GSEB Solutions for Class 6 Mathematics – Perimeter and Area
GSEB SolutionsMathsScience
Activity
Solution 1(1):
(Length of each box = 1 cm)
Perimeter of the square = 5 + 5 + 5 + 5 = 20 cm
And area of the square = 5 × 5 = 25 sq. cm
Perimeter of the rectangle = 4 + 7 + 4 + 7 = 22 cm
And area of the rectangle = 4 × 7 = 28 sq. cm
Solution 1(2):
Area of the triangle = 8 full blocks + 2 more than half blocks + 4 half blocks
∴ Area of the triangle = 8 + 2 + 2 = 12 sq. cm
Area of the Circle = 9 full blocks + 12 more than half blocks
∴ Area of the Circle = 9 + 12 = 21 sq. cm
Exercise
Solution 1(1):
1 m = 100 cm
∴ 1 sq m = 100 cm × 100 cm = 10,000 cm
1 sq m = 10000 sq. cm
Solution 1(2):
1 m × 1 m = 1 sq. m
1 m = 100 cm
∴ 100 cm × 100 cm = 10,000 sq. cm
∴ 1 sq m = 10,000 sq. cm
∴ 40,000 sq. cm = 40,000 ÷ 10000 = 4 sq. m
40000 sq. cm = 4 sq. m
Solution 1(3):
Perimeter of a square = 4 × length
∴ Perimeter of the square of length 4 cm = (4 × 4) cm = 16 cm
Perimeter of a square of length 4 cm = 16 cm.
Solution 1(4):
Perimeter of a rectangle = 2 (length + breadth)
∴ Perimeter of a rectangle of length 3 cm and breadth 2 cm = 2 (3 + 2) m = 10 m.
Perimeter of the rectangle of length 3 m and breadth 2 m is 10 m.
Solution 2(1):
Length of the rectangular field = 45 m
Breadth of the rectangular field = 40 m
Area of a rectangle = length × breadth
Area of the rectangular field = 45 m × 40 m = 1800 sq. m
The cost of tiling 1 sq m field = Rs. 12
The cost of tiling 1800 sq m field = Rs. (1800 × 12)
= Rs. 21600
The cost of tiling the field is Rs. 21600.
Solution 2(2):
Length of the square wall = 12m
Area of a square = length × length
Area of the square wall = 12 m × 12 m = 144 sq. m
The cost of painting 1 sq m of the wall = Rs.10
∴ the cost of painting 144 sq. m wall = Rs. (144 × 10)
∴ the cost of painting 144 sq. m wall = Rs. 1440
The cost of painting the square part of wall is Rs. 1440.
Solution 2(3):
Length of the card paper = 80 cm
Breadth of the card paper = 40 cm
Area of a rectangle = length × breadth
Area of the rectangular card paper = 80 cm × 40 cm
= 3200 sq. cm
Area of a square = length × length
Area of the square to be made from card paper
= length × length
= 10 cm × 10 cm
= 100 sq. cm
Area of one square = 100 sq. cm
Number of squares made from 100 sq. cm card paper
= 1
∴ Number of squares made from 3200 sq. cm card paper = 3200 ÷ 100 = 32
Thus, 32 squares can be made from the card paper.
Solution 2(4):
Length of the dining hall = 15 m
Breadth of the dining hall = 12 cm
Area of a rectangle = length × breadth
Area of the rectangular dining hall = 15 m × 12 m= 180 sq. m
Area of a square = length × length
Area of the square tile = 30 cm × 30 cm = 900 sq. cm
∴ Area of one square tile = 900 sq. cm
But the area of dining hall is in the units of sq. m
Hence, convert the units of the area of the dining hall into sq. cm
Area of the rectangular dining hall = 180 sq. m
1 sq. m = 10,000 sq. cm
∴ 180 sq. m = (180 × 10,000) cm = 18,00,000 sq. cm
Area of one square tile = 900 sq. cm
Number of square tiles needed for 900 sq. cm area = 1
∴ Number of square tiles needed for 18,00,000 sq. cm area = 18,00,000 ÷ 900 = 2,000
2000 tiles are needed for the dining hall.
