Area and Perimeter is the backbone of mensuration in the CDS and OTA Elementary Mathematics paper. Almost every year you will face questions on the area of a triangle, the perimeter of a rectangle, the cost of fencing a field or the region between two circles. The good news is that the whole chapter rests on a short, fixed list of formulas — learn them cleanly and these become some of the quickest marks on the paper.
Why Area and Perimeter Is High-Value in CDS
In the CDS Elementary Mathematics paper you can reliably expect 3 to 5 questions drawn from area, perimeter and the related mensuration of plane figures, and the OTA paper leans on the very same ideas. Across the 38-year span of previous-year papers, this is one of the densest scoring zones in the entire syllabus.
The reason it rewards preparation is simple: these are formula-driven, low-reasoning questions. Once you know that the area of a triangle is ½ × base × height, or that the circumference of a circle is 2πr, the arithmetic that follows is short and mechanical. There is rarely a long chain of logic the way there is in algebra word problems.
The examiner also dresses up the same formulas in real-world clothing — fencing a garden, tiling a floor, the width of a path around a park, the cost of turfing a field. Behind every such story sits one of the basic shapes you already know. Spot the shape, plug in the formula, and bank the mark.
Always note the units. If sides are in metres, area comes in m2 and perimeter in m. A single unit slip — mixing cm with m, or forgetting to square — is the most common way candidates lose an otherwise easy mark.
Area Versus Perimeter: Get the Idea Right
Before formulas, fix the two ideas firmly, because CDS options often differ only in whether they are testing area or perimeter.
- Perimeter is the total length of the boundary of a closed figure. It is measured in units of length — cm, m, km. For any straight-sided figure, it is simply the sum of all the sides.
- Area is the amount of flat surface a figure covers. It is measured in square units — cm2, m2, hectares.
- Circumference is the special name for the perimeter of a circle.
Think of a rectangular garden. The fence you run around its edge is the perimeter; the grass that fills the inside is the area. They answer different questions, so read the stem carefully.
Useful conversions the examiner loves: 1 m = 100 cm, so 1 m2 = 10,000 cm2. Also 1 hectare = 10,000 m2 and 1 are = 100 m2. These show up directly in field and plot problems.
Rectangle and Square
These two are the most frequently tested shapes, so make them automatic.
Rectangle (length l, breadth b): Area = l × b; Perimeter = 2(l + b); Diagonal = √(l2 + b2).
Square (side a): Area = a2; Perimeter = 4a; Diagonal = a√2.
A handy second form for the square: if you know the diagonal d, the area is ½ d2. This single line saves time whenever the diagonal, not the side, is given.
Watch how the examiner links the two figures. A classic trick is to keep the perimeter fixed and ask which shape has the larger area — the answer is always the square, since for a given perimeter the square encloses the maximum rectangular area. Conversely, for a fixed area the square has the smallest perimeter.
A rectangular plot is 40 m long and 30 m broad. Find the cost of fencing it at ₹25 per metre and the cost of levelling it at ₹8 per m2.
Perimeter = 2(40 + 30) = 2 × 70 = 140 m
Fencing cost = 140 × 25 = ₹3,500
Area = 40 × 30 = 1,200 m2
Levelling cost = 1,200 × 8 = ₹9,600
Triangle: Three Ways to Find the Area
The triangle is the single most important shape in this chapter, because every polygon can be split into triangles. CDS gives you three routes to its area, and you choose by what data is supplied.
1. Base-height: Area = ½ × base × height.
2. Heron's formula (three sides a, b, c): s = (a + b + c) ÷ 2, then Area = √[s(s − a)(s − b)(s − c)].
3. Two sides and the included angle: Area = ½ × a × b × sin C.
Perimeter of any triangle is simply a + b + c. For special triangles, learn these shortcuts:
- Equilateral (side a): Area = (√3 ÷ 4) a2; Height = (√3 ÷ 2) a; Perimeter = 3a.
- Right-angled: Area = ½ × (one leg) × (other leg), since the two legs are base and height.
