Well-designed AP 6th Class Maths Guide Chapter 10 Mensuration Exercise 10.1 offers step-by-step explanations to help students understand problem-solving strategies.
Mensuration Class 6 Exercise 10.1 Solutions – 6th Class Maths 10.1 Exercise Solutions
Question 1.
Find the perimeter of each of the following figures.
Solution:
a) Perimeter = 2 cm + 1 cm + 5 cm + 4 cm = 12 cm
b) Perimeter = 23 cm + 35 cm + 40 cm + 35 cm = 133 cm
c) Perimeter = 15 cm + 15 cm + 15 cm + 15 cm = 60 cm
d) Perimeter = 4 cm + 4 cm + 4 cm + 4 cm + 4 cm = 20 cm
e) Perimeter = 4 cm + 0.5 cm + 2.5 cm + 2.5 cm + 0.5 cm + 4 cm + 1 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 round with tape. What is the length of the tape required?
Solution:
Length of the lid = 40 cm
Breadth of the lid = 10 cm
∴ Total length of the tape required
= Perimeter of the rectangular lid
= 2 × [length + breadth]
= 2 × (40 cm + 10 cm)
= 2 × 50 cm = 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 the table top = 2 m 25 cm
Breadth of the table top = 1 m 50 cm
∴ Perimeter of the table top = 2 × [length + breadth]
= 2 × (2 m 25 cm + 1 m 50 cm)
= 2 × (3 m 75 cm)
= 2 × (3.75 cm) = 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 the photograph = 32 cm
Breadth of the photograph = 21 cm
∴ Length of the wooden strip required to frame a photograph = Perimeter of photograph
= 2 × [length + breadth]
= 2 × (32 cm + 21 cm)
= 2 × (53 cm)
= 106 cm = 1 m 6 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:
Length of the rectangular piece of land = 0.7 km
Breadth of the rectangular piece of land = 0.5 km
0.7 km + 0.5 km = 1.2 km
∴ Perimeter of the retangular land = 2 × [length + breadth]
= 2 × [0.7 km + 0.5 km]
= 2 × 1.2 km = 2.4 km
∴ Length of the wire needed to 4 rounds of the land = 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.
Solution:
a) Side of the triangle = 3 cm, 4 cm, 5 cm
∴ Perimeter of the triangle = 3 cm + 4 cm + 5 cm = 12 cm
b) Side of an equilateral triangle = 9 cm
∴ Perimeter of an equilateral triangle = 9 cm + 9 cm + 9 cm = 27 cm
c) Sides of an isosceles triangle = 8 cm, 8 cm, 6 cm
∴ Perimeter of an isosceles triangle = 8 cm + 8 cm + 6 cm = 22 cm
Question 7.
Find the perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.
Solution:
Sides of the given triangle = 10 cm, 14 cm, 15 cm.
∴ 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:
Sides of the regular hexagon = 8 cm
∴ Sides of the regular hexagon = 8 cm, 8 cm, 8 cm, 8 cm, 8 cm, 8 cm
[∴ Regular hexagon has 6 equal sides]
∴ Perimeter of the given regular hexagon = 8 cm + 8 cm + 8 cm + 8 cm + 8 cm + 8 cm = 48 cm
(OR)
Perimeter of regular hexagon = 6 × side = 6 × 8 cm = 48 cm.
Question 9.
Find the side of the square whose perimeter is 20 m.
Solution:
Perimeter of the square = 20 cm
Perimeter of the square = 4 × side
4 × side = 20 m
∴ side = 20 ÷ 4 = 5 m
Question 10.
The perimeter of a regular pentagon is 100 cm. How long is its each side?
Solution:
Perimeter of the regular pentagon = 100 cm
Number of equal sides of the regular pentagon = 5
Perimeter of the regular pentagon = 5 × side
5 × side = 100 cm
side = 100 ÷ 5 = 20 cm
∴ Length of the each side of the given 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) Length of the string = 30 cm
Number of equal sides in a square = 4
∴ Length of each side of the square = 30 cm ÷ 4 = 7.5 cm
b) Length of the string = 30 cm
Number of equal sides in an equilateral triangle = 3
∴ Length of each side of an equilateral triangle = 30 cm ÷ 3 = 10 cm
c) Length of the string = 30 cm
Number of equal sides in an regular hexagon = 6
∴ Length of each side of the regular hexagon = 30 cm ÷ 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 its third side ?
Solution:
Length of two sides of the given triangle
= 12 cm and 14 cm
Perimeter of the triangle = 36 cm
So, length of the third side
= 36 -(12 + 14)
= 36 cm – (12 cm + 14 cm)
= 36 cm – 26 cm = 10 cm
Question 13.
Find the cost of fencing a square park of side 250 m at the rate of ₹ 20 per metre.
Solution:
Length of the side of the square park = 250 m
∴ Perimeter of the square park = 4 × side
= 4 × 250 m
= 1000 m
Rate of fencing = ₹ 20 m
∴ Cost of fencing = 1000 × ₹ 20 = ₹ 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.
Solution:
Length of the rectangular park = 175 m
Perimeter of the park
= 2 × [length + breadth]
= 2 × [175 m + 125 m]
= 2 × 300 m = 600 m
Rate of fencing = ₹ 12 per meter
∴ Cost of fencing of rectangular park = 600 × ₹ 12 = ₹ 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 ?
Solution:
Side of the square park = 75 m
And its perimeter = 4 × side
= 4 × 75 m = 300 m
Length of the rectangular park = 60 m
Breadth of the rectangular park = 45 m
Perimeter of the rectangular park
= 2 × length + breadth
= 2 × [60 m + 45 m]
= 2 × 105 m
= 210 m
Since 210 m < 300 m
So, Bulbul covers less distance.
So, length of the third side
= 36 – (12 + 14)
= 36 cm – (12 cm + 14 cm)
= 36 cm – 26 cm = 10 cm
Question 16.
What is the perimeter of each of the following figures? What do you infer from the answers?
Solution:
a) Perimeter of the given square = 25 cm + 25 cm + 25 cm + 25 cm = 4 × 25 = 100 cm
b) Perimeter of the rectangle = 40 cm + 10 cm + 40 cm + 10 cm = 100 cm
c) Perimeter of the rectangle = 30 cm + 20 cm + 30 cm + 20 cm = 100 cm
d) Perimeter of the given triangle = 30 cm + 30 cm + 40 cm = 100 cm
From the above answers, we conclude that different figures may have equal perimeters.
Question 17.
Avneet buys 9 square paving slabs, each with a side of \(\frac { 1 }{ 2 }\) m. He lays them in the form of a square. 
a) What is the perimeter of his arrangement [(Figure (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 [(Figure (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) The arrangement is in the form of a square of side
= [\(\frac { 1 }{ 2 }\) + \(\frac { 1 }{ 2 }\) + \(\frac { 1 }{ 2 }\)]m = 1\(\frac { 1 }{ 2 }\)m = \(\frac { 3 }{ 2 }\)m
Perimeter of the square arrangement = 4 × \(\frac { 3 }{ 2 }\) = 6m
b) Perimeter of cross – arrangement = \(\frac { 1 }{ 2 }\)m + 1 m + 1 m + \(\frac { 1 }{ 2 }\)m + 1m + 1m + 1m + \(\frac { 1 }{ 2 }\)m + 1m + 1m = 10m
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. \(\frac { 1 }{ 2 }\) + \(\frac { 9 }{ 2 }\) + \(\frac { 1 }{ 2 }\) + \(\frac { 9 }{ 2 }\) = \(\frac { 20 }{ 2 }\) = 10 cm.