ICSE Class 8 Maths Chapter 32 Perimeter and Area of Plane Figures PDF Download

Unit 5: Mensuration

Chapter 32: Perimeter And Area Of Plane Figures

32.1 Review

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

Area: The area of a plane figure is the amount of surface enclosed by its sides (boundary).

Common units of perimeter are metre (m), centimetre (cm), decimetre (dm), etc.

1. \(1 \text{ m} = 100 \text{ cm}\) and \(1 \text{ cm} = \frac{1}{100} \text{ m} = 0.01 \text{ m}\)

2. \(1 \text{ m} = 10 \text{ dm}\) and \(1 \text{ dm} = \frac{1}{10} \text{ m} = 0.1 \text{ m}\)

3. \(1 \text{ dm} = 10 \text{ cm}\) and \(1 \text{ cm} = \frac{1}{10} \text{ dm} = 0.1 \text{ dm}\)

4. \(1 \text{ m} = 1000 \text{ mm}\) and \(1 \text{ mm} = \frac{1}{1000} \text{ m} = 0.001 \text{ m}\) and so on.

Common units of area are square metre (sq. m or m²), square centimetre (sq. cm or cm²), square millimetre (sq. mm or mm²), etc.

1. \(1 \text{ m}^2 = 100 \times 100 \text{ cm}^2 = 10,000 \text{ cm}^2\) and \(1 \text{ cm}^2 = \frac{1}{100 \times 100} \text{ m}^2 = 0.0001 \text{ m}^2\)

2. \(1 \text{ m}^2 = 10 \times 10 \text{ dm}^2 = 100 \text{ dm}^2\) and \(1 \text{ dm}^2 = \frac{1}{10 \times 10} \text{ m}^2 = 0.01 \text{ m}^2\)

3. \(1 \text{ dm}^2 = 10 \times 10 \text{ cm}^2 = 100 \text{ cm}^2\) and \(1 \text{ cm}^2 = \frac{1}{100} \text{ dm}^2 = 0.01 \text{ dm}^2\)

4. \(1 \text{ cm}^2 = 10 \times 10 \text{ mm}^2 = 100 \text{ mm}^2\) and \(1 \text{ mm}^2 = \frac{1}{100} \text{ cm}^2 = 0.01 \text{ cm}^2\) and so on.

Teacher's Note

Understanding unit conversions for perimeter and area helps in practical applications like calculating fencing costs for a garden or flooring materials needed for a room renovation.

32.2 Perimeter And Area Of Triangles

1. If a, b and c are the three sides of a triangle; then its

(i) perimeter = a + b + c

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

Where, \(s = \text{semi-perimeter of the triangle} = \frac{a+b+c}{2}\)

2. If one side (base) and the corresponding height (altitude) of the triangle are known, its

area = \(\frac{1}{2} \text{ base} \times \text{height}\)

Any side of the triangle can be taken as its base and the corresponding height means: the length of perpendicular to this side from the opposite vertex.

(i) If BC is taken as base
area = \(\frac{1}{2} \times \text{BC} \times \text{AP}\)

(ii) If AC is taken as base
area = \(\frac{1}{2} \times \text{AC} \times \text{BQ}\)

(iii) If AB is taken as base
area = \(\frac{1}{2} \times \text{AB} \times \text{CR}\)

Also, area = \(\frac{1}{2} \times \text{base} \times \text{height}\) - (i) base = \(\frac{2 \times \text{area}}{\text{height}}\)

(ii) height = \(\frac{2 \times \text{area}}{\text{base}}\)

Example 1

Find the area of a triangle whose sides are 9 cm, 12 cm and 15 cm. Also, find the length of altitude corresponding to the largest side of the triangle.

Solution:

Let a = 9 cm, b = 12 cm and c = 15 cm

\(s = \frac{a+b+c}{2} = \frac{9+12+15}{2} \text{ cm} = \frac{36}{2} \text{ cm} = 18 \text{ cm}\)

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

= \(\sqrt{18(18-9)(18-12)(18-15)}\)

= \(\sqrt{18 \times 9 \times 6 \times 3} = \sqrt{2916} = 54 \text{ cm}^2\) (Ans.)

Also, area of triangle = \(\frac{1}{2} \text{ base} \times \text{ corresponding altitude}\)

\(54 = \frac{1}{2} \times 15 \times h\) [Taking largest side as the base]

\(h = \frac{54 \times 2}{15} \text{ cm} = 7.2 \text{ cm}\) (Ans.)

Example 2

Find the area of an equilateral triangle, whose one side is a cm.

Solution:

\(s = \frac{a+b+c}{2} = \frac{a+a+a}{2} = \frac{3a}{2}\) [Sides of an equilateral triangle are equal]

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

= \(\sqrt{\frac{3a}{2}\left(\frac{3a}{2}-a\right)\left(\frac{3a}{2}-a\right)\left(\frac{3a}{2}-a\right)}\)

= \(\sqrt{\frac{3a}{2} \times \frac{a}{2} \times \frac{a}{2} \times \frac{a}{2}} = \frac{a \times a}{2 \times 2}\sqrt{3} = \frac{\sqrt{3}}{4} a^2\text{cm}^2\) (Ans.)

Example 3

The base of an isosceles triangle is 12 cm and its perimeter is 32 cm. Find the area of the triangle.

Solution:

Let each of the two equal sides of the given isosceles triangle be x cm.

Since, perimeter of the triangle is 32 cm

\(x + x + 12 = 32 \Rightarrow x = 10\)

Hence, the sides of the given isosceles triangle are 10 cm, 10 cm and 12 cm

Let a = 10 cm, b = 10 cm and c = 12 cm

\(s = \frac{a+b+c}{2} = \frac{10+10+12}{2} \text{ cm} = 16 \text{ cm}\)

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

= \(\sqrt{16(16-10)(16-10)(16-12)} \text{ cm}^2\)

= \(\sqrt{16 \times 6 \times 6 \times 4} \text{ cm}^2 = 4 \times 6 \times 2 \text{ cm}^2 = 48 \text{ cm}^2\) (Ans.)

Alternative method:

Draw AD perpendicular to base BC of the given triangle.

Since, the perpendicular from the vertex of an isosceles triangle to its base bisects the base, therefore

BD = CD = \(\frac{\text{BC}}{2} = \frac{12}{2}\) cm = 6 cm.

In right-angled triangle ABD,

AD\(^2\) + BD\(^2\) = AB\(^2\)

AD\(^2\) + 6\(^2\) = 10\(^2\)

AD\(^2\) = 100 - 36 = 64 and AD = \(\sqrt{64}\) cm = 8 cm

Now the base BC of the given triangle is 12 cm and its height AD is 8 cm

Area of the triangle = \(\frac{1}{2} \times \text{base} \times \text{height}\)

= \(\frac{1}{2} \times 12 \text{ cm} \times 8 \text{ cm} = 48 \text{ cm}^2\) (Ans.)

Test Yourself

1. In case of a triangle; if:

(a) base = 6 m and height = 4 m; area = ......................................................................

(b) area = 67.2 m² and height = 12 m; base = ......................................................................

(c) area = 15.36 cm² and base = 4.8 cm; height = ......................................................................

2. If each side of an equilateral triangle is 8 cm;

its area = ......................................................................

3. The area of an equilateral triangle, with side a cm is numerically equal to its perimeter, then ..................... = ..................... - \(\Rightarrow\) ..................... = ........................ - \(\Rightarrow\) a ........................

4. In the given figure, BC = 20 cm, AD = 15 cm and AC = 25 cm. We have:

\(\frac{1}{2} \times \text{BC} \times \text{AD} = \ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots,\)

........................ = ..................... and BE = ..................... = .......................

5. In the given figure

(a) BD = ................... and AD = ...................

(b) area of triangle ABC = ..................... = .......................

Teacher's Note

Triangle area calculations are essential in construction and land surveying, where irregular plots of land often need to be divided into triangular sections for proper measurement and valuation.

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