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Frank Learning Maths Class 6 Solutions Chapter 10 Perimeter and Area
Welcome to NCTB Solution. Here with this post we are going to help 6th class students for the Solutions of Frank Learning Maths Class 6 Book, Chapter 10 Perimeter and Area. Here students can easily find step by step solutions of all the problems for Perimeter and Area. Here students will find solutions for Exercise 10.1, 10.2 and 10.3 and also get solutions for Chapter Check-up questions.
Exercise wise proper solutions. All the problem are solved with easily understandable methods so that all the students can understand easily. Here in this post all Mathematic solutions are based on the CBSE latest curriculum.
Perimeter and Area Exercise 10.1 Solution
Question no – (1)
Solution :
(a) Length = 14.8 metre,
Breadth = 5.5 metre.
As we know that,
Perimeter of rectangle = 2 (length + breadth)
Now, Perimeter of the rectangle,
= (14.8 + 14.8 + 5.5 + 5.5) cm
= 40.6m
Thus, the perimeter of the rectangle is 40.6 metre.
(b) Length = 16 m 50 cm,
Breadth = 11 m 25 cm
First, 16 m 50 cm = 1650 cm,
= 11 m 25 cm = 1125 cm
As we know that,
Perimeter of rectangle = 2 (length + breadth)
∴ Perimeter,
= 2 × (1650 + 1125)
= 5550 cm
Therefore, the perimeter of the rectangle is 5550 cm.
Question no – (2)
Solution :
(a) Square having length of each side as 5.6 m.
As we know that,
Perimeter of square = 4 × side
Now, the Perimeter,
= (5.6 × 4)
= 22.4 m
Thus, the perimeter of the square is 22.4 m.
(b) Square having length of each side as 24.5 cm.
As we know that,
Perimeter of square = 4 × side
∴ Perimeter,
= (24.5 × 4) cm
= 98 cm
Thus, the perimeter of a square is 98 cm.
Question no – (3)
Solution :
As per the given question,
A man takes two rounds of a regular pentagon of side 125 m.
∴ The perimeter of the pentagon,
= 125 × 5
= 625 m
∴ Distance covered by the man,
= 625 × 2
= 1250 metre.
Hence, the distance covered by him is 1250 metre.
Question no – (4)
Solution :
(a) Triangle having sides 10 cm, 12 cm and 15 cm
As we know that,
Perimeter of triangle = Side + Side + Side (Sum of 3 sides)
∴ Perimeter of triangle
= (10 + 12 + 15)
= 37 cm
Thus, the the perimeter of the triangle is 37 cm.
(b) Triangle having each side as 14 cm.
As we know that,
Perimeter of triangle = Side + Side + Side (Sum of 3 sides)
∴ Perimeter of triangle
= (14 × 3)
= 42 cm
Thus, the perimeter of triangle is 42 cm.
(c) Isosceles triangle having equal sides as 40 cm and third side as 20 cm.
As we know that,
Perimeter of triangle = Side + Side + Side (Sum of 3 sides)
∴ Perimeter of triangle
= (40 + 40 + 20)
= 100 cm
Therefore, the perimeter of isosceles triangle is 100 cm.
Question no – (5)
Solution :
According to the question,
Rectangular lawn 10 m long and 4 m wide.
leaving a gap of = 1 m
Now, (4 + 4 + 10 – 1) (according to the question)
= 17 m
Thus, the length of fencing is 17 m.
Question no – (6)
Solution :
As per the question,
Side of a square whose perimeter is = 124 cm.
∴ Length of side,
= 124/4
= 31 cm
Hence, the the side of square is 31 cm.
Question no – (7)
Solution :
According to the question,
∴ Side of square field,
= 60/4
= 15 m
∴ Side of pentagon field,
= 60/5
= 12 m
∴ Side of triangle field,
= 60/3
= 20 m
Question no – (9)
Solution :
Given in the question,
The perimeter of the design is = 28 cm.
