Haryana State Board HBSE 6th Class Maths Solutions Chapter 10 Mensuration Ex 10.1 Textbook Exercise Questions and Answers.

## Haryana Board 6th Class Maths Solutions Chapter 10 Mensuration Exercise 10.1

Question 1.

Find the perimeter of each of the following figures :

Solution:

(a) Perimeter = 5 cm + 1 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 = 3 cm + 3 cm + 3 cm + 3 cm + 3 cm = 15 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 = 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.

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 rectangular box (l) = 40 cm

Breadth of the rectangular box (b) = 10 cm

Perimeter of the rectangular box

= 2 x (l + b)

= 2 x (40 + 10) cm

= 2 x 50 = 100 cm

Hence, length of the tape required = 100 cm or 1m.

Question 3.

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

Solution:

Length of the table-top (l)

= 2 m 25 cm

= 2.25 m

Breadth of the table-top (b)

= 1 m 50 cm

= 1.50 m

Perimeter of the table-top

= 2 x (l + b)

= 2 x (2.25 + 1.50) m

= 2 x 3.75

= 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 ?

Solution:

Length of the frame (l)

= 32 cm

Breadth of the frame (b)

= 21 cm

Perimeter of the frame

= 2 x (l + b)

= 2 x (32 cm + 21 cm)

= 2 x 53 cm

= 106 cm.

Hence, length of the wooden strip required to frame a 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 wire needed ?

Solution:

Length (l) = 0.7 km

Breadth (b) = 0.5 km

Perimeter – 2 x (l + b)

= 2 x (0.7 + 0.5) km

= 2 x 1.2 = 2.4 km

Length of wire needed to fence with 4 rows = 4 x 2.4 = 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.

Solution:

(a) Perimeter of the triangle

= 3 cm + 4 cm + 5 cm = 12 cm.

(b) One side of an equilateral

Δ = 9 cm

.’. Perimeter of the triangle

= 3 x length of one side

= 3 x 9 cm = 27 cm.

(c) Perimeter of isosceles triangle

= 8 cm + 8 cm + 6 cm

= 22 cm.

Question 7.

Find the peri-meter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.

Solution:

Perimeter of the triangle

= 10 cm + 14 cm + 15 cm

= 39 cm.

Question 8.

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

Solution:

One side of a regular hexagon = 8 m

Perimeter of regular hexagon = 6 x one side

= 6 x 8 m

= 48 m.

Question 9.

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

Solution:

Let, one side of a square = am

Perimeter of the square = 20 m

4 x a = 20

a = \(\frac{20}{4}\) = 5 cm

Hence, one side of the square = 5 m.

Question 10.

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

Solution:

Let one side of the regular pentagon = a cm

Perimeter of the regular pentagon = 100 cm

5 x a = 100

=> a = \(\frac{100}{5}\) = 20 cm.

Hence, each side of a regular 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 ?

Solution:

(a) Let one side of a square = a cm

Perimeter of the square = 30 cm

4 x a = 30 cm

=> a = \(\frac{30}{4}\) cm = 7.5 cm

Hence, length of each side of the square = 7.5 cm.

(b) Let each side of the equilateral Δ = a cm

Perimeter of the equilateral Δ = 30 cm

3 x a = 30 cm

a = \(\frac{30}{3}\)

= 10 cm

Hence, each side of the equilateral Δ = 10 cm.

(c) Let each side of the regular hexagon

= a cm

Perimeter of the regular hexagon = 30 cm

6 x a = 30 cm

=> a = \(\frac{30}{6}\) = 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 the third side ?

Solution:

Let the third side of a triangle = x cm

Perimeter of the triangle = 36 cm

.-. 14 cm + 12 cm + x cm = 36 cm

=> 26 cm + x cm = 36 cm

x = 36 – 26 = 10 cm

Hence, the 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 Rs. 20 per metre.

Solution:

Each side of the square park = 250 m

Perimeter of the square park

= 4 x 250 m

= 1000 m

Cost of one metre of wire = Rs. 20

Cost of fencing the square park = Rs. 20 x 1000

= Rs. 20,000.

Question 14.

Find the cost of fencing a rectangular park of length 175 m and breadth 125 m at the rate of Rs. 12 per metre.

Solution:

Length of the rectangular park (l) = 175 m

Breadth of the rectangular park (b) = 125 m

Perimeter of the rectangular park . = 2 x (l + b)

= 2 x (175 + 125) m

= 2 x 300

= 600 m

Cost of one metre of wire = Rs. 12

.•. Cost of fencing the rectangular park

= Rs. 12 x 600

= Rs. 7200.

Question 15.

Sweety runs around a square park of side 75 m. Bulbul runs around a rectangle with length 60 metre and breadth 45 metre. Who covers less distance ?

Solution:

Each side of the square park = 75 m

Perimeter of the square park = 4 x 75 = 300 m

.’. Distance covered by Sweety = 300 m

Now, length of the rectangular park (l)

= 60 m

breadth of the rectangular park (b) = 45 m

Perimeter of the rectangular park = 2 x (l+ b)

= 2 x (60 + 45)

= 2 x 105 = 210 m

Distance covered by Bulbul = 210 m

Hence, Bulbul covers less distance.

Question 16.

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

Solution:

(a) One side of the square = 25 cm

Perimeter of the square

=4 x 25 = 100 cm.

(b) Length of the rectangle (l) = 40 cm

Breadth of the rectangle (b) = 10 cm

Perimeter of the rectangle = 2 x (l + b)

= 2 x (40 + 10) cm

= 2 x 50 = 100 cm.

(c) Length of the rectangle (l) = 30 cm

Breadth of the rectangle (b) = 20 cm

.’. Perimeter of the rectangle

= 2 x (l + b)

= 2 x (30 + 20) cm

= 2 x 50 = 100 cm.

(d) Sides of the triangle are 40 cm, 30 cm and 30 cm.

Perimeter of the triangle

= 40 cm + 30 cm + 30 cm

= 100 cm.

We infer from the answers that all the figures have the same perimeter.

Question 17.

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

(a) What is the perameter of his arrangement [Fig. 10.24(a)]?

(b) Shari does not like his arrangement.

She gets him to lay them out like a cross [Fig. 10.24(b)]. What is the perimeter of her arrangement ?

(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, they cannot be broken).

Solution: