Perimeter and circumference
Perimeter = the total length around a shape. Just add all the sides. For a triangle with sides a, b, c: perimeter = a + b + c.
The perimeter of a circle has a special name: circumference. Measure the rim of any round thing and divide by its diameter. You always get about 3.14. We call this number π (pi).
- Circumference = π × d = 2πr (d = diameter, r = radius).
- For sums we use π ≈ 22/7 or 3.14.
Why π is irrational
π = 3.14159265… Its decimal digits go on forever and never fall into a repeating pattern. So π cannot be written as a fraction p/q of two whole numbers. Such a number is called irrational.
So 22/7 is not π. It is only a close value (22/7 = 3.142857…, while π = 3.141592…). The proof that π is irrational (by Lambert, 1761) is beyond Class 9; you only need to know the fact and its meaning.
Heron's formula: area of a triangle from its three sides
Area = ½ × base × height needs the height. Often we only know the three sides. Then use Heron's formula.
- Find the semi-perimeter (half the perimeter): s = (a + b + c) ÷ 2.
- Find s − a, s − b and s − c.
- Area = √(s(s − a)(s − b)(s − c)).
Quick check: all of s − a, s − b, s − c must be positive. If one is zero or negative, no triangle can be made with those sides.
Any polygon: draw a diagonal to cut it into triangles, use Heron's formula for each, and add.
Brahmagupta's formula for cyclic quadrilaterals
A cyclic quadrilateral is a four-sided shape whose four corners all lie on one circle. Every rectangle and square is cyclic.
The Indian mathematician Brahmagupta (7th century CE) gave its area from the four sides a, b, c, d:
- s = (a + b + c + d) ÷ 2
- Area = √((s − a)(s − b)(s − c)(s − d))
Notice: if you make side d = 0, the shape becomes a triangle and you get Heron's formula back! Warning: the formula is exact only for cyclic quadrilaterals.
Arc length
An arc is a part of the circle's edge. If the arc makes an angle θ at the centre, it is θ/360 of the whole circle.
Arc length l = (θ/360) × 2πr.
Example: a 90° arc is 90/360 = 1/4 of the circumference.
Area of a circle and a sector
Area of a circle = πr². A sector is a slice of the circle between two radii, like a pizza slice.
- Area of sector = (θ/360) × πr²
- Perimeter of sector = 2r + arc length
- Area of sector can also be written as ½ × l × r (l = arc length).
Think of the slice as a fraction: angle ÷ 360 tells you what fraction of the circle you have.
Board exam pattern
Mensuration carries 14 marks in CBSE Class 9 (2026-27). Expect: Heron's formula on a triangle or a field split into two triangles (3–4 marks), arc length and sector area (2–3 marks), MCQs on π and circumference, and case-study questions on parks and fields.
Try it at home
Take a steel plate or a bangle. Wrap a thread around its rim, then straighten it. Now measure the diameter with a ruler. Divide thread length by diameter. You should get close to 3.14. Try 3 round things. Then cut a paper triangle, measure its sides, and find its area by Heron's formula. Check with a grid (count squares).
Key formulas and definitions
- Perimeter of triangle = a + b + c
- Semi-perimeter s = (a + b + c) ÷ 2
- Heron: Area = √(s(s − a)(s − b)(s − c))
- Brahmagupta (cyclic quadrilateral): Area = √((s − a)(s − b)(s − c)(s − d)), s = (a + b + c + d) ÷ 2
- Circumference = 2πr = πd
- Arc length = (θ/360) × 2πr
- Area of circle = πr²; area of sector = (θ/360) × πr² = ½ l r
Worked examples
1. A wheel has diameter 70 cm. How many full turns does it make to cover 1.1 km? (π = 22/7)
Circumference = πd = 22/7 × 70 = 220 cm. 1.1 km = 110000 cm. Turns = 110000 ÷ 220 = 500.
2. Find the area of a triangle with sides 13 cm, 14 cm and 15 cm.
s = (13 + 14 + 15) ÷ 2 = 21. s − a = 8, s − b = 7, s − c = 6. Area = √(21 × 8 × 7 × 6) = √7056 = 84 cm².
3. An isosceles triangle has sides 5 cm, 5 cm and 6 cm. Find its area.
s = 8. Area = √(8 × 3 × 3 × 2) = √144 = 12 cm². Check: height = √(25 − 9) = 4, so ½ × 6 × 4 = 12 ✔.
4. A quadrilateral ABCD has AB = 3, BC = 4, CD = 4, DA = 5 (in m) and diagonal AC = 5 m. Find its area.
Triangle ABC (3, 4, 5): s = 6, area = √(6 × 3 × 2 × 1) = 6 m². Triangle ACD (5, 4, 5): s = 7, area = √(7 × 2 × 3 × 2) = √84 ≈ 9.17 m². Total ≈ 15.17 m².
5. A cyclic quadrilateral has sides 7, 15, 20 and 24 cm. Find its area.
s = (7 + 15 + 20 + 24) ÷ 2 = 33. (s−a)(s−b)(s−c)(s−d) = 26 × 18 × 13 × 9 = 54756. Area = √54756 = 234 cm².
6. A sector has radius 21 cm and angle 120°. Find its arc length, area and perimeter. (π = 22/7)
Fraction = 120/360 = 1/3. Circumference = 2 × 22/7 × 21 = 132, so arc = 44 cm. Circle area = 22/7 × 441 = 1386, so sector area = 462 cm². Perimeter = 21 + 21 + 44 = 86 cm.
7. A clock's minute hand is 14 cm long. What area does it sweep in 5 minutes? (π = 22/7)
In 60 min it turns 360°, so in 5 min it turns 30°. Area = 30/360 × 22/7 × 14 × 14 = 1/12 × 616 ≈ 51.33 cm².
Common mistakes
- Using the full perimeter instead of the semi-perimeter s in Heron's formula.
- Forgetting the square root at the end, or forgetting to write the unit as cm² (area) and cm (perimeter, arc).
- Using Brahmagupta's formula for any quadrilateral. It is exact only when all four corners lie on a circle.
- Writing the perimeter of a sector as only the arc length. You must add the two radii: 2r + l.