solution 2(5):
Length of the rectangular cloth = 8 m
Breadth of the rectangular cloth = 6 cm
Area of a rectangle = length × breadth
Area of the rectangular cloth = 8 m × 6 m = 48 sq. m
Area of a square = length × length
Area of the square piece of cloth = 20 cm × 20 cm = 400 sq. cm
∴ Area of one square piece of cloth = 400 sq. cm
But the unit of area of the cloth is sq. m
Hence, convert the area of the rectangular cloth into units of sq. cm
Area of rectangular cloth = 48 sq. m
1 sq. m = 10,000 sq. cm
∴ 48 sq. m = (48 × 10,000) cm = 48,00,000 sq. cm
Area of one square piece of cloth = 400 sq. cm
Number of square pieces formed from 400 sq. cm of the cloth = 1
∴ Number of square pieces formed from 48,00,000 sq. cm of the cloth = 48,00,000 ÷ 400 = 1200
Thus, 1200 square pieces of cloth can be formed.
Practice – 1
Solution 1:
Given,
Length of the drawing paper = 25 cm
Perimeter of a square = 4 × length
Hence,
perimeter of the square drawing paper = 4 × 25 = 100
Therefore, the perimeter of the square drawing paper is 100 cm.
Solution 2:
Given,
Length of the square = 14 cm
Perimeter of a square = 4 × length
Hence, perimeter of the square = 4 × 14 = 56 cm
Therefore, perimeter of the square having length 14 cm is 56 cm.
Solution 3:
Given,
Length of the square piece of cloth = 8 metres
Perimeter of the square piece of cloth = 4 × length
Hence, perimeter of the square = 4 × 8 = 32 metres
Therefore, perimeter of the square piece of the cloth is 32 metres.
Solution 4:
Given,
Length of the square plot = 18 metres
Perimeter of a square = 4 × length
Hence, perimeter of the square plot = 4 × 18= 72 metres
Therefore, perimeter of the square plot is 72 metres.
Solution 5:
Length of the square space = 9 metres (given)
Perimeter of a square = 4 × length
Hence, perimeter of the square space = 4 × 9 = 36 metres
Therefore, perimeter of the square space is 36 metres.
Solution 6:
Given,
Length of the square tile = 10cm
Perimeter of a square = 4 × length
Hence, perimeter of the square tile = 4 × 10 = 40 cm
Therefore, perimeter of the square tile is 40 cm.
Practice – 2
Solution 1(1):
Length of the rectangle = 18 cm
Breadth of the rectangle = 16 cm
Perimeter of a rectangle = 2 (length + breadth)
Hence,
perimeter of the rectangle = 2 (18 + 16) = 2(34) = 68 cm
Therefore, perimeter of the rectangle is 68 cm.
Solution 1(2):
Length of the rectangular wooden piece = 40 cm
Breadth of the rectangular wooden piece = 30 cm
Perimeter of a rectangle = 2 (length + breadth)
Hence,
perimeter of the rectangular wooden piece = 2 (40 + 30)
= 2 (70)
= 140 cm
Therefore, perimeter of the rectangular wooden piece is 140 cm.
Solution 1(3):
Length of the computer lab = 15 m
Breadth of the computer lab = 13 m
Perimeter of a rectangle = 2 (length + breadth)
Hence, perimeter of the rectangular computer lab
= 2 (15 + 13)
= 2 (28) = 56 m
Therefore, perimeter of the computer lab is 56 m.
Solution 1(4):
Length of the playground = 30 m
Breadth of the playground = 25 m
Perimeter of a rectangle = 2 (length + breadth)
Hence, perimeter of the playground = 2 (30 + 25)
= 2 (55)
= 110 m
Therefore, perimeter of the playground is 110 m.
Solution 1(5):
Length of the field = 45 m
Breadth of the field = 35 m
Perimeter of a rectangle = 2 (length + breadth)
Hence, perimeter of the field = 2 (45 + 35)
= 2 (80)
= 160 m
Therefore, perimeter of the field is 160 m.