Let, the length of each side is x cm
∴ 14x = 28
∴ x = 2 cm
Therefore, the length of each side of the hexagon is 2 cm.
Question no – (10)
Solution :
Length of each side = 25 m ….(according to the question)
∴ Length of 3 sides,
= (25 × 3)
= 75 m
∴ Total cost,
= (75 × 10)
= 750 rupees
Therefore, the total cost will be 750 rupees.
Question no – (11)
Solution :
According to the question,
The cost of fencing the ground is 5000 rupees at the rate of 50 rupees per metre
Now the perimeter,
= 5000/50
= 100 m
∴ Length
= 100 × 4/5
= 80 m
∴ Breadth,
= 100 × 1/5
= 20 m
Question no – (12)
Solution :
The perimeter of the square park,
= (50 × 40)
= 200 m
∴ Parul covers,
= (200 × 2)
= 400 m
The perimeter of the rectangular park,
= 2(100 + 75)
= 350 m
∴ Hiten covers,
= (350 × 3)
= 1050 m
Therefore, Parul covers less distance.
Question no – (13)
Solution :
According to the question,
Each side of the square field = 20 m
∴ Perimeter of the square field,
= (20 × 4)
= 80 m
∴ Length of the rectangular field,
= 80/2 × 3/4
= 30 m
∴ Breadth of the rectangular field,
= 80/2 × 1/4
= 10 m
Question no – (14)
Solution :
As per the question
The perimeter of a regular heptagon is 91 cm.
∴ Each side of the heptagon,
= 91/7
= 13 cm
Thus, the length of its one side is 13 cm.
Question no – (15)
Solution :
From the question we get,
Triangle having perimeter as 90 cm and the lengths of two sides as 9 cm and 40 cm.
∴ The third side,
= {90 – (9 + 40)} cm
= {90 – 49}
= 41 cm
Therefore, the third side of the triangle is 41 cm.
Question no – (16)
Solution :
Required Figure :
∴ Perimeter,
= (4 × 3 × 2) cm
= 24 cm
Required Figure :
∴ Perimeter,
= (5 × 2 × 4) cm
= 40 cm
Required Figure :
∴ Perimeter,
= (18 + 18 + 2 + 2)
= 40 cm
Question no – (17)
Solution :
As per the given question,
Geeta walks 5 times around a square field along its edge.
She walks a total distance of = 2.2 km
Now,
∴ Walking 5 times Gita covers the distance of 2.2 km,
= 2200 m
∴ Walking 1 times Gita covers the distance of,
= 2200/5
= 440 m
∴ Side of the field
= 440/4
= 110 m
Thus, the length of the side of the field is 110 m.
Question no – (18)
Solution :
Let, the breadth be x km
Length = 2x km
= 2x + 2x + x + x = 6
= 6x = 6
= x = 1 km
∴ Breadth = 1 km,
∴ Length = 2 km
Question no – (19)
Solution :
Required Figure :
∴ Perimeter,
= (4 + 2 + 1 + 1 + 1 + 2 + 4) + (4 + 2 + 1 + 1 + 1 + 2 + 4) + (4 + 2 + 1 + 1 + 1 + 2 + 4)
= 3 × (4 + 2 + 1 + 1 + 1 + 2 + 4)
= 3 × 15
= 45 units
Thus, the perimeter of the figure will be 45 units.
Question no – (20)
Solution :
Perimeter of each display,
= (1.5 + 1.5 + 1 + 1)
= 5 m
∴ The number of displays those can be framed,
= 100/5
= 20
Remaining display
= (24 – 20)
= 4
∴ Required Length of strip,
= (4 × 5)
= 20 m
Try These 9 (ii) :
Question no – (1)
Solution :
As per the question we know,
A rectangular field having length 85 m and breadth 65 m.
