Maharashtra Board Class 7 Math Solution Chapter 12 – Perimeter and Area
Balbharati Maharashtra Board Class 7 Math Solution Chapter 12: Perimeter and Area. Marathi or English Medium Students of Class 7 get here Perimeter and Area full Exercise Solution.
Std |
Maharashtra Class 7 |
Subject |
Math Solution |
Chapter |
Perimeter and Area |
Practice Set 44
1.) If the length and breadth of a rectangle are doubled, how many times the perimeter of the old rectangle will that of the new rectangle be?
ANS:
Let,
Length of a rectangle be x.
Breadth of a rectangle be y.
If we doubled the length of a rectangle is 2x.
Also we doubled the breadth of a rectangle is 2y.
We know,
Perimeter of a rectangle = 2 x (L + B)
Perimeter of a rectangle 1 = 2 x ( x + y)
Now,
Perimeter of a rectangle 2 = 2 x (2x + 2y)
Perimeter of a rectangle 2 = 4 x (x + y)
From above we conclude that,
The perimeter of the old rectangle will doubled of that the new rectangle.
2.) If the side of a square is tripled, how many times the perimeter of the first square will that of the new square be?
ANS:
We know,
Perimeter of a square = 4 x side
Let,
Square 1 has side x.
Perimeter of a square 1 = 4 x X
Perimeter of a square 1 = 4x
Now,
The side of a square is tripled
The side of a square = 3x
Perimeter of a square = 4 x side
Perimeter of a square 2 = 4 x 3x
Perimeter of a square 2 = 12x
From above we conclude that,
The perimeter of the first square will be 3 times that of the new square.
3.) Given alongside is the diagram of a playground. It shows the length of its sides. Find the perimeter of the playground.
ANS:
Perimeter is the sum of all outside boundaries of a figure.
Given that a playground having outside dimensions.
We are finding the perimeter of a playground.
Perimeter of a playground = Sum of all outer sides.
We are adding all side we get following sides.
Perimeter of a playground = (10 + 15 + 5 + 15 + 15 + 30)
Perimeter of a playground = 90 m.
4.) As shown in the figure, four napkins all of the same size were made from a square piece of cloth of length 1 m. What length of lace will be required to trim all four sides of all the napkins?
ANS:
Given that,
Square piece of cloth of length 1 m.
From this we have to make napkins all of the same size.
We have to find length of lace will be required to trim all four sides of all the napkins.
When we cut square piece we get 4 napkins.
Each side of napkin = 1 m / 4 = 0.25 m
Perimeter of Each napkin = 4 x side
Perimeter of Each napkin = 4 x 0.25
Perimeter of Each napkin = 1 m
There are 4 napkins.
Perimeter of 4 napkins = 4m
We have to trim all four sides of a napkins from both sides by lace.
Length of lace will be required = 4 x 2 = 8m
Practice Set 45
1.) If the side of a square is 12 cm, find its area.
ANS:
Given that,
Side of a square is 12 cm.
We have to find area of a square.
We know,
Area of Square = side x side
Area of Square = 12 m x 12 m
Area of Square = 144 m^{2}
2.) If the length of a rectangle is 15 cm and breadth is 5 cm, find its area.
ANS:
Given that,
Length of a rectangle is 15 cm.
Breadth of a rectangle is 5 cm.
We have to find area of a rectangle.
We know,
Area of a rectangle = Length x Breadth
Area of a rectangle = 15 cm x 5 cm
Area of a rectangle = 75 cm^{2}
3.) The area of a rectangle is 102 sqcm. If its length is 17 cm, what is its perimeter?
ANS:
Given that,
The area of a rectangle is 102 sqcm.
Also given that length of a rectangle is 17 cm.
We have to find perimeter of a rectangle.
We know,
Area of a rectangle = Length x Breadth
102= 17 x Breadth
Breadth = 102 / 17
Breadth = 6 cm.
Also we know that,
Perimeter of rectangle = 2 x (Length + Breadth)
Perimeter of rectangle = 2 x (17 + 6)
Perimeter of rectangle = 2 x 23
Perimeter of rectangle = 46 cm.
4.) If the side of a square is tripled, how many times will its area be as compared to the area of the original square?
ANS:
We know,
Area of a square = Side x Side
Let,
Side of a square is x.
Then,
Area of a square = Side x Side
Area of a square 1 = x^{2}
Now,
Given that
Side of a square is tripled.
Side of a square is 3x.
Then,
Area of a square = Side x Side
Area of a square 2 = 3x x 3x
Area of a square 2 = 9x^{2}
Area of Original Square is 9 times of new square.
Practice set 46
1.) A page of a calendar is 45 cm long and 26 cm wide. What is its area?
ANS:
Given that,
A page of a calendar is 45 cm long and 26 cm wide.
Length of page of calendar = 45 cm
Breadth of a page of calendar = 26 cm
Page of calendar have rectangular shape.
