Exercise 10.1
Page No 212:
Question 1:
Find the perimeter of each of the following figures: |  |
(a)
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(b)
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 |  |
(c)
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(d)
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 |  |
(e)
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(f)
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Answer:
Perimeter of a polygon is equal to the sum of the lengths of all sides of that polygon.(a) Perimeter = (4 + 2 +1 + 5) cm = 12 cm
(b) Perimeter = (23 + 35 + 40 + 35) cm = 133 cm
(c) Perimeter = (15 + 15 + 15 + 15) cm = 60 cm
(d) Perimeter = (4 + 4 + 4 + 4 + 4) cm = 20 cm
(e) Perimeter = (1 + 4 + 0.5 + 2.5 + 2.5 + 0.5 + 4) cm = 15 cm
(f) Perimeter = (1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3 + 4 +
1 + 3 + 2 + 3 + 4) = 52 cm
Q 2 - Ex 10.1 - Mensuration - NCERT Maths Class 6th - Chapter 10
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?Answer:
Length (l) of rectangular box = 40 cmBreadth (b) of rectangular box = 10 cm
Length of tape required = Perimeter of rectangular box
= 2 (l + b) = 2(40 + 10) = 100 cm
Q 3 - Ex 10.1 - Mensuration - NCERT Maths Class 6th - Chapter 10
Question 3:
A table-top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the table-top?Answer:
Length (l) of table-top = 2 m 25 cm = 2 + 0.25 = 2.25 mBreadth (b) of table-top = 1 m 50 cm = 1 + 0.50 = 1 .50 m
Perimeter of table-top = 2 (l + b)
= 2 × (2.25 + 1.50)
= 2 × 3.75 = 7.5 m
Q 4 - Ex 10.1 - Mensuration - NCERT Maths Class 6th - Chapter 10
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?Answer:
Length (l) of photograph = 32 cmBreadth (b) of photograph = 21 cm
Length of wooden strip required = Perimeter of Photograph
= 2 × (l + b)
= 2 × (32 + 21) = 2 × 53 = 106 cm
Q 5 - Ex 10.1 - Mensuration - NCERT Maths Class 6th - Chapter 10
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?Answer:
Length (l) of land = 0.7 kmBreadth (b) of land = 0.5 km
Perimeter = 2 × (l + b)
= 2 × (0.7 + 0.5) = 2 × 1.2 = 2.4 km
Length of wire required = 4 × 2.4 = 9.6 km
Q 6 - Ex 10.1 - Mensuration - NCERT Maths Class 6th - Chapter 10
Page No 213:
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.
Answer:
(a) Perimeter = (3 + 4 + 5) cm = 12 cm(b) Perimeter of an equilateral triangle = 3 × Side of triangle
= (3 × 9) cm = 27 cm
(c) Perimeter = (2 × 8) + 6 = 22 cm
Q 7 - Ex 10.1 - Mensuration - NCERT Maths Class 6th - Chapter 10
Question 7:
Find the perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.Answer:
Perimeter of triangle = Sum of the lengths of all sides of the trianglePerimeter = 10 + 14 + 15 = 39 cm
Q 8 - Ex 10.1 - Mensuration - NCERT Maths Class 6th - Chapter 10
Question 8:
Find the perimeter of a regular hexagon with each side measuring 8 m.Answer:
Perimeter of regular hexagon = 6 × Side of regular hexagonPerimeter of regular hexagon = 6 × 8 = 48 m
Q 9 - Ex 10.1 - Mensuration - NCERT Maths Class 6th - Chapter 10
Question 9:
Find the side of the square whose perimeter is 20 m.Answer:
Perimeter of square = 4 × Side20 = 4 × Side
Side =
Question 10:
The perimeter of a regular pentagon is 100 cm. How long is its each side?Answer:
Perimeter of regular pentagon = 5 × Length of side100 = 5 × Side
Side =
= 20 cmQ 11 - Ex 10.1 - Mensuration - NCERT Maths Class 6th - Chapter 10
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?
Answer:
(a) Perimeter = 4 × Side30 = 4 × Side
Side =
(b) Perimeter = 3 × Side
30 = 3 × Side
Side =
(c) Perimeter = 6 × Side
30 = 6 × Side
Side =
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?Answer:
Perimeter of triangle = Sum of all sides of the triangle36 = 12 + 14 + Side
36 = 26 + Side
Side = 36 − 26 = 10 cm
Hence, the third side of the triangle is 10 cm.
Q 13 - Ex 10.1 - Mensuration - NCERT Maths Class 6th - Chapter 10
Question 13:
Find the cost of fencing a square park of side 250 m at the rate of Rs 20 per metre.Answer:
Length of fence required = Perimeter of the square park= 4 × Side
= 4 × 250 = 1000 m
Cost for fencing 1 m of square park = Rs 20
Cost for fencing 1000 m of square park = 1000 × 20
= Rs 20000
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.Answer:
Length (l) of rectangular park = 175 mBreadth (b) of rectangular park = 125 m
Length of wire required for fencing the park = Perimeter of the park
= 2 × (l + b)
= 2 × (175 + 125)
= 2 × 300
= 600 m
Cost for fencing 1 m of the park = Rs 12
Cost for fencing 600 m of the square park = 600 × 12
= Rs 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 lessdistance?
Answer:
Distance covered by Sweety = 4 × Side of square park= 4 × 75 = 300 m
Distance covered by Bulbul = 2 × (60 + 45)
= 2 × 105 = 210 m
Therefore, Bulbul covers less distance.
Question 16:
What is the perimeter of each of the following figures? What do you infer from the answers? |  |  |
(a)
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(b)
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(c)
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 | ||
(d)
| ||
Answer:
(a) Perimeter of square = 4 × 25 = 100 cm(b) Perimeter of rectangle = 2 × (10 + 40) = 100 cm
(c) Perimeter of rectangle = 2 × (20 + 30) = 100 cm
(d) Perimeter of triangle = 30 + 30 + 40 = 100 cm
It can be inferred that all the figures have the same perimeter.
Question 17:
Avneet buys 9 square paving slabs, each with a side of
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.)
Answer:
(a) Side of square =
Perimeter of square =
(b) Perimeter of cross = 0.5 + 1 + 1 + 0.5 + 1 + 1 + 0.5 + 1 + 1
+ 0.5 + 1 + 1 = 10 m
(c) The arrangement in the form of a cross has a greater perimeter.
(d) Arrangements with perimeters greater than 10 m cannot be determined.
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