As we know that,
Perimeter of rectangle = 2 (length + breadth)
∴ Perimeter,
= 2 (85 + 65)
= 300 m
Hence, the perimeter of a rectangular field is 300 m.
Question no – (2)
Solution :
As per the question,
Perimeter is 90 cm and sides are in the ratio of 5 : 4.
Since, perimeter = 90 cm
∴ Length + breadth = 45 cm
∴ Length = 45 × 5/(5+ 4)
= 45 × 5/9
= 25 cm
Therefore, the length of the rectangle will be 25 cm
Question no – (3)
Solution :
According to the question,
Length = 10 m
Breadth = 7 m,
Cost of fencing = Rs 25 per metre
∴ Perimeter,
= (10 + 10 + 7 + 7)
= 34 m
∴ Total Cost of fencing,
= (34 × 25)
= 850 rupees.
Perimeter and Area Exercise 10.2 Solution
Question no – (1)
Solution :
Table – (a)
Rectangle | Length (m) | Breadth (m) | Area (sq m) |
(i) | 10 | 8 | 80 |
(ii) | 40 | 10 | 400 |
(iii) | 80 | 20 | 1600 |
Table – (b) :
Square | Side (m) | Area (sq m) |
(i) | 25 | 625 |
(ii) | 3.7 | 13.69 |
(iii) | 0.35 | 0.1225 |
Question no – (2)
Solution :
(a) length = 14 m 25 cm, Breadth = 2 m 30 cm
As we know that,
Area of rectangle = (Length x Breadth) square units.
∴ The area of rectangle,
= (14.25 × 2.3) m^{2}
= 32.775 m^{2}
Thus, the area of rectangle is 32.775 m^{2}
(b) Length = 10 m 10 cm, Breadth = 3 m 30 cm
As we know that,
Area of rectangle = (Length x Breadth) square units.
∴ The area of rectangle,
= (10.10 × 3.30) m^{2}
= 33.33 m^{2}
Hence, the area of rectangle is 33.33 m^{2}
(c) Length = 8 m 45 cm, Breadth = 6 m 25 cm
As we know that,
Area of rectangle = (Length x Breadth) square units.
∴ Area of rectangle,
= (8.45 × 6.25) m^{2}
= 52.8125 m^{2}
Therefore, the area of rectangle is 52.8125 m^{2}
Question no – (3)
Solution :
(a) Let Length = L unit, breadth B unit
∴ A = BL square unit
Now,
= A^{1} = (2B) × (2L) = 4BL square unit
∴ 4A
∴ The new area becomes four times the original one.
(b) Let, Length = L unit, breadth B unit
= A = BL,
= A^{1} = (L/2) × (B/2)
= BL/4
= A/4
Therefore, the Area becomes one-fourth.
(c) Let, Length = L unit, breadth B unit
∴ A = BL,
= A^{1 }= (3B) × (3L)
= 9BL = 9A
Hence, the Area becomes 9 times.
(d) Let, Length = L unit, breadth B unit
∴ A = BL,
= A^{1} = (2L) × (B/2)
= LB = A
Therefore, Area does not change.
Question no – (4)
Solution :
In the given question,
Breadth is 48 cm
The area is 1728 sq. cm.
Now, 48 × Length
= 1728
∴ Length,
= 1728/48
= 36 cm
Thus, the length of a rectangle is 36 cm.
Question no – (5)
Solution :
The perimeter of the square = 48 cm …(according to the question)
∴ The side of the square
= 48/4
= 12 cm
Therefore, the Area will be 144 cm^{2}
Question no – (6)
Solution :
From the question we get,
A rectangle has an area equal to that of a square of side = 35 cm.
The breadth of the rectangle is = 25 cm,
∴ Area of the square,
= 35 × 35
= 1225 cm^{2}
∴ Length of the rectangle,
= 1225/25
= 49 cm
Hence, the length of the rectangle is 49 cm.
Question no – (7)
Solution :
As per the given question,
The length of a rectangle is 80 m and the breadth is 3/4 of its length.
