NCERT Class 6 Maths Chapter 10 Mensuration Exercise 10.1

Chapter 10 Mensuration Exercise 10.1

Question 1: Find the perimeter of each of the following figures:

Question 1 figure

Answer:
a)
= 5 cm + 1 cm + 2 cm + 4 cm
= 12 cm
Therefore, the perimeter of the figure is 12 cm.

b)
= 40 cm + 35 cm + 23 cm + 35 cm
= 133 cm
Therefore, the perimeter of the figure is 133 cm.

c)
= 15 cm + 15 cm + 15 cm + 15 cm
= 60 cm
Therefore, the perimeter of the figure is 60 cm.

d)
= 4 cm + 4 cm + 4 cm + 4 cm + 4 cm
= 20 cm
Therefore, the perimeter of the figure is 20 cm.

e)
= 1 cm + 4 cm + 4 cm + 0.5 cm + 0.5 cm + 2.5 cm + 2.5 cm
= 15 cm
Therefore, the perimeter of the figure is 15 cm.

f)
= 1 cm + 4 cm + 1 cm + 4 cm + 3 cm + 3 cm + 2 cm + 3 cm + 3 cm + 2 cm + 4 cm + 1 cm + 3 cm + 2 cm + 3 cm + 2 cm + 3 cm + 3 cm + 4 cm + 1 cm
= 52 cm
Therefore, the perimeter of the figure is 52 cm.

Question 2: The lid of a rectangular box of sides 40 cm by 10 cm is sealed all round with tape. What is the length of the tape required?
Answer:

Given
Length of lid of rectangular box = 40 cm
Breadth of lid of rectangular box = 10 cm

To find: The length of tape required

= Perimeter of rectangle = 2 × (l + b)
= 2 × (40 + 10)
= 2 × 50
= 100 cm

Therefore, the length of the tape required is 100 cm or 1 m.

Question 3: A table-top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the table-top?
Answer:

Given
Length of table-top = 2 m 25 cm = 225 cm
Breadth of table-top = 1 m 50 cm = 150 cm

To find: The perimeter of the table-top

= Perimeter of rectangle = 2 × (l + b)
= 2 × (225 + 150)
= 2 × 375
= 750 cm

Therefore, the perimeter of the table-top is 750 cm or 7.5 m.

Question 4: What is the length of the wooden strip required to frame a photograph of length and breadth 32 cm and 21 cm respectively?
Answer:

Given
Length of photograph = 32 cm
Breadth of photograph = 21 cm

To find: The length of the wooden strip required to frame a photograph

= 2 × (32 + 21)
= 2 × 53
= 106 cm

Therefore, the length of the wooden strip required to frame a photograph is 106 cm.

Question 5: A rectangular piece of land measures 0.7 km by 0.5 km. Each side is to be fenced with 4 rows of wires. What is the length of the wire needed?
Answer:

Given
Length of rectangular piece of land = 0.7 km = 700 m
Breadth of rectangular piece of land = 0.5 km = 500 m

To find: The length of the wire required to fence a rectangular piece of land

= 2 × (l + b)
= 2 × (700 + 500)
= 2 × 1200
= 2400 m

Therefore, the length of the wire required to fence a rectangular piece of land is 2400 m or 2.4 km.

Question 6: Find the perimeter of each of the following shapes:
a) A triangle of sides 3 cm, 4 cm and 5 cm.
b) An equilateral triangle of side 9 cm.
c) An isosceles triangle with equal sides 8 cm each and third side 6 cm.
Answer:

a)
Perimeter of triangle = side + side + side
= 3 cm + 4 cm + 5 cm
= 12 cm
Therefore, the perimeter of the triangle is 12 cm.

b)
Perimeter of an equilateral triangle = 3 × side
= 3 × 9
= 27 cm
Therefore, the perimeter of an equilateral triangle is 27 cm.

c)
Perimeter of isosceles triangle = 2a + b
= 2 × 8 + 6
= 22 cm
Therefore, the perimeter of an isosceles triangle is 24 cm.

Question 7: Find the perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.
Answer:

Given
Length of first side of triangle = 10 cm
Length of second side of triangle = 14 cm
Length of third side of triangle = 15 cm

To find: Perimeter of triangle
= Perimeter of triangle = side + side + side
= 10 cm + 14 cm + 15 cm
= 39 cm

Therefore, the perimeter of a triangle is 39 cm.