We know,
Area of a rectangle = Length x Breadth
Area of rectangular Page of calendar = Length x Breadth
Area of rectangular Page of calendar = 45 x 26
Area of rectangular Page of calendar = 1,170 cm^{2}
2.) What is the area of a triangle with base 4.8 cm and height 3.6 cm?
ANS:
Given that,
Base of a triangle = 4.8 cm
Height of a triangle = 3.6 cm
We know,
Area of a triangle = ½ x Base x Height
Area of a triangle = ½ x 4.8 x 3.6
Area of a triangle = 2.4 x 3.6
Area of a triangle = 8.64 cm^{2}
3.) What is the value of a rectangular plot of land 75.5 m long and 30.5 m broad at the rate of 1000 rupees per square metre?
ANS:
Given that,
Rectangular plot of land 75.5 m long and 30.5 m broad.
Length of a rectangular plot = 75.5 m
Breadth of rectangular plot = 30.5 m
We know,
Area of rectangular Plot = Length x Breadth
Area of rectangular Plot = 75.5 x 30.5
Area of rectangular Plot = 2302.75 m^{2}
We have to find the value of a plot at the rate of 1000 rupees per square metre.
Value of a plot = Area of rectangular Plot x 1000
Value of a plot = 2302.75 m^{2 }x 1000
Value of a rectangular plot =Rs. 23, 02,750
4.) A rectangular hall is 12 m long and 6 m broad. Its flooring is to be made of square tiles of side 30 cm. How many tiles will fit in the entire hall? How many would be required if tiles of side 15 cm were used?
ANS:
Given that,
A rectangular hall is 12 m long and 6 m broad
Length of rectangular hall = 12 m.
Breadth of a rectangular hall = 6 m.
We know,
We know,
Area of rectangular hall= Length x Breadth
Area of rectangular hall = 12 x 6
Area of rectangular hall = 72 m^{2}
Now,
In this rectangular hall flooring is to be made of square tiles of side 30 cm.
Given that
Side of square tile = 30 cm = 0.3 m
Area of a square tile = side x side
Area of a square tile = 0.3 x 0.3
Area of a square tile = 0.09 m^{2}
We have to find total number of tiles in a hall.
We are dividing total area by area of square tile.
Total number of tiles in a hall = 72 / 0.09
Total number of tiles in a hall = 800
Now,
If side of square tile is 15 cm,
Then
Side of square tile = 15 cm = 0.15 m
Area of a square tile = side x side
Area of a square tile = 0.15 x 0.15
Area of a square tile = 0.0225 m^{2}
We have to find total number of tiles in a hall.
We are dividing total area by area of square tile.
Total number of tiles in a hall = 72 / 0.0225
Total number of tiles in a hall = 3200
Square tiles of side 30 cm is 800 required and for 15 cm is 3200 square tiles are required.
5.) Find the perimeter and area of a garden with measures as shown in the figure alongside.
ANS:
We know,
Perimeter is the sum of all outside boundaries of a figure.
There are total 12 outside boundaries of a figure.
Perimeter of a given garden = 12 x (side)
Perimeter of a given garden = 12 x 13
Perimeter of a given garden = 156 m
Now,
Area of garden =
There are 5 square shape inside garden.
We are finding area of 1 square and multiply it by 5.
Area of square = Side x Side
Area of square = 13 x 13
Area of square = 169 m^{2}
Now,
Area of total garden = 5 x 169
Area of total garden = 845 m^{2}
Perimeter and area of a garden is 156 m and 845 m^{2 }respectively.
Practice Set 47
1.) Find the total surface area of cubes having the following sides.