Length = 80 m,
∴ Breadth = (80 × 3/4)
= 60 m
∴ Area = (80 × 60)
= 4800 m^{2}
Therefore, the area of the rectangle will be 4800 m^{2}
Question no – (8)
Solution :
The perimeter of the rectangular plot,
= 2 (160 + 82)
= 2 × 242
= 484 m
∴ Side of the square,
= 484/4
= 121 m
∴ Area of the square,
= 121 × 121
= 14641 m^{2}
Area of the rectangular plot,
= 160 × 82
= 13120
∴ 14641 – 13120
= 1521 m^{2}
Therefore, the area of the square will greater by 1621 m^{2}
Question no – (9)
Solution :
According to given question,
Rectangular plot of land 500 m long and 200 m wide
∴ Area of the land
= (500 × 200)
= 1090000 m^{2}
∴ The total cost,
= 100000 × 8
= 800000 rupees
Thus, the cost of tiling will be 800000 rupees.
Question no – (10)
Solution :
As per the question,
A blackboard is 2 m 25 cm long and 1 m 15 cm broad.
∴ Area of the blackboard,
= (2.25 × 1.15)
= 25875 m^{2}
∴ Total cost,
= (2.5875 × 4/100) rupees
= 0.1035 rupees
Therefore, the cost of painting the blackboard 0.1035 rupees.
Question no – (11)
Solution :
From the given data,
A road 90 m long and 10 m wide is to be paved with bricks having length and breadth as 20 cm and 7.5 cm respectively
∴ The area of the road = (9000 × 1000) cm^{2}
The area of each brick = (20 × 7.5) cm^{2}
∴ No of bricks needed
= 9000 × 1000 × 10/20 × 7.5
= 60000
Hence, 60000 bricks will be required to pave the road.
Question no – (12)
Solution :
As per the given question,
A room is 14 m long and 12 m wide.
A square carpet of side 10 m is to be laid on its floor.
∴ The area of the floor,
= (14 × 12) m^{2}
= 168 m^{2}
∴ The area of the carpet,
= (10 × 12) m^{2}
= 168 m^{2}
∴ Area of floor, which is not carpeted is,
= (168 – 100)
= 68 m^{2}
Thus, the area of the floor which is not carpeted will be 68 m^{2}
Question no – (13)
Solution :
From the question we get,
A paper sheet is 300 cm long and 150 cm wide.
Envelopes from small sheets measuring 10 cm and 30 cm.
∴ Total numbers of envelopes,
= 300 × 150/30 × 10
= 150
Thus, 150 envelopes can be made from the big sheet.
Question no – (14)
Solution :
Let, the courtyard be a square
∴ Then the length of a side
= 20 × 9/10
= 18 m
∴ The original area
= 18 × 18
= 324 m^{2}
Thus, the correct area of the courtyard is 324 m^{2}
Question no – (16)
Solution :
First, Area of half page,
= 1/2 (15 × 24)
= 180 cm^{2}
Now, The cost of 10 cm^{2} is = 300 rupees
The cost of 1 cm2 is = 30 rupees
∴ The cost of 18 cm2 is,
= 30 × 18
= 540 rupees
Thus, the company has to pay 540 rupees for it.
Question no – (17)
Solution :
In the given question,
Perimeter of a square garden is = 48 m.
A small flower bed covers = 18 sq m area
(a) The area of the garden that is not covered by the flower bed,
The required are,
= (48 – 18)
= 30m²
(b) Fractional part of the garden is covered by flower bed,
= 18/48
= 3/8
(c) The ratio of the area covered by the flower bed and the remaining area,
= 18 : 30
= 3 : 5
Question no – (18)
Solution :
First, The area of the room,
= 15 × 10
= 150 m^{2}
Now, The area of the room covered by blue tiles = 75 m^{2}
∴ No of blue tiles required,
= 75 × 100 × 100/15 × 10
= 5000
The area of the room covered by white tiles = 75 m^{2}
∴ No of white tiles required,
= 75 × 100 × 100/20 × 15
= 2500
Question no – (19)
Solution :
From the question we get,
A wire is cut into several small pieces.
