MP Board Class 6th Maths Solutions Chapter 10 Mensuration Ex 10.1

MP Board Class 6th Maths Solutions Chapter 10 Mensuration Ex 10.1

Question 1.
Find the perimeter of each of the following figures:

Solution:
(a) Perimeter = Sum of all the sides
= 4 cm + 2 cm + 1 cm + 5 cm = 12 cm

(b) Perimeter = Sum of all the sides
= 23 cm + 35cm + 40 cm + 35cm = 133 cm

(c) Perimeter = Sum of all the sides
= 15 cm + 15 cm + 15 cm + 15 cm = 60 cm

(d) Perimeter = Sum of all the sides
= 4 cm + 4 cm + 4 cm + 4 cm + 4 cm = 20 cm

(e) Perimeter = Sum of all the sides
= 1 cm + 4 cm + 0.5 cm + 2.5 cm + 2.5 cm + 0.5 cm + 4 cm = 15 cm

(f) Perimeter = Sum of all the sides
= 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 round with tape. What is the length of the tape required?
Solution:
Total length of tape required
= Perimeter of rectangle
= 2 (40 +10) cm = 2 × 50 cm = 100 cm = 1 m

Thus, the total length of tape required is 100 cm or 1 m.

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 table-top = 2 m 25 cm = 2.25 m
Breadth of table-top = 1 m 50 cm = 1.50 m
Perimeter of table-top = 2 × (length + breadth)
= 2 × (2.25 + 1.50) m
= 2 × 3.75 m = 7.50 m
Thus, perimeter of table-top is 7.5 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 wooden strip
= Perimeter of photograph
= 2 × (length + breadth)
= 2 (32 + 21) cm = 2 × 53 cm = 106 cm
Thus, the length of the wooden strip required is 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?
Solution:
Since 4 rows of wires are needed. Therefore, the total length of wire is equal to 4 times the perimeter of land.
Perimeter of land = 2 × (length + breadth)
= 2 × (0.7 + 0.5) km = 2 × 1.2 km = 2.4 km
= 2.4 × 1000 m = 2400 m
Thus, the length of wire
= 4 × 2400 m = 9600 m = 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 ∆ABC
= AB + BC + CA
= 3 cm+ 5 cm+ 4 cm
= 12 cm

(b) Perimeter of equilateral ∆ABC
= 3 × side
= 3 × 9 cm
= 27 cm

(c) Perimeter of ∆ABC
= AB + BC + CA
= 8 cm + 6 cm + 8 cm
= 22 cm

Question 7.
Find the perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.
Solution:
Perimeter of triangle
= Sum of all three sides
= 10 cm + 14 cm + 15 cm = 39 cm
Thus, perimeter of triangle is 39 cm.

Question 8.
Find the perimeter of a regular hexagon with each side measuring 8 m.
Solution:
Perimeter of a regular hexagon
= 6 × length of one side
= 6 × 8m
= 48m
Thus, the perimeter of the regular hexagon is 48 m.

Question 9.
Find the side of the square whose perimeter is 20 m.
Solution:
Perimeter of square = 4 × side
⇒ 20 m = 4 × side ⇒ side = $$\frac{20}{4}$$ m = 5m
Thus, the side of square is 5 m.

Question 10.
The perimeter of a regular pentagon is 100 cm. How long is its each side?
Solution:
Perimeter of regular pentagon = 5 × side
⇒ 100 cm = 5 × side ⇒ side = $$\frac{100}{5}$$ cm = 20 cm
Thus, the side of regular pentagon is 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:
Length of string = Perimeter of each shape
(a) Perimeter of square = 4 × side
⇒ 30 cm = 4 × side ⇒ side = $$\frac{30}{4}$$ cm = 7.5 cm
Thus, the length of each side of square will be 7.5 cm.

(b) Perimeter of equilateral triangle = 3 × side
⇒ 30 cm = 3 × side ⇒ side = $$\frac{30}{3}$$ cm = 10 cm
Thus, the length of each side of equilateral triangle will be 10 cm.

(c) Perimeter of regular hexagon = 6 × side
⇒ 30 cm = 6 × side ⇒ side = $$\frac{30}{6}$$ cm = 5 cm
Thus, the length of each side of regular hexagon will be 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?
Solution:
Let the length of third side be x cm. Length of other two sides are 12 cm and 14 cm.
Now, perimeter of triangle = 36 cm
⇒ 12 + 14 + x = 36 ⇒ 26 + x = 36
⇒ x = 36 – 26 ⇒ x = 10
Thus, the length of third side is 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:
Side of square park = 250 m
Perimeter of square park = 4 × side
= 4 × 250 m = 1000 m
Since, cost of fencing for 1 metre = Rs. 20
Therefore, cost of fencing for 1000 metres
= Rs. 20 × 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 rectangular park = 175 m
Breadth of rectangular park = 125 m
Perimeter of park = 2 × (length + breadth)
= 2 × (175 + 125) m = 2 × 300 m = 600 m
Since, cost of fencing park for 1 metre = Rs. 12
Therefore, cost of fencing park for 600 m = Rs. 12 × 600 = Rs. 7,200

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:
Distance covered by Sweety
= Perimeter of square park = 4 × side
= 4 × 75 m = 300 m
Thus, distance covered by Sweety is 300 m.
Now, distance covered by Bulbul
= Perimeter of rectangular park
= 2 × (length + breadth)
= 2 × (60 + 45) m = 2 × 105 m = 210 m
Thus, Bulbul covers a distance of 210 m.
So, 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) Perimeter of square = 4 × side
= 4 × 25 cm = 100 cm

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

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

(d) Perimeter of triangle = Sum of all sides
= 30 cm + 30 cm + 40 cm
= 100 cm
Thus, all the figures have same perimeter.

Question 17.
Avneet buys 9 square paving slabs, each with side of $$\frac{1}{2}$$ m. He lays them in the form of a square.
(a) What is the perimeter of his arrangement [see 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 [see fig. (ii)]?
(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 i.e. they cannot be broken.)

Solution:
(a) Side of one small square = $$\frac{1}{2}$$ m
∴ Side of given square = $$\frac{3}{2}$$ m
Perimeter of square = 4 × side
= 4 × $$\frac{3}{2}$$ m = 6 m

(b) Perimeter of given figure
= Sum of all sides = 20 × $$\frac{1}{2}$$ m = 10 m

(c) The arrangement cross has greater perimeter.
(d) It is not possible to determine the arrangement with perimeter greater than 10 m.