Solution 1(6):
Length of the rectangular garden = 20 m
Breadth of the rectangular garden = 1700 cm
∴ Breadth of the rectangular garden = 1700 ÷ 100 = 17
∴ Breadth of the rectangular garden = 17 m
Perimeter of a rectangle = 2 (length + breadth)
Hence, perimeter of the field = 2 (20 + 17)
= 2 (37)
= 74 m
Therefore, perimeter of the rectangular garden is 74 m.
Solution 2:
(1) Perimeter of square = 4 × length = 4 × 6 = 24 m
(2) Perimeter of rectangle = 2( length + breadth)
Perimeter of rectangle = 2 ( 7 + 5) = 24 cm
(3) Perimeter of square = 4 × length = 4 × 8 = 32 cm
(4) Perimeter of rectangle = 2( length + breadth)
Perimeter of rectangle = 2 ( 8 + 7) =2 (15) = 30 cm
(5) Perimeter of square = 4 × length = 4 × 12 = 48 m
(6) Perimeter of rectangle = 2( length + breadth)
Perimeter of rectangle = 2 ( 16 + 14) = 2 (30) = 60 cm
(7) Perimeter of square = 4 × length = 4 × 15 = 60 cm
(8) Perimeter of rectangle = 2( length + breadth)
Perimeter of rectangle = 2 ( 14 + 13)= 2 (27) = 54 cm
(9) Perimeter of square = 4 × length = 4 × 17 = 68 m
(10) Perimeter of rectangle = 2( length + breadth)
Perimeter of rectangle = 2 (20 + 10)= 2 (30)= 60 m
Practice – 3
Solution 1:
To find the cost of drawing stripes of lime surrounding the garden, find the perimeter of the square garden. Length of the square garden = 35 m (given)
Perimeter of the square garden = 4 × length
Hence,
perimeter of the square garden = (4 × 35) m = 140 m
Cost of drawing 1 metre of stripes = Rs. 7
∴ Cost of drawing stripes of lime surrounding the square garden = Rs. (140 × 7) = Rs. 980
Therefore, the cost of drawing stripes of lime is Rs. 980.
Solution 2:
To find the cost of fixing coloured tiles on the boundary of the classroom, find the perimeter of the classroom.
Length of the square classroom = 15 m (given)
Perimeter of the square classroom = 4 × length
Hence,
perimeter of the square classroom = (4 × 15) m = 60 m
Cost of fixing coloured tiles for 1 m = Rs. 20
∴ Cost of fixing coloured tiles on the boundary of Rs. (60 × 20) =Rs. 1200
Therefore, the cost of fixing coloured tiles on the boundary is Rs. 1200.
Solution 3:
To find the cost of wire fencing surrounding the rectangular plot, find the perimeter of the plot.
Given,
Length of the rectangular plot = 40 m
Breadth of the rectangular plot = 30 m
Perimeter of the rectangle = 2(length + breadth)
Hence, the perimeter of the rectangular plot = 2(40 + 30) m
= 2 (70) m
= 140 m
Cost of wire fencing for 1 m = Rs. 20
∴ Cost of wire fencing surrounding the given plot
= Rs. (140 × 20)
= Rs. 2800
Therefore, the cost of wire fencing is Rs. 2800.
Solution 4:
To find the cost to cover the rectangular table with a stripe, find the perimeter of the table.
Given,
Length of the rectangular table = 4 m
Breadth of the rectangular table = 3 m
Perimeter of the rectangle = 2(length + breadth)
Hence, perimeter of the rectangular table = 2(4 + 3) m
= 2 (7) m
= 14 m
Cost to cover the table with a stripe for 1 m = Rs. 5
∴ Total Cost to cover the table with a stripe = Rs. (14 × 5)
= Rs. 70
Therefore, the cost to cover the table with a stripe is Rs. 70
Solution 5:
To find the cost of knitting a stripe surrounding the carpet, find the perimeter of the carpet.
Perimeter of the carpet = perimeter of the prayer hall.