Each of the small pieces is bent into a square of side = 2 cm.
The total area of the small squares is = 28 square cm
Area of each squares
= 2 × 2
= 4 cm^{2}
No of squares,
= 28/4
= 7
∴ Perimeter of each squares,
= 4 × 2
= 8 cm
∴ Perimeter of 7 square,
= 8 × 7
= 56 cm
Therefore, the length of the wire is 56 cm.
Question no – (20)
Solution :
In the question we get,
Tiles length and breadth are = 12 cm and 5 cm
Now, (a) 100 cm and 144 cm
∴ No of tiles,
= 100 × 144/12 × 5
= 240
(b) 70 cm and 36 cm.
∴ No of tiles,
= 70 × 36/12 × 5
= 42
Perimeter and Area Exercise 10.3 Solution
Question no – (6)
Solution :
Given, Length = 15 cm
And, Breadth = 12 cm
(a) Will the perimeter of the sheet increase or decrease?
Perimeter (Before)
= 2 (15 + 12)
= 54 cm
Perimeter = (now)
= 12 + 15 + 12 + 15 + 5 + 5
= 64 cm
∴ Perimeter will increase
(b) Will the area of the sheet increase or decrease?
= Area (before)
= 12 × 15 = 180 cm^{2}
Area (Now)
= 180 – 25
= 155cm^{2}
∴ Area will decrease
Question no – (7)
Solution :
Perimeter |
Area | |
Figure – (a) | 18 | 14 |
Figure -(b) | 28 | 13 |
Figure -(c) | 27 | 13 |
Question no – (8)
Solution :
Figure – (a)
= (10 × 2) + 910 × 2)
= 40 square unit
Figure – (b)
= (7 × 7) × 5
= 245 square units
Figure – (c)
= (4 × 1) + (5 × 1)
= 9 square units
Perimeter and Area Chapter Check-up Solution :
Question no – (1)
Solution :
As per the given question,
The perimeter of a rectangular park is 370 m if its length is 170 m.
∴ 2 (170 + breath) = 370 m
or, 9170 + breath) = 370 ÷ 2 = 185
or, breath = 185 – 170 = 15 m
∴ The area of the park,
= 170 × 15
= 2550 m^{2}
Therefore, the area of the park will be 2550 m^{2}
Question no – (2)
Solution :
According to the question,
Perimeter of the square park
= 200 m
∴ Side of the square ark,
= 200 ÷ 4
= 50 m^{2}
∴ Area of the square,
= 50 × 50
= 2500 m^{2}
Thus, the area of the square will be 2500 m^{2}
Question no – (3)
Solution :
From the question we get,
Rectangular park of length 175 m and breadth 125 m at the rate of 12 per meter.
∴ Perimeter,
= 2 (175 + 125)
= 2 × 300
= 600 m
∴ Cost,
= 600 × 12
= 7200 rupees
Thus, the cost of fencing a rectangular will be 7200 rupees.
Question no – (4)
Solution :
(a) What is the perimeter of his arrangement [Fig (i)]?
= Perimeter,
= 1/2 × 3 × 4
= 6 m
(b) Shari does not like his arrangement. She gets him to lay them out like a cross. What is the perimeter of her arrangement [Fig (ii)]?
= 1/2 × 5 × 4
= 10 m
(c) Which has greater perimeter?
= The fraction of cross has greater perimeter.
(d) Avneet wonders if there is a way of getting an even greater perimeter. Can you find a way of doing this?
= No there is no such arrangement.