- Isosceles (equal sides a, base b): Area = (b ÷ 4) × √(4a2 − b2).
In Heron's formula, candidates forget to halve the perimeter to get s, or take the square root too early. Compute s first, then the full product inside the root, and only then take the square root once at the end.
Parallelogram and Rhombus
These two share a family resemblance — a rhombus is just a parallelogram with all sides equal — but their area formulas differ in a useful way.
Parallelogram (base b, height h): Area = base × height = b × h; Perimeter = 2(sum of two adjacent sides).
Rhombus (diagonals d1, d2): Area = ½ × d1 × d2; Perimeter = 4 × side.
The height of a parallelogram is the perpendicular distance between the two parallel sides, not the slanted side — a distinction the examiner exploits constantly. If you are given the slant side, you usually need a right triangle or the sine of an angle to recover the true height.
For a rhombus, the diagonals bisect each other at right angles. So each half-diagonal forms a right triangle with the side: side = √[(d1÷2)2 + (d2÷2)2]. This lets you move freely between the side and the diagonals.
The diagonals of a rhombus are 16 cm and 12 cm. Find its area and perimeter.
Area = ½ × 16 × 12 = 96 cm2
Half-diagonals are 8 cm and 6 cm, so side = √(82 + 62) = √(64 + 36) = √100 = 10 cm
Perimeter = 4 × 10 = 40 cm
Trapezium
A trapezium has exactly one pair of parallel sides. Its area depends on those two parallel sides and the perpendicular distance between them.
Trapezium (parallel sides a and b, height h): Area = ½ × (a + b) × h.
Read the formula in plain words: take the average of the two parallel sides, then multiply by the height. The perimeter is just the sum of all four sides, which you add directly when each is given.
A frequent CDS twist gives you the slant sides instead of the height. In that case drop perpendiculars from the ends of the shorter parallel side, forming right triangles at the corners, and use Pythagoras to recover the height before applying the area formula.
If a figure has one pair of parallel sides and looks irregular, it is almost certainly a trapezium — reach straight for ½(a + b)h rather than trying to split it into awkward pieces.
Circle: Circumference, Area and Sectors
The circle brings π into play. Use π = 22 ÷ 7 when the data are multiples of 7, and π = 3.14 otherwise — the examiner usually arranges the numbers to make one of these clean.
Circle (radius r): Circumference = 2πr; Area = πr2.
Sector (angle θ°): Arc length = (θ ÷ 360) × 2πr; Sector area = (θ ÷ 360) × πr2.
Semicircle: Area = ½ πr2; Perimeter = πr + 2r (curved part plus the diameter).
Two cautions worth a separate note. First, the perimeter of a semicircle is not half the circumference — you must add the straight diameter back. Second, a sector is a fraction of the whole circle, and that fraction is the angle out of 360°, so a quarter circle uses θ = 90°.
Confusing radius and diameter is the number-one error in circle problems. The formulas use the radius. If the question gives the diameter, halve it first — otherwise your area is four times too big.
A classic application is the area between two concentric circles, called an annulus or a circular ring: Area = π(R2 − r2), where R is the outer radius and r the inner radius. This is the model for the width of a circular track or a metal washer.
Combined and Shaded Figures
Higher-value CDS questions rarely show a single clean shape. Instead they give a composite figure — a rectangle with a semicircular end, a square with a quarter-circle cut from a corner, or a path running around a garden. The method is always the same and never fails.
Decompose, then add or subtract. Break any complicated figure into rectangles, triangles and circular parts whose areas you know. Add the areas that are present; subtract the areas that are cut away or shaded out.
For a uniform path of width w running outside a rectangular plot of dimensions l × b, the outer rectangle measures (l + 2w) × (b + 2w). The path's area is the outer area minus the inner area. If the path runs inside the plot, the inner rectangle shrinks by 2w on each dimension instead. Drawing the figure and labelling both rectangles prevents the off-by-one-width slip that catches so many candidates.
