ICSE Class 9 Maths Chapter 20 Area and Perimeter of Plane Figures

Unit 6: Mensuration

Area And Perimeter Of Plane Figures

20.1 Introduction

1. Perimeter

The perimeter of a plane figure is the length of its boundary.

The unit of perimeter is the same as the unit of length, i.e. cm, m, etc.

2. Area

The area of a plane figure is the measure of the surface enclosed by its boundary.

The unit of area is cm² (square centimetre); m² (square metre), etc.

Students should know the difference between "square metre" and "metre square". x square metre means an area and x metre square means a square each of whose sides is x metre long and so its area = x x x = x² square metre.

20.2 Area And Perimeter Of Triangles

Area = \(\frac{1}{2}\) base × corresponding height (altitude).

= \(\frac{1}{2}\) BC × AD

1. Corresponding height (or altitude) of a triangle means length of perpendicular from the opposite vertex to the base.

2. In a triangle, any of its sides can be considered as base e.g.

(i) If side AC is taken as the base, the length of perpendicular BE is the corresponding height (altitude).

Area = \(\frac{1}{2}\) base × height

= \(\frac{1}{2}\) AC × BE

(ii) If side AB is taken as the base, the length of perpendicular CF is the corresponding height.

Area = \(\frac{1}{2}\) AB × CF

Heron's formula:

If a, b and c are three sides of a triangle, then its perimeter (2s) = a + b + c

and semi-perimeter (s) = \(\frac{a + b + c}{2}\)

Area of the triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\)

Remember:

Example 1

Find the area of a triangle:

(i) whose height is 6 cm and base is 10 cm.

(ii) whose three sides are 17 cm, 8 cm and 15 cm long. Also, in part (ii) of this question; calculate the length of the altitude corresponding to the largest side of the triangle.

Solution:

(i) Area of triangle = \(\frac{1}{2}\) base × height

= \(\frac{1}{2}\) × 10 × 6 cm² = 30 cm²

(ii) Let a = 17 cm, b = 8 cm and c = 15 cm

s = \(\frac{a + b + c}{2}\) = \(\frac{17 + 8 + 15}{2}\) cm = 20 cm

Area = \(\sqrt{s(s-a)(s-b)(s-c)}\)

= \(\sqrt{20(20-17)(20-8)(20-15)}\) = 60 cm²

Since, the largest side of the triangle is 17 cm

and \(\frac{1}{2}\) × base × altitude = area

\(\frac{1}{2}\) × 17 × alt. = 60 ∴ alt. = \(\frac{60 \times 2}{17}\) cm = 7.06 cm

Teacher's Note

Understanding perimeter and area helps in real-world situations like fencing a garden or calculating flooring materials needed for a room.

20.3 Some Special Types Of Triangles

1. Equilateral Triangle:

Let the length of each side of an equilateral triangle be a unit; then, its perimeter = 3 × its side = 3a

and its area = \(\frac{\sqrt{3}}{4}\) × (side)² = \(\frac{\sqrt{3}}{4}\) × a²

Example 2

The area of an equilateral triangle is numerically equal to its perimeter. Find a side of the triangle. Take \(\sqrt{3}\) = 1.73.

Solution:

Given: Area = Perimeter (Numerically)

\(\frac{\sqrt{3}}{4}\) (side)² = 3 side i.e. side = \(\frac{3 \times 4}{\sqrt{3}}\)

side = 4\(\sqrt{3}\) = 4 × 1.73 unit = 6.92 unit

Example 3

Calculate the area of an equilateral triangle, whose height is 20 cm.

Solution:

Let ABC be the given equilateral triangle and AD is perpendicular to base BC; then clearly; AD = 20 cm

If each side of the given triangle be a cm; then AB = a cm

and, BD = \(\frac{1}{2}\) BC (In equilateral triangle, perpendicular from vertex bisects the base)

= \(\frac{1}{2}\) a cm

In right-angled triangle ABD:

AD² + BD² = AB² ∴ (20)² + \(\left(\frac{a}{2}\right)^2\) = a² (Pythagoras Theorem)

On simplifying, we get: a² = 400 × \(\frac{4}{3}\) = \(\frac{1600}{3}\)

Area of the triangle = \(\frac{\sqrt{3}}{4}\) a²

= \(\frac{\sqrt{3}}{4}\) × \(\frac{1600}{3}\) cm² = 230.9 cm²

2. Isosceles Triangle:

Example 4

Find the area of an isosceles triangle whose equal sides are 5 cm each and base is 6 cm.

Solution:

In an isosceles triangle ABC, let AB = AC = 5 cm and BC = 6 cm. Draw AD perpendicular to BC. Since, the perpendicular from the vertex to the base of an isosceles triangle bisects the base, therefore

BD = CD = \(\frac{1}{2}\) × 6 cm = 3 cm

Applying Pythagoras Theorem in triangle ABD, we get:

AD² = AB² - BD²

= 5² - 3² = 25 - 9 = 16 ∴ AD = 4 cm

Area of triangle = \(\frac{1}{2}\) base × height

= \(\frac{1}{2}\) BC × AD = \(\frac{1}{2}\) × 6 × 4 cm² = 12 cm²

Alternative method:

Since, the sides of the given isosceles triangle are 5 cm, 5 cm and 6 cm

s = \(\frac{5 + 5 + 6}{2}\) cm = 8 cm

and area of triangle = \(\sqrt{8(8-5)(8-5)(8-6)}\) cm²

= \(\sqrt{8 \times 3 \times 3 \times 2}\) cm² = 12 cm²

Third method:

Area of an isosceles triangle

= \(\frac{1}{4}\) × b × \(\sqrt{4a^2 - b^2}\); where, a = length of each equal side and, b = length of base.

= \(\frac{1}{4}\) × 6 × \(\sqrt{4 \times 5^2 - 6^2}\) = 12 cm²

Teacher's Note

Isosceles and equilateral triangles appear in architecture and design, such as roof trusses and decorative patterns in buildings.

20.4 Area And Perimeter Of Quadrilaterals

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