Question no – (5)
Solution :
First, The total area,
= 5 × 4
= 20 m^{2}
∴ Area of flower beds,
= 1 × 1 × 5
= 5 m^{2}
∴ Remaining Area,
= 20 – 5
= 15 m^{2}
Hence, the area of the remaining part of the land is 15 m^{2}
Question no – (6)
Solution :
Figure – (a)
= (2 × 2) + (4 × 3)
= 4 + 12
= 16m^{2}
Figure – (b)
= {(3 + 2 + 3) × (2)} + (2 × 2) + (2 × 1)
= 16 + 4 + 2
= 22 m^{2}
Figure – (c)
= (3 × 3) + (2 × 2) + (5 × 3)
= 9 + 4 + 15
= 28 m^{2}
Question no – (7)
Solution :
In the question we get,
Length of a rectangular field is 250 m and width is 150 m.
Anuradha runs around this field 3 times
The perimeter of the field,
= 2 (250 + 150)
= 800 m
∴ Running 3 rounds, Anuradha covers,
= (800 × 3)
= 2400 m
∴ To cover distance of 4 km she has to run,
= 4000/800
= 5 rounds
Question no – (8)
Solution :
Given in the question,
A rectangular path of 60 m length and 3 m width
Covered by square tiles of side = 25 cm
∴ The number of tiles along width,
= 3 × 100/25
= 12
∴ The number of rows,
= 60 × 100/25
= 240
∴ The number of tiles,
= 12 × 240
= 2880
Thus, 2880 tiles is used to make this path.
Question no – (9)
Solution :
In the question,
Square slabs each with side = 90 cm
Floor of area = 81 sq m
∴ The number of square slabs,
= 81 × 100 × 100/90 × 90
= 100
Hence, 100 square slab will be needed.
Question no – (10)
Solution :
(a) the perimeter of shaded portion to the perimeter of the whole design.
= Perimeter of the shaded region,
= (60 × 4) + (40 × 6)
= 240 + 240
= 480 cm
Perimeter of the whole region,
= (60 × 8) (40 × 10)
= 480 + 400
= 880 cm
Their ratio,
= 480 : 880
= 6 : 11
(b) the area of the shaded portion to the area of the unshaded portion.
= Ratio of area of the shaded and unshaded region = (No of shaded tiles) : (no of unshaded tiles)
= 6 : 14
= 3 : 7
Question no – (11)
Solution :
(a) shaded portion I to shaded portion II?
= (5 × 5) : {(5 × 5) + (5 × 3)}
= 25 : 40
= 5 : 8
(b) shaded portion II to shaded portion III?
= {(5 × 5) + (5 × 3)} : (5 × 7)
= 40 : 35
= 8 : 7
(c) shaded portions I and II taken together and shaded portion III?
= [(5 × 5) + {(5 × 5) + (5 × 3)}] : (5 × 7)
= [25 + 40] : 35
= 65 : 35
= 13 : 7
Question no – (12)
Solution :
In the question,
The floor of a hall is completely covered by 25 carpets,
Each measure = 40 m × 2.5 m.
∴ The area of the floor,
= (40 × 2.5 × 25)
= 100 × 25
= 2500 m^{2}
Thus, the area of the floor will be 2500 m^{2}
Question no – (13)
Solution :
(a) side is doubled
Let, side be a and area be A
= A = a^{2}
And, A^{1} = 2a × 2a^{2}
= 4a^{2}
(b) side is halved
Now, A = a^{2}, A^{1} = a/2 × a/2 = a^{2}/4
= A : A^{1} a^{2} : a2/4
= 4 : 1
Question no – (14)
Solution :
Given in the question,
Tiles measuring 2 m by 1 m
Rectangular hall 12 m long and 8 m broad
∴ No of tiles,
= 12 × 18/2 × 1
= 108
∴ Cost of tiles,
= 108 × 250
= 27000 rupees
Therefore, the cost of tiles at the rate of 250 per ten tiles will be 27000 rupees.
Precious Chapter Solution :