NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.1

# NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.1

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.1 are the part of NCERT Solutions for Class 6 Maths (Rationalised Contents). Here you can find the NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.1.

## NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.1 (Rationalised Contents)

### Ex 10.1 Class 6 Maths Question 1.

Find the perimeter of each of the following figures:

Solution:
(a) Perimeter of the given figure
= 4 cm + 2 cm + 1 cm + 5 cm = 12 cm
(b) Perimeter of the given figure
= 40 cm + 35 cm + 23 cm + 35 cm
= 133 cm
(c) Perimeter of the given figure
= 15 cm + 15 cm + 15 cm + 15 cm = 15 cm × 4 = 60 cm
(d) Perimeter of the given figure
= 4 cm + 4 cm + 4 cm + 4 cm + 4 cm – 4 cm × 5 = 20 cm
(e) Perimeter of the given figure
= 4 cm + 0.5 cm + 2.5 cm + 2.5 cm + 0.5 cm + 4 cm + 1 cm
= 15 cm
(f) Perimeter of the given figure

= 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 + 1 cm + 3 cm + 2 cm + 3 cm = 52 cm

### Ex 10.1 Class 6 Maths 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?

Solution:
The length of the required tape = Perimeter of the rectangular lid
= 2 [length + breadth] = 2 × [40 + 10]
= 2 × 50 = 100 cm

## Ex 10.1 Class 6 Maths Question 3.

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

Solution:
Length of the table-top = 2 m 25 cm = 2 × 100 cm + 25 cm = 225 cm
Breadth of the table-top = 1 m 50 cm = 1 × 100 cm + 50 cm = 150 cm
Perimeter of the table-top = 2 [length + breadth]
= 2 [225 cm + 150 cm]
= 2 × 375 cm
= 750 cm
= 7 m 50 cm

### Ex 10.1 Class 6 Maths 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 photograph = 32 cm
Breadth of the photograph = 21 cm
The length of the wooden strip = Perimeter of the photograph

= 2 [32 cm + 21 cm]
= 2 × 53 cm = 106 cm
Hence, the required length of the wooden strip is 106 cm or 1 m 6 cm.

### Ex 10.1 Class 6 Maths 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:
Length of the rectangular piece of land = 0.7 km = 0.7 × 1000 m = 700 m
Breadth of the rectangular piece of land = 0.5 km = 0.5 × 1000 m = 500 m
Perimeter of the rectangular piece land
= 2 [700 m + 500 m]
= 2400 m
Length of wire needed in 4 rounds of the land = 4 × 2400 = 9600 m = 9.6 km.

### Ex 10.1 Class 6 Maths 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.

Solution:
(a) We know that the perimeter of the given triangle = Sum of all sides of the triangle
Perimeter of the triangle = 3 cm + 4 cm + 5 cm = 12 cm
(b) We know that the perimeter of the given triangle
= Sum of all the sides of the triangle
= (9 + 9 + 9) cm = 3 × 9 = 27 cm
(c) Perimeter of the given isosceles triangle
= Sum of all the sides of the triangle
= (8 + 8 + 6) cm = 22 cm

### Ex 10.1 Class 6 Maths Question 7.

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

Solution:
Perimeter of a triangle = Sum of all the sides of the triangle
= 10 cm + 14 cm + 15 cm
= 39 cm

### Ex 10.1 Class 6 Maths Question 8.

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

Solution:

Here, side of the hexagon = 8 m
Perimeter of a regular hexagon = 6 × side = 6 × 8 m = 48 m

### Ex 10.1 Class 6 Maths Question 9.

Find the side of the square whose perimeter is 20 m.

Solution:
Perimeter of the given square = 20 m
Perimeter of a square = 4 × side
20 = 4 × side
side = 20 ÷ 4 = 5 m

### Ex 10.1 Class 6 Maths Question 10.

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

Solution:
Here, perimeter of the regular pentagon = 100 cm
Number of sides in a regular pentagon = 5
Length of each side of the pentagon = Perimeter ÷ Number of sides
= 100 cm ÷ 5 = 20 cm

### Ex 10.1 Class 6 Maths 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) Length of string = 30 cm = Perimeter of the square
Number of equal sides in a square = 4
Length of each side of the square = 30 cm ÷ 4 = 7.5 cm

(b) Length of string = 30 cm = Perimeter of the equilateral triangle
Number of equal sides in an equilateral triangle = 3
Length of each side of the equilateral triangle = 30 cm ÷ 3 = 10 cm

(c) Length of string = 30 cm = Perimeter of the regular hexagon
Number of equal sides in a regular hexagon = 6
Length of each side of the regular hexagon = 30 cm ÷ 6 = 5 cm

### Ex 10.1 Class 6 Maths 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?

Solution:
Perimeter of the triangle = 36 cm
Lengths of the two sides of the triangle are 12 cm and 14 cm.
Length of the third side of the triangle = 36 – (12 + 14) cm
= (36 – 26) cm = 10 cm

### Ex 10.1 Class 6 Maths Question 13.

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

Solution:
Length of each side of a square = 250 m
Perimeter of the square = 250 m × 4 = 1000 m
Rate of fencing = ₹20 per m
Cost of fencing = ₹20 × 1000 = ₹20,000

### Ex 10.1 Class 6 Maths 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.

Solution:
Length of the rectangular park = 175 m
Breadth of the rectangular park = 125 m
Perimeter of the rectangular park = 2 [length + breadth]
= 2 [175 m + 125 m]
= 2 × 300 m = 600 m
Rate of fencing = ₹ 12 per metre

Cost of fencing = ₹12 × 600 = ₹7200

### Ex 10.1 Class 6 Maths 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:
Each side of the square park = 75 m
Perimeter of the square park = 4 × 75 m = 300 m
Perimeter of the rectangular park = 2 [length + breadth]
= 2 [60 m + 45 m]
= 2 × 105 m = 210 m
Since 210 m < 300 m.
Therefore, Bulbul covers less distance.

### Ex 10.1 Class 6 Maths Question 16.

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

Solution:
(a) Perimeter of the square = 4 × side = 4 × 25 cm = 100 cm
(b) Perimeter of the rectangle = 2 [length + breadth] = 2 [30 cm + 20 cm] = 2 × 50 cm = 100 cm
(c) Perimeter of the rectangle = 2 [length + breadth] = 2 [40 cm + 10 cm] = 2 × 50 cm = 100 cm
(d) Perimeter of the triangle = Sum of all sides = 30 cm + 30 cm + 40 cm = 100 cm From the above answers, we conclude that different figures may have equal perimeters.

### Ex 10.1 Class 6 Maths Question 17.

Avneet buys 9 square paving slabs, each with a side of ½ 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 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) His arrangement is in the form of a square of side
(b) Perimeter of arrangement in cross shape
(c) Since 10 m > 6 m
Cross-arrangement has greater perimeter.
(d) Total number of tiles = 9
We have the following arrangement
The above arrangement will also have the greater perimeter, i.e. 10 m.

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