NCERT Solutions for Class 6 Maths Exercise 10.1

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NCERT Solutions for Class 6 Maths Mensuration

Class –VI Mathematics

(Ex. 10.1)

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

Answer:

(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 + 35 cm + 40 cm + 35 cm = 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

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NCERT Solutions for Class 6 Maths Exercise 10.1

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: Total length of tape required = Perimeter of rectangle

= 2 ( length + breadth)

= 2 ( 40 + 10)

= 2 x 50

= 100 cm = 1 m

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

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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 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 x (length + breadth)

= 2 x (2.25 + 1.50)

= 2 x 3.75 = 7.50 m

Thus, perimeter of table top is 7.5 m.

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NCERT Solutions for Class 6 Maths Exercise 10.1

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 of wooden strip = Perimeter of photograph

Perimeter of photograph = 2 x (length + breadth)

= 2 (32 + 21)

= 2 x 53 cm = 106 cm

Thus, the length of the wooden strip required is equal to 106 cm.

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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: Since the 4 rows of wires are needed. Therefore the total length of wires is equal to 4 times the perimeter of rectangle.

Perimeter of field = 2 x (length + breadth) = 2 x (0.7 + 0.5) = 2 x 1.2 = 2.4 km

= 2.4 x 1000 m = 2400 m

Thus, the length of wire = 4 x 2400 = 9600 m = 9.6 m

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NCERT Solutions for Class 6 Maths Exercise 10.1

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 of {tex}\Delta {\text{ABC}}{/tex} = AB + BC + CA

= 3 cm + 5 cm = 4 cm

= 12 cm

(b) Perimeter of equilateral ABC = 3 x side

= 3 x 9 cm

= 27 cm

(c) Perimeter of {tex}\Delta {\text{ABC}}{/tex} = AB + BC + CA

= 8 cm + 6 cm + 8 cm

= 22 cm

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NCERT Solutions for Class 6 Maths Exercise 10.1

Question 7.Find the perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.

Answer:

Perimeter of triangle = Sum of all three sides

= 10 cm + 14 cm + 15 cm

= 39 cm

Thus, perimeter of triangle is 39 cm.

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

Answer:

Perimeter of Hexagon = 6 x length of one side

= 6 x 8 m

= 48 m

Thus, the perimeter of hexagon is 48 m.

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NCERT Solutions for Class 6 Maths Exercise 10.1

Question 9.Find the side of the square whose perimeter is 20 m.

Answer: Perimeter of square = 4 x side

{tex} \Rightarrow {/tex} 20 = 4 x side

{tex} \Rightarrow {/tex} side = {tex}\frac{{20}}{4}{/tex} = 5 cm

Thus, the side of square is 5 cm.

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Question 10.The perimeter of a regular pentagon is 100 cm. How long is its each side?

Answer:

Perimeter of regular pentagon = 100 cm

{tex} \Rightarrow {/tex} 5 x side = 100 cm

{tex} \Rightarrow {/tex} side = {tex}\frac{{100}}{5}{/tex} = 20 cm

Thus, the side of regular pentagon is 20 cm.

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NCERT Solutions for Class 6 Maths Exercise 10.1

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:

Length of string = Perimeter of each figure

(a)Perimeter of square = 30 cm

{tex} \Rightarrow {/tex} 4 x side = 30 cm

{tex} \Rightarrow {/tex} side = {tex}\frac{{30}}{4}{/tex} = 7.5 cm

Thus, the length of each side of square is 7.5 cm.

(b)Perimeter of equilateral triangle = 30 cm

{tex} \Rightarrow {/tex} 3 x side = 30 cm

{tex} \Rightarrow {/tex} side = {tex}\frac{{30}}{3}{/tex} = 10 cm

Thus, the length of each side of equilateral triangle is 10 cm.

(c)Perimeter of hexagon = 30 cm

{tex} \Rightarrow {/tex} 6 x side = 30 cm

{tex} \Rightarrow {/tex} side = {tex}\frac{{30}}{6}{/tex} = 5 cm

Thus, the side of each side of hexagon is 5 cm.

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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?

Answer:

Let the length of third side be {tex}x{/tex} cm.

Length of other two side are 12 cm and 14 cm.

Now, Perimeter of triangle = 36 cm

{tex} \Rightarrow {/tex} {tex}12 + 14 + x = 36{/tex}

{tex} \Rightarrow {/tex} {tex}26 + x = 36{/tex}

{tex} \Rightarrow {/tex} {tex}x = 36 – 26{/tex}

{tex} \Rightarrow {/tex} {tex}x = 10{/tex} cm

Thus, the length of third side is 10 cm.

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NCERT Solutions for Class 6 Maths Exercise 10.1

Question 13.Find the cost of fencing a square park of side 250 m at the rate of ` 20 per meter.

Answer:

Side of square = 250 m

Perimeter of square = 4 x side

= 4 x 250 = 1000 m

Since, cost of fencing of per meter = Rs. 20

Therefore, cost of fencing of 1000 meters = 20 x 1000 = = Rs. 20,000

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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 meter.

Answer:

Length of rectangular park = 175 m

Breadth of rectangular park = 125 m

Perimeter of park = 2 x (length + breadth)

= 2 x (175 + 125)

= 2 x 300 = 600 m

Since, cost of fencing park per meter = = Rs. 12

Therefore, cost of fencing park of 600 m = 12 x 600 = Rs. 7,200

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NCERT Solutions for Class 6 Maths Exercise 10.1

Question 15Sweety runs around a square park of side 75 m. Bulbul runs around a rectangular park with length of 60 m and breadth 45 m. Who covers less distance?

Answer:

Distance covered by Sweety = Perimeter of square park

Perimeter of square = 4 x side

= 4 x 75 = 300 m

Thus, distance covered by Sweety is 300 m.

Now, distance covered by Bulbul = Perimeter of rectangular park

Perimeter of rectangular park = 2 x (length + breadth)

= 2 x (60 + 45)

= 2 x 105 = 210 m

Thus, Bulbul covers the distance of 210 m.

And Bulbul covers less distance.

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Question 16.What is the perimeter of each of the following figures? What do you infer from the answer?

Answer:

(a) Perimeter of square = 4 x side

= 4 x 25 = 100 cm

(b) Perimeter of rectangle = 2 x (length + breadth)

= 2 x (40 + 10)

= 2 x 50 = 100 cm

(c) Perimeter of rectangle = 2 x (length + breadth)

= 2 x (30 + 20)

= 2 x 50 = 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.

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NCERT Solutions for Class 6 Maths Exercise 10.1

Question 17.Avneet buys 9 square paving slabs, each with a side {tex}\frac{1}{2}{/tex} m. He lays them in the form of a square

(a)What is the perimeter of his arrangement?

(b)Shari does not like his arrangement. She gets him to lay them out like a cross. What is the perimeter of her arrangement?

(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) 6 m

(b) 10 m

(c) Second arrangement has greater perimeter.

(d) Yes, if all the squares are arranged in row, the perimeter be 10 cm.

NCERT Solutions for Class 6 Maths Exercise 10.1

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