## NCERT SOLUTIONS FOR CLASS 6 MATHS EXERCISE 10.1

Find here step by step NCERT Solutions for Class 6 Maths Exercise 10.1 Chapter 10 Mensuration. NCERT Solutions for Class 6 Maths Exercise 10.1 deals with the Problems on Perimeter. In this exercise we will learn to find the Perimeter of a Triangle, the Perimeter of a Square, the Perimeter of a Rectangle and the Perimeter of a Regular Polygon.

## Class 6 Mensuration | How to find Perimeter

Perimeter is the distance covered along the boundary forming a closed figure when we go round the figure once. Perimeter of a closed figure can be find by adding the length of all its sides.

So, Perimeter of a triangle is the sum of its three sides. Perimeter of a Quadrilateral is the sum of its four sides.

Therefore, Perimeter of a Polygon is the sum of all its sides.

(i) Perimeter of a rectangle = 2 × (length + breadth)
(ii) Perimeter of a square = 4 × length of its side
(iii) Perimeter of an equilateral triangle = 3 × length of a side

### Exercise 10.1Finding Perimeter

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

(a) Sides of the figure are 4 cm, 2 cm, 1 cm and 5 cm,

Therefore, Perimeter of the figure = (4 + 2 + 1 + 5) cm
= 12 cm.

(b) Side of the figure are 23 cm, 35 cm, 40 cm and 35 cm,

Therefore, Perimeter of the figure = (23 + 35 + 40 + 35) cm = 133 cm.

(c) The figure has 4 equal sides of 15 cm,

Therefore, Perimeter of the figure = 4×15 cm = 60.

(d) The figure has 5 equal sides of 4 cm,

Therefore, Perimeter of the figure = 5×4 cm = 20 cm.

(e) Sides of the figure = 1 cm, 4cm, 0.5 cm, 2.5 cm, 2.5 cm, 0.5 cm and 4 cm,

Therefore, Perimeter of the figure = {1 + 2(4 + 0.5 + 2.5)} cm = {1 + 2×7} cm = 15 cm.

(f) The figure has 20 sides in which,

4 sides are equal to 1 cm, 4 sides are equal to 2 cm, 4 sides are equal to 4 cm and remaining 8 sides are equal to 3 cm,

Therefore, Perimeter of the figure = {4×(1 + 2 + 4) + 8×3} cm = {4×7 + 24} cm = {28 + 24} cm = 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?

Length of the rectangular box = 40 cm,
Breadth of the rectangular box = 10 cm,

Length of the tape required to seal the lid of the box will be the perimeter of
the box,

Now, perimeter of the rectangular box
= 2 × (length + breadth)
= 2 × (40 cm + 10 cm)
= 2 × 50 cm = 100 cm

Therefore, the length of the tape required is 100 cm.

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

Length of the table top = 2 m 25 cm = 2.25 m,
Breadth of the table top = 1 m 50 cm = 1.50 m,

Then, perimeter of the table top
= 2 × (length + breadth)
= 2 × (2.25 m + 1.50 m)
= 2 × 3.75 m = 7.50 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?

Length of the photograph = 32 cm,
Breadth of the photograph = 21 cm,

Length of the wooden strip required to frame the photograph will be the perimeter of the photograph,

Now, perimeter of the photograph
= 2 × (length + breadth)
= 2 × (32 cm + 21 cm)
= 2 × 53 cm = 106 cm,

So, the length of the wooden strip required to frame the photograph = 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?

Length of the rectangular land = 0.7 km,
Breadth of the rectangular park = 0.5 km,

So, totak length of the wire required to fence the field with one row of wire will be the perimeter of the field,

Now, perimeter of the rectangular field
= 2 × (length + breadth)
= 2 × (0.7 km + 0.5 km)
= 2 × 1.2 km = 2.4 km,

Therefore, length wire required to fence the field with 4 rows of wires = 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 and third side 6 cm.

