ICSE Class 7 Maths Chapter 31 Perimeter and Area

Unit 5 - Mensuration

Chapter 31 - Perimeter And Area

For Plane Figure

31.1 Perimeter

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

Thus, the perimeter of the given figure (quadrilateral) = AB + BC + CD + DA

31.2 Area

The area of a plane figure is the amount of surface enclosed by its sides. In the figure, given above, the shaded portion shows its area.

31.3 Perimeter And Area Of Some Special Figures

1. Rectangle

A rectangle is a four sided closed figure with opposite sides equal and each angle 90 degrees.

In general, the longer side of a rectangle is called its length and is denoted by letter 'l' whereas, the shorter side is called its breadth and is denoted by letter 'b'.

Perimeter, P = Length of its boundary

= l + b + l + b

= 2l + 2b

Therefore, P = 2(l + b)

And, area, A = its length × its breadth

Therefore, A = l × b

2. Square

A square is a four sided closed figure with all its sides equal and each angle of 90 degrees.

Clearly, its perimeter P = Length of its boundary

= l + l + l + l

Therefore, P = 4l

And, its area, A = length × breadth

= l × l

Therefore, A = l²

31.4 Units Of Perimeter And Area

If the sides are in centimetre (cm), the unit of perimeter is also in centimetre and the unit of area is square centimetre (cm²).

Similarly, if the sides are in metre (m), the unit of perimeter is also in metre and the unit of area is square metre (m²).

1 m = 100 cm and 1 m² = 100 cm × 100 cm = 10,000 cm²

1 cm = 1/100 m and 1 cm² = 1/10,000 m²

1 cm = 10 mm and 1 cm² = 10 mm × 10 mm = 100 mm²

1 mm = 1/10 cm and 1 mm² = 1/100 cm²

Greater units used for area (usually, for the area of land) are Are and Hectare, such that:

1 Are = 100 m²

and 1 Hectare = 100 Are = 100 × 100 m² = 10,000 m².

Example 1

In a rectangle:

1. length = 10 cm and breadth = 6 cm, find its area and its perimeter.

2. area = 240 cm² and length = 20 cm, find its breadth and perimeter.

3. length = 8 cm and breadth = 8 cm, find its area and perimeter.

Solution

1. Given: l = 10 cm and b = 6 cm

Area = l × b = 10 cm × 6 cm = 60 cm²

And, perimeter = 2 (l + b) = 2(10 + 6) cm = 2 × 16 cm = 32 cm

2. Given: A = 240 cm² and l = 20 cm

Since A = l × b, breadth, b = A / l = 240 / 20 cm = 12 cm

and, perimeter = 2 (l + b) = 2(20 + 12) cm = 2 × 32 cm = 64 cm

3. Here l = 8 cm and b = 8 cm. Since l = b, its a square.

Area = l² = 8 cm × 8 cm = 64 cm²

and, perimeter = 4 l = 4 × 8 cm = 32 cm

Example 2

The length of a rectangular field is 200 m and its width is 100 m.

Find: 1. the cost of ploughing it at the rate of rupees 10 per m².

2. the cost of fencing it with wire at the rate of rupees 15 per metre.

Solution

1. For ploughing, we need to calculate the area (A).

Since, l = 200 m and b = 100 m

Area of the field = 200 m × 100 m = 20,000 m²

And, cost of ploughing the field = Area × Rate = 20,000 × rupees 10 = rupees 2,00,000

2. Length of fence = Perimeter = 2 (l + b) = 2(200 + 100) m = 600 m

Cost of fencing = Length of fence × Rate = 600 × rupees 15 = rupees 9,000

Example 3

Find the area and the perimeter of the given figure. All measurements are in cm and the angle at each vertex is 90 degrees.

Solution

For such figures, first of all draw dotted lines to divide the figure in convenient parts of squares and rectangles.

As shown in the figure, the three parts obtained are marked as (1), (2) and (3).

Now, find the area of each part.

Area of rectangle shown by part (1) = 2 cm × 1 cm = 2 cm²

Area of rectangle shown by part (2) = 2 cm × 3 cm = 6 cm²

Area of rectangle shown by part (3) = 2 cm × 1 cm = 2 cm²

Total required area = 2 cm² + 6 cm² + 2 cm² = 10 cm²

For finding the perimeter, we have to add the outer boundary lines (not the dotted lines).

For this, the simplest way is to start adding the sides, starting from any point of its boundary and then reach to the same point again.

Here, if we start from A and move to right (in the anticlockwise direction), we get:

Perimeter = (2 + 1 + 2 + 1 + 2 + 1 + 2 + 1 + 2 + 1 + 2 + 1) cm = 18 cm

Example 4

Use the informations given in the adjoining figure to find the area of the shaded portion. Every measurement, given in the figure, is in metre.

Solution

Area of the shaded portion along the length = 3 m × 30 m = 90 m²

Area of the shaded portion along the width = 5 m × 20 m = 100 m²

The portion ABCD with AB = 5 m and BC = 3 m is common to both the shaded portions, one along the length and the other along the width. So, this portion has been taken twice. Since area of this portion ABCD = 5 m × 3 m = 15 m².

The area of the shaded portion = 90 m² + 100 m² - 15 m² = 175 m²

Example 5

Find the side of the square, whose area is 441 sq. cm.

Solution

We know that, area of a square = (side)²

Since (side)² = 441 cm²

Therefore, side = \(\sqrt{441}\) cm = 21 cm

Example 6

Area of a square is 100 Hectare. Find its each side and perimeter.

Solution

If the side of a square is l m, its area = l² sq. m. = (m²)

Given, area of the square = 100 Hectare

Therefore, l² = 100 × 10,000 m² (since 1 Hectare = 10,000 m²)

Since, l = \(\sqrt{100 \times 10,000}\) m = 1,000 m

Therefore, Side of the given square = 1,000 m

Also, perimeter, P = 4 l = 4 × 1,000 m = 4,000 m

Example 7

If length of a rectangle is 40 cm and its perimeter is 130 cm; find its breadth and area.

Solution

Given, perimeter = 130 cm

Therefore, 2l + 2b = 130 cm (Since, P = 2l + 2b)

Therefore, 2 × 40 cm + 2b = 130 cm

Therefore, 2b = 130 cm - 80 cm = 50 cm

Breadth, b = 50 / 2 cm = 25 cm

And, area = l × b = 40 cm × 25 cm = 1,000 cm²

Teacher's Note

Understanding perimeter and area helps us calculate fencing needs for gardens and flooring costs for rooms - practical skills used by contractors and architects daily.

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