NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.1, are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.1.
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.1
Question 1.
Find the perimeter of each of the following figures:
Solution :
(a) Perimeter
= 5 cm + I cm + 2 cm + 4 cm
= 12 cm
(b) Perimeter
= 40 cm + 35 cm + 23 cm + 35 cm
= 133 cm
(c) Perimeter
= 15 cm + 15 cm + 15 cm+ 15 cm
= 60 cm
(d) Perimeter
= 4 cm + 4 cm + 4 cm + 4 cm + 4 cm
= 20 cm
(e) Perimeter
= 1 cm + 4 cm + 0.5 cm + 2.5 cm + 2.5 cm + 0.5 cm + 4 cm
= 15 cm
(f) Perimeter = 1 cm + 3 cm + 2 cm + 3 cm + 4 cm + 1 cm + 3 cm + 2 cm + 3 cm + 4 cm + 1 cm + 3 cm + 2 cm + 3 cm + 4 cm + 1 cm + 3 cm + 2 cm + 3 cm + 4 cm
= 52 cm
Question 2.
The lid of a rectangular box of sides 40 cm by 10 cm is sealed all around with tape. What is the length of the tape required?
Solution :
Length of the tape required
= Perimeter of the rectangular box = 2 × (Length + Breadth)
= 2 × (40 cm + 10 cm)
= 2 × (50 cm)
= 100 cm = l m.
Question 3.
A table-top measures 2 m 25 cm by l m 50 cm. What is the perimeter of the table-top?
Solution :
Perimeter of the table-top
= 2 × (Length + Breadth)
= 2 × (2 m 25 cm + 1 m 50 cm)
= 2 × (2.25 m+1.50 m)
= 2 × (3.75 m)
= 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?
Solution :
Length of the wooden strip required = 2 × (Length + Breadth)
= 2 × (32 cm + 21 cm)
= 2 × (53 cm)
= 106 cm = 1.06 m.
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?
Solution :
Perimeter of the rectangle
= 2 × (Length + Breadth )
= 2 × (0.7 km + 0.5 km)
= 2 × (1.2km)
= 2.4 km
Length of the wire needed
= 4 × Perimeter of the rectangle = 4 × (2.4 km)
= 9.6 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 a mi third side 6 cm.
Solution :
(a) Perimeter of the triangle = 3 cm + 4 cm + 5 cm = 12 cm
(b) Perimeter of the equilateral triangle = 3 × Length of a side = 3 × (9 cm) = 27 cm
(c) Perimeter of the isosceles triangle = 8 cm + 8 cm + 6 cm = 22 cm.
Question 7.
Find the perimeter of a triangle with sides measuring 10 cm, 14 cm, and 15 cm.
Solution :
Perimeter of the triangle
= Sum of the lengths of its three sides
= 10 cm + 14 cm + 15 cm
= 39 cm.
Question 8.
Find the perimeter of a regular hexagon with each side measuring 8 m.
Solution :
Perimeter of the regular hexagon
= 6 × Length of a side
= 6 × (8m)
= 48 m.
Question 9.
Find the side of the square whose perimeter is 20 m.
Solution :
Perimeter of the square
= 4 × Length of a side
⇒ Length of one side
Question 10.
The perimeter of a regular pentagon is 100 cm. How long is its every side?
Solution :
Perimeter of the regular pentagon = 5 × Length of a side
⇒ Length of one (each) side
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?
Solution :
(a) Perimeter of the square = 4 × Length of a side
⇒ Length of a side
(b) Perimeter of the equilateral triangle
= 3 × Length of a side
⇒ Length of a side
(c) Perimeter of the regular hexagon = 6 × Length of a side.
⇒ Length of a side
Question 12.
Two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is the third side?
Solution :
Perimeter of a triangle = Sum of the lengths of its three sides
⇒ 36 cm = 12 cm + 14 cm + Length of the third side
⇒ 36 cm = 26 cm + Length of the third side
⇒ Length of the third side = 36 cm – 26 cm = 10 cm.
Question 13.
Find the cost of fencing a square park of side 250 m at the rate of ₹ 20 per meter.
Solution :
Perimeter of the square park = 4 × Length of a side = 4 × (250m)
= 1000 m
∴ Cost of fencing the square park at the rate of?
20 per metre = ₹ 1000 × 20
= ₹ 20,000
Question 14.
Find the cost of fencing a rectangular park of length 175 m and breadth 125 m at the rate of? 12 per meter.
Solution :
Perimeter of the rectangular park = 2 × (Length + Breadth)
= 2 × (175m + 125 m)
= 2 × (300 m)
= 600 m
Cost of fencing the rectangular park at the rate of ?
12 per metre = ₹ 600 × 12
= ₹ 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?
Solution :
Perimeter of the square park = 4 × Length of a side = 4 × (75 m)
= 300 m
Perimeter of the rectangular park
= 2 × (Length + Breadth)
= 2 × (60 m + 45 m)
= 2 × (105 m)
= 210 m.
Since, the perimeter of the rectangular park is less than the perimeter of the square park, therefore. Bulbul covers les
s distance.
Question 16.
What is the perimeter of each of the following figures? What do you infer from the answers?
Solution :
(a) Perimeter
= Sum of the lengths of all the sides
= 25 cm + 25 cm + 25 cm + 25 cm
= 100 cm.
(b) Perimeter
= Sum of the lengths of all the sides
= 40 cm + 10 cm + 40 cm + 10 cm
= 100 cm.
(c) Perimeter
= Sum of the lengths of all the sides
= 30 cm + 20 cm + 30 cm + 20 cm
= 100 cm.
(d) Perimeter
= Sum of the lengths of all the sides
= 30 cm + 40 cm + 30 cm
= 100 cm.
The inference from the answers. All the figures have the same perimeter.
Question 17.
Avneet buys 9 square paving slabs, each with a side of (frac { 1 }{ 2 }) m. He lays them in the form of a square.
(a) What is the perimeter of his arrangement (Fig. 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. ii)?
(c) Which has a 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 slabs must meet along complete edges i.e. they cannot be broken.)
Solution :
(a) Perimeter of his arrangement = 4 × Length of one side
(b) Perimeter of her arrangement = Sum of the lengths of all the sides
(c) Cross has greater perimeter.
(d) Yes ! there is a way of getting an even greater perimeter. It is shown below:
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