Question 8: Find the perimeter of a regular hexagon with each side measuring 8 m.
Answer:

Given
Length of one side of regular hexagon = 8 m

To find: The perimeter of a regular hexagon
= Perimeter of regular hexagon = 6 × length of one side
= 6 × 8 m
= 48 m
Therefore, the perimeter of a regular hexagon is 48 m.

Question 9: Find the side of the square whose perimeter is 20 m.
Answer:

Given
Perimeter of square = 20 m

To find: Side of square
Let the side of square be x.
= Perimeter of square = 4 × side
= 20 cm = 4 × x
= x = 20/4
= x = 5 m

Therefore, the length of side of square whose perimeter is 20 m is 5 m.

Question 10: The perimeter of a regular pentagon is 100 cm. How long is its each side?
Answer:

Given
Perimeter of regular pentagon = 100 cm

To find: Length of each side
Let the side of regular pentagon be x.
= Perimeter of regular pentagon = 5 × side
= 100 cm = 5 × x
= x = 100/5
= x = 20 cm

Therefore, the length of side of regular pentagon whose perimeter is 100 cm is 20 cm.

Question 11: A piece of string is 30 cm long. What will be the length of each side if the string is used to form:
a) a square?
b) an equilateral triangle?
c) a regular hexagon?
Answer:

a)
Perimeter of square = 30 cm
Let the length of each side of square be x.
= Perimeter of square = 4 × side
= 30 = 4 × x
= x = 30/4
= x = 7.5 cm

b)
Perimeter of equilateral triangle = 30 cm
Let the length of each side of square be x.
= Perimeter of equilateral triangle = 3 × side
= 30 = 3 × x
= x = 30/3
= x = 10 cm

c)
Perimeter of regular hexagon = 30 cm
Let the length of each side of square be x.
= Perimeter of square = 6 × side
= 30 = 6 × x
= x = 30/6
= x = 5 cm

Question 12: Two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is its third side?
Answer:

Given
Length of first side of triangle = 12 cm
Length of second side of triangle = 14 cm
Perimeter of triangle = 36 cm

To find: Length of third side of triangle.
Let the third side of triangle be x.
= 12 + 14 + x = 36
= 26 + x = 36
= x = 36 - 26
= x = 10 cm
Therefore, the length of the third side of triangle is 10 cm.

Question 13: Find the cost of fencing a square park of side 250 m at the rate of ₹20 per metre.
Answer:

Given
Length of side of a square park = 250 m
Cost of fencing per metre = ₹20

To find: The cost of fencing the square park
Perimeter of square = 4 × side
= 4 × 250
= 1000

Therefore, the perimeter of square park is 1000 m.

Cost of fencing per metre = ₹20
Cost of fencing 1000 m = ?
= 1000 × 20
= ₹20000

Therefore, the cost of fencing the square park is ₹20000.

Question 14: Find the cost of fencing a rectangular park of length 175 m and breadth 125 m at the rate of ₹12 per metre.
Answer:

Given
Length of rectangular park = 175 m
Breadth of rectangular park = 125 m
Cost of fencing per metre = ₹12

To find: The cost of fencing the rectangular park
Perimeter of rectangle = 2 × (l + b)
= 2 × (175 + 125)
= 2 × 300
= 600 m

Therefore, the perimeter of the rectangular park is 600 m.

Cost of fencing per metre = ₹12
Cost of fencing 600 metre = ?
= 600 × 12
= ₹7200

Therefore, the cost of fencing the rectangular park is ₹7200.

Question 15: Sweety runs around a square park of side 75 m. Bulbul runs around a rectangular park with length 60 m and breadth 45 m. Who covers less distance?
Answer:

Sweety
Length of side of square = 75 metre
Perimeter of square park = ?
Perimeter of square = 4 × side
= 4 × 75 = 296 metre

Therefore, Sweety runs 296 metre.

Bulbul
Length of rectangular park = 60 metre
Breadth of rectangular park = 45 metre
Perimeter of rectangular park = ?
Perimeter of rectangle = 2 × (l + b)
= 2 × (60 + 45)
= 2 × 105 = 210 metre

Therefore, Bulbul runs 210 metre.

Compare
296 metre > 210 metre
Therefore, Bulbul covers less distance.

Question 16: What is the perimeter of each of the following figures? What do you infer from the answers?

Question 16 figure

Answer:
a)
Perimeter of square = 4 × side
= 4 × 25
= 100 cm

b)
Perimeter of rectangle = 2 × (l + b)
= 2 × (40 + 10)
= 2 × 50
= 100 cm

c) Perimeter of rectangle = 2 × (l + b)
= 2 × (30 + 20)
= 2 × 50
= 100 cm

d)
Perimeter of triangle = side + side + side
= 30 + 30 + 40
= 100 cm

We can conclude that the perimeter of a), b), c) and d) are 100 cm.

