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Frank Learning Maths Class 5 Solutions Chapter 11 Perimeter Area and Volume
Welcome to NCTB Solution. Here with this post we are going to help 5th class students for the Solutions of Frank Learning Maths Class 5 Book, Chapter 11 Perimeter Area and Volume. Here students can easily find step by step solutions of all the problems for Perimeter Area and Volume. Here students will find solutions for Exercise 11.1, 11.2, 11.3, 11.4, 11.5 and 11.6. Exercise wise proper solutions for every problems. 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 Area and Volume Exercise 11.1 Solution
Question no – (1)
Solution :
Figure – (a)
Given measures = 18, 30, 17, 22 m
∴ Perimeter,
= (18 + 30 + 17 + 22)
= 87 m
Figure – (b)
Given measures = 1.2, 80, 1.2, 80 m
∴ Perimeter,
= (1.2 + 80 + 1.2 + 80)
= 120 + 80 + 120 + 80
= 400 cm
Figure – (c)
Given measures = 8.5, 7, 8.5, 7 cm
∴ Perimeter,
= (8.5 + 7 + 8.5 + 7)
= 31 cm
Figure – (d)
Given measures = 4.5, 3.2, 3.2, 5.4, 4 cm
∴ Perimeter,
= (4.5 + 3.2 + 3.2 + 5.4 + 4)
= 21.2 cm
Figure – (e)
Given measures = 4.8, 4.8, 4.8, 4.8, 4.8, 4.8 cm
∴ Perimeter,
= (4.8 + 4.8 + 4.8 + 4.8 + 4.8 + 4.8)
= 28.8 cm
Figure – (f)
Given measures = 3, 1, 2, 3, 2, 1, 3, 1, 2, 3, 2, 1 m
∴ Perimeter,
= 3 + 1 + 2 + 3 + 2 + 1 + 3 + 1 + 2 + 3 + 2 + 1
= 24 m
Question no – (2)
Solution :
Dear students, as we know that,
Perimeter of rectangle = 2 (length + breadth)
So, now,
(a) Given,
Length = 12 mm,
Breadth = 10 mm
∴ Perimeter,
= 2 (12 + 10)
= 2 × 22
= 44 cm
So, the perimeter is 44 cm
(b) Given,
Length = 9.5 cm,
Breadth = 6 cm
∴ Perimeter,
= 2 (9.5 + 6)
= 2 × 15.5
= 31 cm
So, the perimeter is 31 cm
(c) Given,
Length = 7 m,
Breadth = 10.5 m
∴ Perimeter,
= 2 (7 + 10.5)
= 2 × 17.5
= 35 m
So, the perimeter is 35 m.
(d) Given,
Length = 6.2 cm,
Breadth = 4.3 cm
∴ Perimeter,
= 2 (6.2 + 4.3)
= 2 × 10.5
= 21 cm
So, the perimeter is 21 cm.
Question no – (3)
Solution :
Dear students as we know that,
Perimeter of square = 4 × side
(a) Given side = 8 cm
∴ Perimeter,
= 4 × 8
= 32 cm
So, perimeter is 32 cm
(b) Given side = 15 mm
∴ Perimeter,
= 4 × 15
= 60 mm
So, perimeter is 60 mm
(c) Given side = 4.3 m
∴ Perimeter,
= 4 × 4.3
= 17.2 m
So, perimeter is 17.2 m
(d) Given side = 8.5 cm
∴ Perimeter,
= 4 × 8.5
= 34 cm
So, perimeter is 34 cm.
Question no – (4)
Solution :
(a) In the given question,
The length of rectangular field = 110 m
The breadth of rectangular field = 85 m
First, Perimeter of the field,
= 2 (110 + 85)
= 2 × 195
= 390 m
∴ Total cost,
= 25 × 390
= 9750 Rs.
So, the cost to fence the field is Rs. 9750
(b) In the given question,
The side of a square is = 11 cm
Hence, Perimeter,
= 4 × 11 ….(Perimeter of square = 4 × side)
= 44 cm
∴ The length become,
= 3 × 44
= 132 cm
So, 132 cm length will you cover.
(c) Given
Lengths are = 10 cm, 12 cm and 25 cm.
Perimeter,
= (10 + 12 + 25)
= 47 cm
So, the perimeter of a triangle is 47 cm.
(d) In the figure,
Edge is = 25 cm
∴ Perimeter of the table,
= 6 × 25
= 150 cm
So, he needs a 150 cm strip.
(e) Given,
Equilateral triangle side = 4 cm
Then perimeter,
= 3 × 4
= 12 cm
So, the perimeter of an equilateral triangle is 12 cm.
(f) Given,
Pentagon each side = 3 inches.
∴ Perimeter,
= 3 × 5
= 15 inches
So, the perimeter of a regular pentagon is 15 inches.
(g) We know hexagon has = 6 side
Given hexagon is perimeter = 42 cm
∴ Each side
= 42/6
= 7 cm
So, the length of each side will be 7 cm.
Question number – (5)
Solution :
Let,
AF = 8, FE = 13
ED = 5, CD = 6
∴ BC = (8 – 5) = 3 cm
AB = (13 – 6) = 7
∴ Perimeter of the plot
= 8 + 13 + 5 + 6 + 3 + 7
= 42 cm
So, 42 cm wire needed for fencing it.
Perimeter Area and Volume Exercise 11.3 Solution
Question no – (1)
Solution :
Dear students, as we know that,
Area of a Square = Side × Side
(a) Given,
Side = 7 mm
∴ Area = 7 × 7
= 49 mm^{2}
(b) Given,
Side = 13 cm
∴ Area = 13 × 13
= 169 cm^{2}
(c) Given,
Side = 4.5 cm
∴ Area = 4.5 × 4.5
= 20.25 cm^{2}
(d) Given,
Side = 17 cm
∴ Area = 17 × 17
= 289 cm^{2}
(e) Given,
Side = 5.2 m
∴ Area = 5.2 × 5.2
= 27.04 cm^{2}
(f) Given,
Side = 24 m
∴ Area = 24 × 24
= 576 m^{2}
Question no – (2)
Solution :
Dear students, as we know that,
Area of rectangle = Length × Breadth
So now,
(a) l = 7 mm, b = 4 mm
∴ Area = 7 × 4
= 28 mm^{2}
(b) l = 3.5 cm, b = 1.5 cm
∴ Area = 3.5 × 1.5
= 5.25 cm^{2}
(c) l = 12 cm , b = 9 cm
∴ Area = 12 × 9
= 108 cm^{2}
(d) l = 19 m, b = 14 m
∴ Area = 19 × 14
= 266 m^{2}
(e) l = 4.8 m, b = 3.2 m
∴ Area = 4.8 × 3.2
= 15.36 m^{2}
(f) l = 35 m, b = 12 m
∴ Area = 35 × 12
= 420 m^{2}
Question no – (3)
Solution :
Figure – (a)
It’s a square,
So, area = 1.5 × 1.5
= 2.25 cm^{2}
Figure – (b)
It’s a rectangle
So, Area = 3.2 × 7
= 22.4 cm^{2}
Figure – (c)
Area of ABCD = 5 × 3 = 15
Area of DEFE = 12 × 8 = 96
———————————————-
Total area = 111 cm^{2}
Figure – (d)
Area of ABCD = 7 × 4 = 28
Area of CEGF = 3 × 10 = 30
———————————————-
Total area = 58 cm^{2}
Figure – (e)
= Area of CDEF = 1 × 15 = 15
= Area of ABGH = 7 × 6 = 42
= 75 cm^{2}
Question no – (4)
Solution :
Length | Breadth | Area | Perimeter | |
(a) | 3 cm | 18 sq cm | 2 (6 + 3) = 18 | |
(b) | 9 cm | 4 cm | 36 | 2 (9 + 4) = 26 |
(c) | 120 cm | 3600 sq cm | 2 (120 + 30) = 300 | |
(d) | 15 | 5 cm | 75 | 2 (15 + 5) = 40 |
(e) | 9 cm | 6 cm | 54 | 2 (9 + 6) = 30 |
(f) | 2 m 20 cm | 1 m
6 cm |
35,200 | 2 (220 + 160) = 760 sqm |
Question no – (5)
Solution :
Side | Perimeter | Area | |
(a) | 2 mm | Mm | 4 = sq mm |
(b) | 8 | 32 mm | 64 sq mm |
(c) | 6 | 24 cm | 36 sq cm |
(d) | 25 cm | 100 cm | 625 sq cm |
(e) | 15 cm | 60 cm | 15 × 15 = 225 sq mm |
(f) | 100/10 = 10 | 4 × 10 = 40 cm | 100 sq cm |
Perimeter Area and Volume Exercise 11.4 Solution
Question number – (1)
Solution :
In the question,
Side length of a square play area is = 5.2 m
As we know that,
Area of a Square = Side × Side
∴ Area of the square play area,
= 5.2 × 5.2 sq m
= 27.04 sqm
So, the area is 27.04 sqm.
Question number – (2)
Solution :
In the question,
wall 9 m long and 7 m wide
So here,
Area = Length × Breadth
∴ Area of the wall,
= 9 × 7
= 63 sqm
∴ Total cost for painting,
= 63 × 4.75
= 299.25 Rs.
So, the cost of painting is Rs. 299.25
Question number – (3)
Solution :
In the question we get,
Floor of a room 10 m long and 6 m wide
So here,
Area = Length × Breadth
∴ Area of the Floor,
= 10 × 6
= 60 = sqm
∴ Cost for carpeting,
= 35 × 60
= 2100 Rs.
So, the cost of carpeting the floor is Rs. 2100
Question number – (4)
Solution :
According to rule,
Area of a Square = Side × Side
Area of rectangle = Length × Breadth
Now,
∴ Rectangle area,
= 15 × 7
= 105 sqm
∴ Square area,
= 10 × 10
= 100 sqm
So, rectangle area is Greater.
Question number – (5)
Solution :
In the given question,
Area = 600 sq km
length= 40 km
∴ Breadth= 600/40
= 15 km
So, the breadth of the forest is 15 km.
Question number – (6)
Solution :
(a) the area of the flower bed,
= Area = 21 × 19 …(Area of rectangle = Length × Breadth)
= 339 cm
(b) the cost of laying the flower bed,
= Cost
= 12 × 399
= 4788 Rs.
Question number – (7)
Solution :
In the given question,
Perimeter = 45
Breadth = 9
∴ Length = 45/2 – 9
= 22.5 – 9
= 13.5
∴ Area,
= 9 × 13.5
= 121.5 sq.cm
So, the area of the figure is 121.5 sq.cm.
Question number – (8)
Solution :
According to the given question,
Perimeter,
= 2(3.4 + 2.3)
= 15.64
Area,
= 3.4 × 2.3
= 7.82
(a) Are the perimeters of both the figures same?
= No, they are not same.
(b) Are the areas of both the figures same?
= No also they are not same.
Question number – (9)
Solution :
(a) Perimeter = 2(230 + 660)
= 1780 cm
∴ Area = 230 × 660
= 15800 sqcm
(b) 3 m 50 cm = 350 cm
5 m 6 cm = 500
∴ Perimeter = 2 (350 + 506)
= 1712 cm2
∴ Area = (350 × 506)
= 177100 sqcm
(i) Find the difference in the areas of the rectangles.
= Difference
= 177100 – 151800
= 25300 sq.cm
(ii) Find the difference in the perimeters of the rectangles.
= Difference
= 1780 – 1712
= 68 cm
Perimeter Area and Volume Exercise 11.6 Solution
Question no – (1)
Solution :
(a) l = 7 cm, b = 4 cm, h = 3 cm
∴ Volume
= 7 × 4 × 3
= 84 cu cm
(b) 1= 10 cm, b = 7.5 cm, h = 2.8 cm
∴ Volume
= 10 × 7.5 × 2.8
= 210 cum cm
(c) l= 15 cm, b = 12 cm, h = 9 cm
∴ Volume
= 15 × 12 × 9
= 1620 cu cm
(d) l = 5.1 cm, b= 3.5 cm, h = 2.6 cm
Volume
= 5.1 × 3.5 × 2.6
= 46.41
Question no – (2)
Solution :
(a) In the question
The side is = 4.2 cm
As we know,
Volume of a cube = Side × Side × Side
∴ Volume,
= 4.2 × 4.2 × 4.2
= 74.088 cu cm
So, the volume is 74.088 cu cm.
(b) In the question
The side is = 5 cm
As we know,
Volume of a cube = Side × Side × Side
∴ Volume,
= 5 × 5 × 5
= 125 cu cm
So, the volume is 125 cu.cm
(c) In the question
The side is = 7.5 cm
As we know,
Volume of a cube = Side × Side × Side
∴ Volume,
= 7.5 × 7.5 × 7.5
= 421.875 cu cm
So, the volume is 421.875 cu.cm
(d) In the question
The side is = 11 cm
As we know,
Volume of a cube = Side × Side × Side
∴ Volume,
= 11 × 11 × 11
= 1331 cu cm
So, the volume is 1331 cu.cm
Question no – (5)
Solution :
(a) In the question we get,
A swimming pool is = 25 m long, 15 m wide and 10 m deep,
Its volume = ?
As we know,
Volume = length × breadth × height
∴ Volume of the swimming pool,
= 25 × 15 × 10 cu cm
= 3750 cu cm
So, the volume of the swimming pool is 3750 cu.cm
(b) From the question we get,
A brick is 10.2 cm long, 6.5 cm wide and 3 cm high.
Volume of the brick = ?
As we know,
Volume = length × breadth × height
∴ Volume of the brick
= 10.2 × 6.5 × 3
= 198.9 cu cm
So, the volume of the brick is 198.9 cu cm.
(c) Given in the question,
Each side of the cube is = 20 cm
As we know that,
Volume of a cube = Side × Side × Side
∴ Volume of the cube,
= 20 × 20 × 20
= 8000 cu cm
So, the volume of the cube is 8000 cu cm.
(d) In the given question,
Rectangular piece of wood has length =1.2 m,
breadth =78 cm
high =13 cm
As we know,
Volume = length × breadth × height
∴ Volume of the piece of word,
= 120 × 78 × 13 cu cm
= 121680 cu cm
So, the volume of the piece of wood is 121680 cu cm.
(e) Volume of the box = 30 × 15 × 20 …(Volume = length × breadth × height)
= 9000 cu cm
∴ Volume of each box,
= 5 × 3 × 3
= 45
∴ No of books can be fit,
= 9000/45
= 200
(f) In the given question,
Cupboard has a volume of = 8 cu m.
length is = 4 m
height is = 2 m
We know that,
Breadth in volume = Volume /Length × Height
∴ Breadth of the cupboard
= 8/4 × 2
= 8/8
= 1 m
So, breadth of the cupboard is 1 m.
(g) In the given question,
A box has a volume of = 1176 cu cm.
Length = 12 cm
Breadth = 7 cm
We know that,
Height = Volume/Length × Breadth
∴ Height of the box,
= 1176/12 × 7
= 1170/84
= 14 cm
So, the height of the box is 14 cm.
(h) If container forms a cube,
So, its height length breadth are same.
So, Height = 3√343 = 7
Thus, the height of the cube is 7 cm.
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