Length of the prayer hall = 19 m (given)
Breadth of the prayer hall = 17 m (given)
Perimeter of a rectangle = 2 (length + breadth)
Hence,
perimeter of the rectangular prayer hall
= 2(19 + 17) m = 2 (36) m = 72 m
Cost of knitting 1 m stripe = Rs. 30
∴ Cost of knitting 72 m stripe = Rs. (72 × 30) = Rs. 2160
Therefore, the cost of knitting a stripe surrounding the carpet is Rs. 2160.
Solution 6:
To find the cost of knitting a stripe surrounding the cloth, find the perimeter of the cloth.
Given,
Length of the cloth = 8 m
Breadth of the cloth = 300 cm
But 100 cm = 1 m,
∴ 300 cm = 300÷ 100 = 3 m.
Perimeter of the rectangle = 2(length + breadth)
Hence, perimeter of the rectangular prayer hall
= 2(8 + 3) m = 2 (11) m = 22 m
Cost of knitting 1 m stripe = Rs. 30
∴ Cost of knitting 22 m stripe = Rs. (22 × 30) = Rs. 660
Therefore, the cost of knitting a stripe surrounding the cloth is Rs. 660.
Practice – 4
Solution 1:
Given,
Length of the square = 5 cm
Area of the square = length × length
∴ Area of the square = 5cm × 5 cm= 25 sq. cm
Therefore, the area of the square is 25 sq. cm
Solution 2:
Given,
Length of the square ground = 15
Area of the square = length × length
∴ Area of the ground = 15 m × 15 m = 225 sq. m
Therefore, the area of the square ground is 225 sq. m
Solution 3:
Given,
length of the square piece of cloth = 8 m
Area of the square = length × length
∴ Area of the square piece of cloth = 8 m × 8 m = 64 sq. m
Therefore, the area of the square piece of cloth is 64 sq. m
Solution 4:
Given,
Length of the square plot = 35 m
Area of the square = length × length
∴ Area of the square plot = 35 m × 35 m = 1225 sq. m
Therefore, the area of the square plot is 1225 sq. m
Solution 5:
Given,
Length of the square handkerchief = 20 cm
Area of the square = length × length
∴ Area of the square handkerchief = 20 cm × 20 cm
Area of the square handkerchief = 400 sq. cm
Therefore, the area of the square plot is 400 sq. cm
Practice – 5
Solution 1(1):
Given,
Length of the rectangle = 19 cm
Breadth of the rectangle = 17 cm
Area of the rectangle = length × breadth
∴ Area of the given rectangle = 19 cm × 17 cm= 323 sq. cm
Therefore, the area of the rectangle is 323 sq. cm
Solution 1(2):
Given,
Length of the top of the table = 90 cm
Breadth of the top of the table = 70 cm
Area of the rectangle = length × breadth
∴ Area of the top of the table = 90 cm × 70 cm
= 6300 sq. cm
Therefore, the area of the rectangle top of the table is 6300 sq. cm
Solution 1(3):
Length of the ground = 55 m (given)
Breadth of the ground = 50 m (given)
Area of the rectangle = length × breadth
∴ Area of the ground = 55 m × 50 m = 2750 sq. m
Therefore, the area of the rectangle ground = 2750 sq. m
Solution 1(4):
Given,
Length of the rectangular floor of the library = 10 m
Breadth of the floor of the library = 4 m
Area of the rectangle = length × breadth
∴ Area of the floor of the library = 10 m × 4 m = 40 sq. m
Therefore, the area of the rectangle floor of the library = 40 sq. m
Solution 1(5):
Given,
Length of the rectangular classroom = 12 m
Breadth of the classroom = 8 m
Area of a rectangle = length × breadth
∴ Area of the classroom = 12 m × 8 m = 96 sq. m
Therefore, the area of the classroom = 96 sq. m
Solution 2:
(1) Area of square = length × length
= 11 m × 11m
= 121 sq. m
(2) Area of rectangle = length × breadth
= 14 m × 12 m
= 168 sq. m
(3) Area of square = length × length
= 15 m × 15 m
= 225 sq. m
(4) Area of rectangle = length × breadth
= 26 cm × 24 cm
= 624 sq. cm
(5) Area of square = length × length
= 12 m × 12m
= 144 sq. m
(6) Area of rectangle = length × breadth
= 17 m × 16 m
= 272 sq. m
(7) Area of square = length × length
= 13 cm × 13 cm
= 169 sq. cm
(8) Area of rectangle = length × breadth
= 21 cm × 19 c m
= 399 sq. cm
Practice – 6
Solution 1:
1 m = 100 cm
1 sq.m = 10,000 sq. cm
∴ 5 sq.m = 50,000 sq. cm
(ii) 20 sq. m
1 m = 100 cm
1 sq. m = 10,000 sq. cm
∴ 20 sq. m = 2,00,000 sq. cm
(iii) 30 sq. m
1 m = 100 cm
1 sq. m = 10,000 sq. cm
∴ 30 sq.m = 3,00,000 sq. cm
Solution 2:
Practice – 7
Solution 1:
To find the cost of leveling the school ground, find its area.