Question 17: Avneet buys 9 square paving slabs, each with a side of 1/2 m. He lays them in the form of a square.

Question 17 figure


a) What is the perimeter of his arrangement [Fig 10.7(i)]?
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 10.7 (ii)]?
c) Which 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? (The paving labs must meet along complete edges i.e. they cannot be broken.)
Answer:

a) Side of square = 3 × side
= 3 × 1/2
= 3/2 metre

Perimeter of square = 4 × side
= 4 × 3/2
= 12/2
= 6 metre

Therefore, the perimeter of Avneet’s arrangement is 6 metres.

b) Perimeter of cross arrangement
= 0.5 + 1 + 1 + 0.5 + 1 + 1 + 0.5 + 1 + 1 + 0.5 + 1 + 1
= 10 metre

Therefore, the perimeter of Shari’s arrangement is 10 metres.

c) The arrangement in the form of cross has greater perimeter.

d) Perimeters greater than 10 metre cannot be determined.

No comments:

Post a Comment

Also read!

Click on any of the below text for additional links:

Class 6

Class 6 Computer Kips Solutions
Class 6 Gul Mohar English Solutions
→ Class 6 Map Pointing Solutions
Class 6 NCERT Civics Solutions
Class 6 NCERT Geography Solutions
Class 6 NCERT History Solutions
Class 6 NCERT Maths Solutions
Class 6 NCERT Science Solutions

Class 7

Class 7 Computer Kips Solutions
Class 7 Gul Mohar English Solutions
English Grammar Class 7
English Writing Skills Class 7
Class 7 Map Pointing Solutions
Class 7 NCERT Civics Solutions
Class 7 NCERT Civics Notes
Class 7 NCERT Civics Extra Questions
Class 7 NCERT Geography Solutions
Class 7 NCERT Geography Notes
Class 7 NCERT Geography Extra Questions
Class 7 NCERT History Solutions
Class 7 NCERT History Notes
Class 7 NCERT History Extra Questions
Class 7 NCERT Maths Solutions
Class 7 NCERT Science Solutions
Class 7 NCERT Science Notes
Class 7 NCERT Science Extra Questions

Class 8

Class 8 Kips Artificial Intelligence Solutions
Class 8 Gul Mohar English Solutions
English Grammar Class 8
English Writing Skills Class 8
Class 8 Map Pointing Solutions
Class 8 NCERT Civics Solutions
Class 8 NCERT Civics Notes
Class 8 NCERT Civics Extra Questions
Class 8 NCERT Geography Solutions
Class 8 NCERT Geography Notes
Class 8 NCERT Geography Extra Questions
Class 8 NCERT History Solutions
Class 8 NCERT History Notes
Class 8 NCERT History Extra Questions
Class 8 NCERT Maths Solutions
Class 8 NCERT Science Solutions
Class 8 NCERT Science Notes
Class 8 NCERT Science Extra Questions
Class 8 NCERT Science Exemplar
Class 8 NCERT Science MCQs

Class 9

Class 9 Kips Artificial Intelligence Solutions
Class 9 Map Pointing Solutions
Class 9 NCERT Civics Solutions
Class 9 NCERT Geography Solutions
Class 9 NCERT History Solutions
Class 9 NCERT Maths Solutions
Class 9 NCERT Science Solutions

Class 10

Class 10 Kips Artificial Intelligence Solutions
Class 10 NCERT English Solutions
→ Class 10 Map Pointing Solutions
Class 10 NCERT Civics Solutions
Class 10 NCERT Civics Notes
Class 10 NCERT Civics Extra Questions
Class 10 NCERT Geography Solutions
Class 10 NCERT Geography Notes
Class 10 NCERT Geography Extra Questions
Class 10 NCERT History Solutions
→ Class 10 NCERT History Notes
Class 10 NCERT History Extra Questions
Class 10 NCERT Maths Solutions
Class 10 NCERT Science Solutions
Class 10 NCERT Science Notes
Class 10 NCERT Science Extra Questions

Computer Languages

Python Basics

Extra Activities

Extra Activities

Extra Knowledge

Extra Information
General Knowledge
Historical Places in India
Latest technology
Physical Sciences
Facts

• Mathematics quick links
Mathematical Terms
Maths Tricks

Home Top