A rectangular lawn 50 m by 30 m has a 2 m wide gravel path running all around it on the outside. Find the area of the path.
Inner (lawn) area = 50 × 30 = 1,500 m2
Outer dimensions = (50 + 2×2) × (30 + 2×2) = 54 × 34
Outer area = 54 × 34 = 1,836 m2
Path area = 1,836 − 1,500 = 336 m2
Effect of Changing Dimensions
A favourite CDS theme asks how area changes when you scale the sides. These come up every year and are pure reasoning once you see the pattern, so they cost almost no time.
If every length in a figure is multiplied by a factor k, then the perimeter is multiplied by k, but the area is multiplied by k2.
So if the side of a square is doubled, the perimeter doubles but the area becomes four times as large. If a circle's radius is tripled, the circumference triples while the area grows nine-fold. The square relationship between length scaling and area is the heart of this entire question type.
For percentage versions, remember the product rule for two-dimensional change. If the length increases by x% and the breadth by y%, the net change in area is given by x + y + (xy ÷ 100) percent, using a minus sign for any decrease. This compact formula answers the common stem about a percentage change in the area of a rectangle in one line.
When sides increase by 10% and 20%, area change = 10 + 20 + (10×20)÷100 = 32% increase. Verify the sign — a decrease enters the formula as a negative number.
PYQ Practice and Quick Recap
Work through the previous-year style question below exactly as you would in the exam, then use the recap to lock everything in.
Q. The circumference of a circle is 88 cm. What is its area? (Take π = 22 ÷ 7)
Answer: 2πr = 88, so r = 88 ÷ (2 × 22÷7) = 88 × 7 ÷ 44 = 14 cm. Area = πr2 = (22÷7) × 14 × 14 = 22 × 28 = 616 cm2.
Notice how the question handed you the circumference, expecting you to recover the radius first — a two-step pattern that recurs constantly. Always ask what the given quantity lets you find next.
- Perimeter is boundary length; area is surface covered — check which the stem wants.
- Rectangle: lb and 2(l+b). Square: a2 and 4a, with diagonal a√2.
- Triangle: ½×base×height, or Heron's √[s(s−a)(s−b)(s−c)]. Equilateral area = (√3÷4)a2.
- Rhombus: ½d1d2. Trapezium: ½(a+b)h. Parallelogram: base × height.
- Circle: 2πr and πr2; sector uses θ÷360; ring = π(R2−r2).
- Scale lengths by k: perimeter ×k, area ×k2. Decompose combined figures, then add or subtract.
Frequently asked questions
How many questions on Area and Perimeter come in the CDS exam?
Across recent papers and the 38-year topic-wise record, you can expect roughly 3 to 5 questions every year from area, perimeter and related plane-figure mensuration in the CDS Elementary Mathematics paper, with the OTA paper using the same formulas. It is one of the highest-yield scoring areas in the syllabus.
Which value of π should I use, 22/7 or 3.14?
Use π = 22 ÷ 7 when the radius or diameter is a multiple of 7, because the 7 cancels cleanly and leaves whole numbers. Use 3.14 in other cases. The examiner almost always picks numbers that make one of these two convenient, so a quick glance at the data tells you which to choose.
What is the fastest way to handle a combined or shaded figure?
Decompose it. Break the figure into rectangles, triangles, circles and semicircles whose areas you already know. Add the areas of the parts that are present and subtract the areas that are cut away or shaded. Drawing and labelling the figure first prevents the common width and overlap errors.
If the side of a square doubles, what happens to its perimeter and area?
The perimeter doubles, but the area becomes four times as large. In general, multiplying every length in a figure by a factor k multiplies the perimeter by k and the area by k squared. This square relationship is the key to every dimension-change question in CDS mensuration.
Why is the perimeter of a semicircle not just half the circumference?
Half the circumference gives only the curved arc, which is πr. A semicircle is a closed figure, so its boundary must also include the straight diameter that caps the flat side. The full perimeter is therefore πr + 2r. Forgetting the diameter is one of the most common slips in circle problems.
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