Length of the school ground = 25 m (given)
Breadth of the school ground = 20 m (given)
Area of a rectangle = length × breadth
∴ Area of the school ground = 25 m × 20 m = 500 sq m
Cost of leveling 1 sq. m area = Rs. 9
Cost of leveling 500 sq. m area = Rs. (500 × 9) = Rs. 4500
Therefore, the cost of leveling the school ground is Rs. 4500.
Solution 2:
To find the cost of dining cloth on the table, find the area of the table.
Length of the square dining table = 4 m (given)
Area of a square = length × length
∴ Area of the given dining table = 4 m × 4 m= 16 sq. m
Cost of preparing 1 sq. m of dining cloth = Rs. 30 per sq. m
Cost of preparing 16 sq. m of dining cloth = Rs. (16 × 30)
= Rs. 480
Therefore, the cost of preparing the dining cloth on the dining table is Rs. 480.
Solution 3:
To find the cost of fixing tiles on the floor, find the area of the swimming pool.
Length of the swimming pool = 16 m
Breadth of the swimming pool = 4 m
Area of a rectangle = length × breadth
∴ Area of the swimming pool = 16 m × 4 m = 64 sq. m
Cost of fixing tiles on the floor for 1 sq. m area = Rs. 22
Cost of fixing tiles for 64 sq. m area = Rs. (64 × 22)
= Rs. 1408
Therefore, the cost of fixing tiles on the floor of the swimming pool is Rs. 1408.
Solution 4:
To find the number of boxes which can be formed in the card paper, first find the area of the card paper.
Length of the card paper = 60 cm (given)
Breadth of the card paper = 40 cm (given)
Area of a rectangle = length × breadth
∴ Area of the card paper = 60 cm × 40 cm
= 2400 sq. cm
Now, find the area of one square box.
Area of a square = length × length
Length of the square box = 5 cm
∴ Area of one square box = 5 cm × 5 cm = 25 sq. cm
One square box covers 25 sq. cm of the card paper.
∴ Number of boxes to cover 2400 sq. cm of area
= 2400 ÷ 25
∴ Number of boxes formed = 96
Therefore, 96 boxes can be formed in the card paper.
Solution 5:
First, we need to find the area of the floor of the prayer hall.
Area of a rectangle = length × breadth
∴ Area of the rectangular prayer hall = 14 m × 11m
∴ Area of the rectangular prayer hall = 154 sq. m
Now, let us find the area of one square tile.
Length of the square tile = 50 cm.
But the units of length of the hall are given in metre.
Hence, let us convert the area of the prayer hall in sq. cm
1 sq. m = 10,000 sq. cm
∴ 154 sq. m = 154 × 10,000 sq. cm
∴ Area of the prayer hall = 15,40,000 sq. cm
Area of the square tile = length × length
Area of the square tile = 50 × 50 = 2500 sq. cm
One square tile covers 2500 sq. cm of the prayer hall floor.
∴ Number of tiles needed to cover 15,40,000 sq. cm of area = 15,40,000 ÷ 25
Therefore, 616 tiles are needed.