NCERT Solutions for Class 9th Maths Chapter 6 Measuring Space: Perimeter and Area
Updated on 2026-09-19
About this chapter
For every circle the ratio of circumference to diameter is the same number, π . So C = 2πr = πd . π is irrational (Lambert, 1761), so no fraction equals it: π ≈ 22/7 but π ≠ 22/7 . Better approximations are 355/113 (Zu Chongzhi) and 3.1416 (Āryabhaṭa); Mādhava gave the first exact formula, π/4 = 1 − 1/3 + 1/5 − 1/7 + … A circle has rotational symmetry, so an arc turning through θ° at the centre is the fraction θ/360 of the whole circle. Hence arc length = 2πr × θ°/360° and area of a sector = πr² × θ°/360° . The half-turn and quarter-turn cases (πr and πr/2) are just θ = 180° and θ = 90°. Area of a rectangle = ab ; of a parallelogram = base × height (cut a triangle off one end and slide it to the other); of a triangle = ½ × base × height (two congruent copies make a parallelogram). The side
- Chapter opening: the 4 × 100 m relay track
- Perimeter of a Shape
- Perimeter of a Circle
- A Closer Look at a 400 m Athletics Track
- Problems, Puzzles, and Paradoxes on Perimeter
- Area of a Parallelogram
- Area of a Triangle
- Squaring a Rectangle
- Area of a Triangle and 6.9 Squaring a Rectangle
- Area of a Circle
Quick revision
| Quantity | Formula | Where it comes from | Watch out for |
|---|---|---|---|
| Circumference of a circle | C = 2πr = πd | π is the fixed C/D ratio of every circle | π ≈ 22/7, but π ≠ 22/7 |
| Length of an arc | l = 2πr × θ°/360° | The arc is the fraction θ/360 of the whole circle | θ must be the angle at the centre |
| Perimeter of a sector | 2πr × θ°/360° + 2r | Curved arc plus the two bounding radii | The two radii are easy to forget |
| Area of a circle | A = πr² | ½ × circumference × radius (Archimedes) | Use the radius, not the diameter |
| Area of a sector | A = πr² × θ°/360° | Same proportion argument as the arc | A sector is not a segment |
| Area of a segment | sector − triangle | The chord slices the triangle off the sector | Minor segment for θ < 180° |
| Area of a triangle | ½ × base × height | Half of the rectangle that encloses it | Height must be perpendicular to that base |
| Heron's formula | √(s(s−a)(s−b)(s−c)), s = ½(a+b+c) | Needs only the three side lengths | Best when no height is given |
| Area of a parallelogram | base × height | Cut a triangle off one end, slide it to the other | The two sides alone are not enough |
| Area of a trapezium | ½(a + b)h | Two copies make a parallelogram of base a+b | a and b are the parallel sides |
| Area of a cyclic 4-gon | √((s−a)(s−b)(s−c)(s−d)) | Brahmagupta; Heron is the case d = 0 | Only for 4-gons with a circumcircle |
| Median of a triangle | splits it into two equal areas | Equal bases BD = DC, same height | The two halves are usually not congruent |
Exercises
- Chapter opening: the 4 × 100 m relay track — In-text Questions Page 118
- Chapter opening: the 4 × 100 m relay track — Think and Reflect Page 118
- In-text Questions — Perimeter of a Shape Page 119
- Think and Reflect — Perimeter of a Shape Page 119
- In-text Questions — Perimeter of a Circle Page 120
- C/D's Adventurous Journey: From Ancient Approximations to the Exact Formula of Mādhava — In-text Questions Page 121
- A Closer Look at a 400 m Athletics Track — Think and Reflect Page 127
- Exercise Set 6.1 — Problems, Puzzles, and Paradoxes on Perimeter Page 129–130
- Think and Reflect — Area of a Parallelogram Page 131
- Think and Reflect — Area of a Parallelogram Page 131–132
- In-text Questions — Area of a Triangle Page 132
- In-text Questions — Area of a Triangle Page 133
- Think and Reflect — Area of a Triangle Page 133
- Think and Reflect — Area of a Triangle Page 134
- Think and Reflect — Squaring a Rectangle Page 142
- Exercise Set 6.2 — Area of a Triangle and 6.9 Squaring a Rectangle Page 142–143
- Think and Reflect — Area of a Circle Page 144
- In-text Questions — Area of a Circle Page 146
- Exercise Set 6.3 — Area of Sector of a Circle Page 148
- Whole chapter — End-of-Chapter Exercises Page 149–153