector with central angle θ:
- arc length = (θ/360) × 2πr;
- sector area = (θ/360) × πr²;
- ring (annulus) area = π(R² − r²).
Worked example: Find the area of a sector with radius 14 cm and angle 90°. (π = 22/7)
(90/360) × (22/7) × 196 = ¼ × 616 = 154 cm².
Worked example: Find the area of a circular ring with outer radius 10 cm and inner radius 6 cm. (Use π = 3.14.)
3.14 × (100 − 36) = 3.14 × 64 = 200.96 cm².
Paths and borders
Path outside a rectangle: a garden of l × b with a path of width w all around.
Path area = (l + 2w)(b + 2w) − lb.
Worked example: A 30 m × 20 m garden has a 2 m path around it outside. Find the path area.
(34 × 24) − 600 = 816 − 600 = 216 m².
Path inside: path area = lb − (l − 2w)(b − 2w).
Two crossing roads of width w through the middle, parallel to the sides:
area = w(l + b) − w².
Worked example: Two roads, each 3 m wide, cross through the middle of a 60 m × 40 m park. Find the road area.
3 × (60 + 40) − 9 = 291 m².
Wheels and revolutions
Distance in one revolution = circumference = 2πr.
Worked example: A wheel of radius 35 cm makes 500 revolutions. How far does it travel? (π = 22/7)
Circumference = 2 × 22/7 × 35 = 220 cm. Distance = 220 × 500 = 1,10,000 cm = 1.1 km.
Percentage change in area
If the length changes by x% and the breadth by y%, the area changes by:
x + y + xy/100 %
Worked example: The length of a rectangle increases by 20% and its breadth decreases by 10%. Area change = 20 − 10 − 2 = +8%.
For a square or circle where the side or radius changes by x%, the area changes by 2x + x²/100 %.
Worked example: The radius of a circle increases by 10%. Area change = 20 + 1 = 21%.
Comparing figures
- For a given perimeter, a circle has the largest area; among rectangles, a square does.
- If a wire bent into a square of side 11 cm is re-bent into a circle, the circle's circumference is 44 cm, so r = 7 cm and the area = 154 cm² (π = 22/7), compared with the square's 121 cm².
Common traps
| Trap | Correct approach |
|---|---|
| Forgetting to subtract the overlap of crossing roads | Subtract w² |
| Mixing cm and m | Convert units first |
| Using the diameter for the radius | Halve the diameter |
Practice
- Find the diagonal of a rectangle measuring 24 cm × 7 cm.
- Find the arc length of a sector with radius 21 cm and angle 60°. (π = 22/7)
- A 50 m × 30 m field has a 2.5 m path inside along its edges. Find the path area.
- The side of a square increases by 30%. Find the percentage increase in area.
- A wheel covers 88 m in 40 revolutions. Find its radius. (π = 22/7)
- The area of an equilateral triangle is 25√3 cm². Find its side.
Answers: 1. 25 cm. 2. 22 cm. 3. 375 m². 4. 69%. 5. 35 cm. 6. 10 cm.
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 .
Get the next SSC CGL guide by email
New guides every week. No spam, unsubscribe any time.