Class 9 Maths Chapter 6: Measuring Space: Perimeter and Area — Notes, Examples & Practice

The perimeter of a polygon is the sum of its side lengths. For a circle the perimeter is called the circumference $C$.

For every circle, big or small, $C \div d$ is the same number — this constant is $\pi$. So $C = \pi d = 2\pi r$. Archimedes trapped $\pi$ between the perimeters of polygons drawn inside and outside a circle; Aryabhata (499 CE) gave $\pi \approx 3.1416$; Zu Chongzhi in China found $\dfrac{355}{113}$.

$\pi$ is irrational, so $\dfrac{22}{7}$ and 3.14 are approximations. Use whichever value the question asks for, and do not round in the middle of a calculation.

An arc with central angle $\theta$ (in degrees) is the fraction $\dfrac{\theta}{360}$ of the full circle, so its length is that fraction of the circumference.

This proportional thinking — 'what fraction of the full turn?' — also gives sector areas later. A quarter-circle arc ($90^\circ$) is $\dfrac{1}{4}$ of the circumference.

Area counts unit squares. A rectangle has area $l \times b$. A parallelogram can be cut and slid into a rectangle with the same base and the perpendicular height, so its area is base × height.

A triangle is half of a parallelogram with the same base and height: area $= \dfrac{1}{2} \times b \times h$. Any side can be the base, as long as $h$ is the perpendicular height to that side.

Puzzles in the book show that shapes with the same perimeter can have very different areas (and vice versa) — so you can never get one from the other without more information.

When you know all three sides $a, b, c$ of a triangle but not its height, use Heron's formula. First compute the semi-perimeter $s = \dfrac{a + b + c}{2}$.

Before using it, check the triangle inequality: each side must be less than the sum of the other two. Otherwise no triangle exists (and one of the brackets $s - a$ becomes zero or negative).

Heron's formula also lets you find a height: once you have the area, $h = \dfrac{2 \times \text{Area}}{\text{base}}$.

Brahmagupta (628 CE) gave the area of a cyclic quadrilateral (all four vertices on a circle) with sides $a, b, c, d$ using its semi-perimeter $s = \dfrac{a + b + c + d}{2}$.

It works only for cyclic quadrilaterals. A square, a rectangle and an isosceles trapezium are always cyclic; for other quadrilaterals you split into triangles instead.

Heron's formula is the special case $d = 0$: the quadrilateral collapses to a triangle (and every triangle is 'cyclic' since it has a circumcircle).

Cut a circle into many thin sectors and arrange them alternately up and down: they form a shape close to a rectangle of length $\pi r$ (half the circumference) and height $r$. So the area is $\pi r^2$.

A sector with angle $\theta$ is the fraction $\dfrac{\theta}{360}$ of the circle. A segment (cut off by a chord) is the sector minus the triangle formed by the two radii and the chord.

For composite figures — a running track, a flower bed with semicircular ends — split into pieces you know, then add or subtract.