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:
