Two-dimensional mensuration is one of the most predictable parts of the CDS maths paper. The syllabus names areas of squares, rectangles, parallelograms, triangles and circles, and of figures that can be split into these. The questions follow a few standard settings: a path around a garden, roads crossing a park, a wheel turning, a sector cut from a circle, a side changed by some percentage.
If you know the formulas and the settings, these are quick marks. The main risks are units (cm against m) and the value of π. CDS questions usually choose radii like 7, 14, 21 or 35 so that π = 22/7 cancels neatly. When the numbers suit 22/7, use it.
Formula table
| Figure | Area | Perimeter |
|---|---|---|
| Rectangle, l by b | l × b | 2(l + b) |
| Square, side a | a², or d²/2 from the diagonal d | 4a |
| Parallelogram | base × height | 2(a + b) |
| Rhombus | ½ × d₁ × d₂ | 4 × side |
| Trapezium | ½ × (a + b) × h | sum of sides |
| Triangle | ½ × base × height | a + b + c |
| Equilateral triangle, side a | (√3/4)a² | 3a |
| Right triangle | ½ × product of the legs | sum of sides |
| Circle, radius r | πr² | 2πr |
| Semicircle | ½πr² | πr + 2r |
| Sector, angle θ | (θ/360) × πr² | 2r + (θ/360) × 2πr |
| Ring, radii R and r | π(R² − r²) | not usually asked |
Heron's formula
For a triangle with sides a, b and c, let s = (a + b + c)/2. Then the area is √(s(s − a)(s − b)(s − c)). Use it when you know three sides and no height. The 13-14-15 triangle (area 84) and the 5-5-6 triangle (area 12) are worth knowing by heart.
Circle facts that save time
- Radius 7: circumference 44, area 154. Radius 14: circumference 88, area 616. Radius 21: circumference 132, area 1,386.
- A square inscribed in a circle of radius r has a diagonal of 2r, so its area is 2r².
- A circle inscribed in a square of side a has radius a/2, so its area is πa²/4.
Standard settings
Paths and borders
- Path of width w outside an l by b rectangle: (l + 2w)(b + 2w) − lb.
- Path of width w inside: lb − (l − 2w)(b − 2w).
- Two crossroads of width w through the middle, parallel to the sides: w(l + b) − w². The square where the roads cross would otherwise be counted twice.
Wheels
In one revolution a wheel covers its circumference. So distance = number of revolutions × 2πr. Watch the units: a wheel radius is usually given in cm, and the distance in km or m.
Percentage change in area
Area depends on two lengths, so percentage changes combine like successive percentages: a% and b% give a + b + ab/100.
- Each side up by 10%: area up by 10 + 10 + 1 = 21%.
- Length up 20%, breadth down 10%: 20 − 10 − 2 = +8%.
- The radius of a circle up by 10%: area up by 21%, and the circumference up by only 10%.
Worked questions
Question 1: A 2 m wide path runs around the outside of a 20 m by 15 m garden. Find its area and the cost of paving it at ₹50 per m².
- Outer rectangle: 24 × 19 = 456 m². Garden: 300 m².
- Path = 156 m². Cost = 156 × 50 = ₹7,800.
Question 2: A rectangular park measures 60 m by 40 m. Two roads, each 3 m wide, run through the middle, one parallel to each side. Find the area of the roads.
- 3 × 60 + 3 × 40 − 3 × 3 = 180 + 120 − 9 = 291 m².
Question 3: Find the area of a triangle with sides 13 cm, 14 cm and 15 cm, and its altitude to the 14 cm side.
- s = 21. Area = √(21 × 8 × 7 × 6) = √7,056 = 84 cm².
- Altitude = 2 × 84 ÷ 14 = 12 cm.
Question 4: A wheel has a diameter of 70 cm. How many revolutions does it make in covering 1.1 km?
- Circumference = 22/7 × 70 = 220 cm.
- 1.1 km = 1,10,000 cm. Revolutions = 1,10,000 ÷ 220 = 500.
Question 5: A circular track lies between two concentric circles of radii 21 m and 14 m. Find its area.
- π(21² − 14²) = 22/7 × (441 − 196) = 22/7 × 245 = 770 m².
Question 6: A sector of a circle of radius 21 cm has an angle of 60°. Find its area and perimeter.
- Area = (60/360) × 22/7 × 441 = 231 cm².
- Arc = (60/360) × 2 × 22/7 × 21 = 22 cm. Perimeter = 22 + 21 + 21 = 64 cm.
Practice set
- A square has a diagonal of 10 cm. Find its area.
- A circle has an area of 616 cm². Find its circumference.
- An equilateral triangle has an area of 16√3 cm². Find its perimeter.
- The radius of a circle increases by 10%. By what percentage does its area increase?
- A trapezium has parallel sides of 12 cm and 18 cm and a height of 8 cm. Find its area.
- A wheel of radius 28 cm makes 1,000 revolutions. How far does it travel?
- A rhombus has diagonals of 18 cm and 24 cm. Find its area and side.
- Find the perimeter of a semicircular protractor of diameter 14 cm.
Answers
- 50 cm². d²/2 = 100/2.
- 88 cm. r² = 196, so r = 14.
- 24 cm. (√3/4)a² = 16√3 gives a² = 64, so a = 8.
- 21%. 10 + 10 + 100/100.
- 120 cm². ½ × 30 × 8.
- 1.76 km. Circumference = 2 × 22/7 × 28 = 176 cm, and 176 × 1,000 = 1,76,000 cm.
- 216 cm² and 15 cm. The half-diagonals are 9 and 12.
- 36 cm. Arc = 22/7 × 7 = 22 cm, plus the 14 cm diameter.
What to do next
- Learn the circle values for radii 7, 14 and 21 so that you never multiply by 22/7 for them
- Draw the path, crossroads and inside-border figures once each, and label the dimensions
- Revise quadrilaterals and polygons for rhombus and trapezium properties
- Continue with surface area and volume
- Solve the area and perimeter questions from the last five CDS papers under time
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 Union Public Service Commission website .
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