Introduction
- Books Name
- CBSE Class 7 Mathematics Book
- Publication
- Param Publication
- Course
- CBSE Class 7
- Subject
- Mathmatics
Introduction
In this chapter we learn about the perimeter and area of a rectangle and a square and circumference (perimeter) of a circle and area of a triangle, a parallelogram and a circle etc.
PERIMETER and area
1. Perimeter
The perimeter of a plane figure is the length of its boundary. The units of perimeter are the same as the units of length, i.e., cm, m, etc.
2. Area
The measurement of the region enclosed by a plane figure is called its area.
3. STANDARD UNITS OF AREA
The units of area are sq cm (written as cm2), sq metres (written as m2), etc.
Conversion of units
Length units Area units
1 cm = 10 mm 1 cm2 = (10 × 10) mm2 = 100 mm2
1 dm = 10 cm 1 dm2 = (10 × 10) cm2 = 100 cm2
1 m = 10 dm 1 m2 = (10 × 10) dm2 = 100 dm2
1 m = 100 cm 1 m2 = (100 × 100) cm2 = 10000 cm2
1 hectare = 10000 m2
4. AREA AND PERIMETER OF A RECTANGLE
Consider a rectangle with length = l units and breadth = b units. Then, we have :
(i) Area of the rectangle = (l × b) sq units
Illustration 1
A door of length 1 m and breadth 0.5 m is on a wall. The length of the wall is 4.5 m and the breadth is 3.6 m as shown in Fig. Find the cost of white washing the wall, if the rate of white washing the wall is Rs 20 per m2.
Solution
We have,
l = Length of the wall = 4.5 m,
b = Breadth of the wall = 3.6 m
∴ Area of the wall = l × b = 4.5 m × 3.6 m = 16.2 m2
Area of the door = 1 m × 0.5 m = 0.5 m2
∴ Area to be white washed = 16.2 m2 – 0.5 m2 = 15.7 m2
Cost of white washing the wall = Rs (15.7 × 20) = Rs 314
Illustration 2
The length and breadth of a playground are 75 m 20 cm and 34 m 80 cm, respectively. Find the cost of levelling it at Rs 1.50 per square metre. How long will a boy take to go three times round the field, if he walks at the rate of 1.5 m/sec.
Solution
We have,
Length of the playground = 75 m 20 cm = 75.20 cm
Breadth of the playground = 34 m 80 cm = 34.80 m .
∴ Area of the playground = 75.20 × 34.80 m2 = 2616.96 m2
∴ Cost of levelling = Rs 2616.96 × 1.50 = Rs 3925.44
Perimeter of the playground = 2 (Length + Breadth)
= 2 (75.20 + 34.80) m = 2 × 110m = 220m
Illustration 3
A room is 9 m long, 8 m broad and 6.5 m high. It has one door of dimensions 2 m ´ 1.5 m and three windows each of dimensions 1.5 m × 1 m. Find the cost of white washing the walls at Rs 3.80 per square metre.
Solution
The dimensions of the room are:
l = Length = 9 m, b = Breadth = 8 m and h = Height = 6.5 m
∴ Area of 4 walls of the room = 2 {(l + b) × h} = 2 {9 + 8) × 6.5}m2 =221m2
Area of 1 door = (2 × 1.5) m2 =3 m2
Area of 3 windows = {3 × (1.5 × l)}m2 = 4.5 m2
∴ Area to be white washed = Area of 4 walls – Area of 1 door – Area of 3 windows
= (221 –3 – 4.5)m2 =213.5m2
So, cost of white washing = Rs (213.5 × 3.80) = Rs 811.30
AREA OF PATHS AND VERANDAHS
- Books Name
- CBSE Class 7 Mathematics Book
- Publication
- Param Publication
- Course
- CBSE Class 7
- Subject
- Mathmatics
AREA OF PATHS AND VERANDAHS
Area of verandah is nothing but the difference between the areas of two rectangles.
Illustration 4
A rectangular field is 45 m × 30 m. A path 1.5 m wide is to be constructed along the sides inside the field. Find the area of the path and its cost of construction at the rate of Rs 1.50 per square metre.
Solution
Length of the field = 45 m
Width of the field = 30 m
Area of the field = 45 × 30 sq m = 1350 sq m
Length of the inner rectangle = 45 – 3 or 42 m
Width of the inner rectangle = 30 – 3 or 27 m
Area of the inner rectangle = 42 × 27 sq m = 1134 sq m
Area of path = 1350 – 1134 or 216 sq m
Cost of constructing 1 sq m of path = Rs 1.50
Cost of constructing 216 sq m of path = Rs 1.50 × 216 = Rs 324
Area of Cross Roads
ABCD is a rectangular piece of ground. Two roads one parallel to the length and the other parallel to the breadth together are called cross roads.
Illustration 5
A field is 75 m long and 32 m wide. Two 3 m wide roads are constructed in the centre of the field one parallel to its length and the other parallel to its breadth. Find (a) area of cross roads (B) the cost of levelling the roads at Rs 1.80 per square metre (C) area of the remaining field.
Solution
(A) Area of the road parallel to the length = 75 × 3 m2 = 225 m2
Area of the road parallel to the breadth = 32 × 3 m2 = 96 m2
Area of shaded portion = 3 × 3 or 9 m2
∴ Area of the roads = (225 + 96 – 9) m2 = 312 m2
(B) Cost of levelling 1 m2 road = Rs 1.80
Cost of levelling 312 m2 road = Rs 312 × 1.80 = Rs 561.60
(C) Area of field = 75 × 32 or 2400 m2
Area of remaining field = (2400 – 312) m2 = 2088 m2
Areas and Perimeter of Triangle
- Books Name
- CBSE Class 7 Mathematics Book
- Publication
- Param Publication
- Course
- CBSE Class 7
- Subject
- Mathmatics
Areas of Parallelogram and Rhombus
- Books Name
- CBSE Class 7 Mathematics Book
- Publication
- Param Publication
- Course
- CBSE Class 7
- Subject
- Mathmatics
Illustration 8
The base of a parallelogram is thrice its height. If the area is 876 cm2, find the base and height of the parallelogram.
Solution
Let the height of the parallelogram be x cm.
Then, base = 3x cm.
∴ Area of the parallelogram= (x × 3x) cm2 = 3x2 cm2
But, area of the parallelogram is given as 867 cm2.
∴ 3x2 = 867
⇒ x2 = 289
⇒ x2 = 172
⇒ x = 17 cm.
Thus, height = 17 cm and base = (3 × 17) cm = 51 cm.
Area related to circle
- Books Name
- CBSE Class 7 Mathematics Book
- Publication
- Param Publication
- Course
- CBSE Class 7
- Subject
- Mathmatics
Area related to circle
Circle : Circle is a path of a moving point, which moves in such a manner that its distance from a fixed point is always equal. The fixed point is called centre of the circle and the fixed distance is called radius of the circle.