**NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.1****NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.2****NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.3**

**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

**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
[length + breadth]

= 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 [length + breadth]

= 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.(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.

**Related Links:**

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**NCERT Solutions for Maths Class 8**

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**NCERT Solutions for Maths Class 10**