(i) 3 cm
ANS:
Given that side of cube = 3 cm
We know,
Total surface area of cube = 6 x side^{2}
Total surface area of cube = 6 x (3)^{2}
Total surface area of cube = 6 x 9
Total surface area of cube = 54 cm^{2}
(ii) 5 cm
ANS:
Given that side of cube = 5 cm
We know,
Total surface area of cube = 6 x side^{2}
Total surface area of cube = 6 x (5)^{2}
Total surface area of cube = 6 x 25
Total surface area of cube = 150 cm^{2}
(iii) 7.2 m
ANS:
Given that side of cube = 7.2 m
We know,
Total surface area of cube = 6 x side^{2}
Total surface area of cube = 6 x (7.2)^{2}
Total surface area of cube = 6 x 51.84
Total surface area of cube = 311.04 m^{2}
(iv) 6.8 m
ANS:
Given that side of cube = 6.8 m
We know,
Total surface area of cube = 6 x side^{2}
Total surface area of cube = 6 x (6.8)^{2}
Total surface area of cube = 6 x 46.24
Total surface area of cube = 277.44 m^{2}
(v) 5.5 m
ANS:
Given that side of cube = 5.5 m
We know,
Total surface area of cube = 6 x side^{2}
Total surface area of cube = 6 x (5.5)^{2}
Total surface area of cube = 6 x 30.25
Total surface area of cube = 181.5 m^{2}
2.) Find the total surface area of the cuboids of length, breadth and height as given below:
(i) 12 cm, 10 cm, 5 cm
ANS:
Given that,
Length of cuboid = 12 cm
Breadth of cuboid = 10 cm
Height of cuboid = 5 cm
We know,
Total surface area of cuboid = 2 x (length × breadth + breadth × height + length × height)
Total surface area of cuboid = 2 x (12 x 10 + 10 x 5 + 12 x 5)
Total surface area of cuboid = 2 x (120 + 50 + 60)
Total surface area of cuboid = 2 x 230
Total surface area of cuboid = 460 cm^{2}
(ii) 5 cm, 3.5 cm, 1.4 cm
ANS:
Given that,
Length of cuboid = 5 cm
Breadth of cuboid = 3.5 cm
Height of cuboid = 1.4 cm
We know,
Total surface area of cuboid = 2 x (length × breadth + breadth × height + length × height)
Total surface area of cuboid = 2 x (5 x 3.5 + 3.5 x 1.4 + 5 x 1.4)
Total surface area of cuboid = 2 x (17.5 + 4.9 + 7)
Total surface area of cuboid = 2 x 29.4
Total surface area of cuboid = 58.8 cm^{2}
(iii) 2.5 cm, 2 m, 2.4 m
ANS:
Given that,
Length of cuboid = 2.5 cm
Breadth of cuboid = 2 cm
Height of cuboid = 2.4 cm
We know,
Total surface area of cuboid = 2 x (length × breadth + breadth × height + length × height)
Total surface area of cuboid = 2 x (2.5 x 2 + 2 x 2.4 + 2.5 x 2.4)
Total surface area of cuboid = 2 x (5 + 4.8 + 6)
Total surface area of cuboid = 2 x 15.8
Total surface area of cuboid = 31.6 cm^{2}
(iv) 8 m, 5 m, 3.5 m
ANS:
Given that,
Length of cuboid = 8 m
Breadth of cuboid = 5 m
Height of cuboid = 3.5 cm
We know,
Total surface area of cuboid = 2 x (length × breadth + breadth × height + length × height)
Total surface area of cuboid = 2 x (8 x 5 + 5 x 3.5 + 8 x 3.5)
Total surface area of cuboid = 2 x (40 + 17.5 + 28)
Total surface area of cuboid = 2 x 85.5
Total surface area of cuboid = 171 m^{2}
3.) A matchbox is 4 cm long, 2.5 cm broad and 1.5 cm in height. Its outer sides are to be covered exactly with craft paper. How much paper will be required to do so?
ANS:
Given that,
A matchbox is 4 cm long, 2.5 cm broad and 1.5 cm in height.
We know,
Match box is cuboid in shape.
Outer side of match box is covered with craft paper.
We have to find how much craft paper required.
I.e. we have to find total surface area of a cuboidal matchbox.
We know,
Total surface area of cuboid = 2 x (length × breadth + breadth × height + length × height)
Here,
Length = 4 cm
Breadth = 2.5 cm
Height = 1.5 cm
Total surface area of cuboidal matchbox = 2 x (length × breadth + breadth × height + length × height)
Total surface area of cuboidal matchbox = 2 x (4 x 2.5 + 2.5 x 1.5 + 1.5 x 4)
Total surface area of cuboidal matchbox = 2 x (10 + 3.75 + 6)
Total surface area of cuboidal matchbox = 2 x 19.75
Total surface area of cuboidal matchbox = 39.5 cm^{2}
Craft paper required is = 39.5 cm^{2}
4.) An open box of length 1.5 m, breadth 1 m, and height 1 m is to be made for use on a trolley for carrying garden waste. How much sheet metal will be required to make this box? The inside and outside surface of the box is to be painted with rust proof paint. At a rate of 150 rupees per sqm, how much will it cost to paint the box?
ANS:
Given that,
An open box of length 1.5 m, breadth 1 m, and height 1 m is to be made for use on a trolley for carrying garden waste.
Box shape is cuboidal.
We have to find how much sheet metal required for this we have to find total surface area of cuboidal box.
We know,
Total surface area of cuboid = 2 x (length × breadth + breadth × height + length × height)
Here,
Length = 1.5 m
Breadth = 1m
Height = 1 m
Total surface area of cuboidal box = 2 x (length × breadth + breadth × height + length × height)
Total surface area of cuboidal box = 2 x (1.5 x 1 + 1 x 1 + 1.5 x 1)
Total surface area of cuboidal box = 2 x (1.5 + 1 + 1.5)
Total surface area of cuboidal box = 2 x 4
Total surface area of cuboidal box = 8 m^{2}
Total sheet metal required is 8 m^{2}
Now,
We have to paint inside and outside of this box at rate of 150 rupees per sq.m
We have to find cost of paint.
Cost of paint = (Total surface area of cuboidal box x 2 ) x 150
Cost of paint = (8 x 2) x 150
Cost of paint = 16 x 150
Cost of paint = Rs.2400