(a) Sides of the triangle are 3 cm, 4 cm and 5 cm,

So, Perimeter of the triangle = 3 cm + 4 cm + 5 cm
= 12 cm

(b) Sides of the eauilateral triangle = 9 cm,

So, Perimeter of the triangle = 3×9 cm
= 27 cm

(c) Equal sides of the isosceles triangle = 8 cm,
Third side of the isosceles triangle = 6 cm,

So, Perimeter of the triangle = 2×8 cm + 6 cm
= 16 cm + 6 cm
= 22 cm.

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

Sides of the triangle are 20 cm, 14 cm and 15 cm

So, perimeter of the triangle = (20 + 14 + 15) cm,
= 39 cm.

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

A regular hexagon has six equal sides,

Side of the regular hexagon = 8 m,

So, the perimeter of the hexagon = 6×8 m
= 48 m.

QUESTION 9
Find the side of the square whose perimeter is 20 m.

A square has four equal sides,

Perimeter of the square = 20 m,

So, the side of the square = m
= 5 m.

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

A regular pentagon has 5 equal sides,

The perimeter of the regular pentagon = 100 cm,

So, length of each side of the pentagon =
= 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?

Length of the string = 30 cm,

(a) If the string is used to form a square, then
Perimeter of the square = 30 cm,

So, length of the square =
= 7.5 cm.

(b) If the string is used to form an equilateral triangle, then
Perimeter of the equilateral triangle = 30 cm,

So, length of the triangle =
= 10 cm.

(c) If the string is used to form a regular hexagon, then
Perimeter of the hexagon = 30 cm,

So, length of the hexagon =
= 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?

Two sides of a triangle are 12 cm and 14 cm,
The perimeter of the triangle = 36 cm,
The third side = ?

Since, perimeter of the triangle = first side + second side + third side
Therefore, third side = perimeter – (first side + second side)

third side = 36 cm – (12 cm + 14 cm)
third side = 36 cm – 26 cm
third side = 10 cm

Third side of the triangle = 10 cm.

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

Side of the square park = 250 m,

So, Perimeter of the square park = 4×250 m
= 1000 m,

Rate of the fencing = ₹ 20 per metre,

Therefore, cost of the fencing = ₹ 20 × 1000
= ₹ 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 metre.

Length of the rectangular park = 175 m,
Breadth of the rectangular park = 125 m,

So, Perimeter of the rectangular park = 2×(175 + 125) m
= 2×300 m
= 600 m,

Rate of the fencing = ₹ 12 per metre,

Therefore, cost of the fencing = ₹ 12 × 600
= ₹ 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?

Side of the square park = 75 m,

Perimeter of the square park = 4×75 m
= 300 m,

So, Sweety covered 300 m during her run,

Length of the rectangular park = 60 m,
Breadth of the rectangular park = 45 m,

Perimeter of the rectangular park = 2×(60 + 45) m
= 2×105 m
= 210 m

So, Bulbul covered 210 m during her run,

210 m < 300 m

Therefore, Bulbul covered less distance than Sweety.

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

(a) Perimeter of the square = 4×25 cm
= 100 cm

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

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

(d) Perimeter of the triangle = 30 cm + 40 cm + 40 cm
= 100 cm

The perimeter of all the figures are equal.

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.
(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.)

Side of a paving slabs = m,

(a) Perimeter of the arrangement in Fig (i) = m
= 6 m.

(b) Perimeter of the arrangement in Fig (ii) = m
= 10 m.

(c) 6 m < 10 m
So, cross i.e., the arrangement in fig (ii) has greater perimeter.

(d) Perimeter greater than 10 m with the given condition is not possible.

## NCERT SOLUTIONS FOR CLASS 6 MATHS CHAPTER 10 MENSURATION

Check NCERT Solutions of other exercises from Class 6 Maths Chapter 10 Mensuration by clicking the links given below.

• Exercise 10.1
• Exercise 10.2
• Exercise 10.3

We hope NCERT Solutions for Class 6 Maths Exercise 10.1 has helped you understand how to find Perimeter of a triangle, a quadrilateral and a regular polygon.

Write in the comment section for any error or any solution related queries from the exercise. [ NCERT solutions for class 6 maths Exercise 10.1 ]