These NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.1 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Exercise 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

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 × 50 = 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?

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 × (length + breadth) = 2 × (2.25 + 1.50) = 2 × 3.75 = 7.5 m

Thus, the perimeter of tabletop 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?

Answer:

Length of wooden strip = Perimeter of photograph

Perimeter of photograph

= 2 x (length + breadth)

= 2 (32 + 21) = 2 × 53cm =106 cm

Thus, the length of the wooden strip required is equal to 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?

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 × (length + breadth)

= 2 × (0.7 + 0.5) = 2 × 1.2 = 2.4 km

= 2.4 × 1000 m = 2400 m

Thus, the length of wire = 4 × 2400 = 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.

Answer:

(a) Perimeter of ΔABC = 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 AABC = 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.

Answer:

Perimeter of triangle

= Sum of all three sides

= 10 cm + 14 cm + 15 cm = 39 cm

Thus, the perimeter of triangle is 39 cm

Question 8.

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

Answer:

Perimeter of Hexagon

= 6 × length of one side

= 6 × 8 m = 48 m

Thus, the perimeter of hexagon is 48 m.

Question 9.

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

Answer:

Perimeter of square = 4 × side

⇒ 20 = 4 × side

⇒ Side = \(\frac{20}{4}\) = 5 cm

Thus, the side of square is 5 cm.

Question 10.

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

Answer:

Perimeter of regular pentagon = 100 cm

⇒ 5 × side = 100 cm

⇒ Side = \(\frac{100}{5}\) = 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?

Answer:

Length of string = Perimeter of each figure

(a) Perimeter of square = 30 cm

⇒ 4 × side = 30 cm

⇒ Side = \(\frac{30}{4}\) = 7.5 cm 4

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

(b) Perimeter of equilateral triangle = 30 cm

⇒ 3 × side = 30 cm

⇒ Side = \(\frac{30}{3}\) = 10 cm

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

(c) Perimeter of hexagon = 30 cm

⇒ 6 × side = 30 cm

⇒ Side = \(\frac{30}{6}\) = 5 cm

Thus, the side of each side of hexagon is 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?

Answer:

Let the length of third side be X cm.

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

Now, Perimeter of triangle = 36 cm

⇒ 12 +14 + X = 36

⇒ 26 + X = 36

⇒ X = 36 – 26 = 10 cm

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 ₹ 20 per metre.

Answer:

Side of square = 250 m

Perimeter of square

= 4 × side = 4 × 250 = 1000 m

Since, cost of fencing of per meter = ₹20

Therefore, the cost of fencing of 1000 meters = 20 × 1000 = ₹ 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.

Answer:

Length of rectangular park = 175 m

Breadth of rectangular park = 125 m

Perimeter of park

= 2 × (length + breadth)

= 2 × (175 + 125) = 2 × 300 = 600 m

Since, the cost of fencing park per meter = ₹ 12

Therefore, the cost of fencing park of 600 m = 12 × 600 = ₹ 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?

Answer:

Distance covered by Sweety

= Perimeter of square park

Perimeter of square = 4 × side

= 4 × 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 × (length + breadth)

= 2 × (60 + 45) = 2 × 105 = 210 m

Thus, Bulbul covers the distance of 210 m and Bulbul covers less distance.

Question 16.

What is the perimeter of each of the following figures? What do you infer from the answers?

Answer:

(a) Perimeter of square

= 4 × side = 4 × 25 = 100 cm

(b) Perimeter of rectangle

= 2 × (length + breadth)

= 2 × (40 + 10) = 2 × 50 = 100 cm

(c) Perimeter of rectangle

= 2 × (length + breadth)

= 2 × (30 + 20) = 2 × 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.

Question 17.

Avneet buys 9 square paving slabs, each with a side of 1/2 m. He lays them in the form of a square.

(a) What is the perimeter of his arrangement [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 [(Fig (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) The arrangement so formed in Fig (i) is a square of side equal to 3 × side of the square slab

∴ Perimeter of his arrangement in Fig (i)

= 4 × ( 3 × side of the square slab)

= 4 × ( 3 × 0.5) m = 4 × 1.5 m = 6 m

(b) The cross arrangement so formed in Fig (ii) by 9 square slabs, has perimeter, which consists of 20 length segments each equal to side of I a square slab i.e. 0.5 m

∴ Perimeter of her arrangement in Fig (ii)

= 20 × ( side of the square slab)

= 20 × 0.5 m = 10 m

(c) ∵ 10m > 6m

∴ Cross arrangement has greater perimeter

(d) ∵ Total number of tiles = 9

∴ The following arrangement will also have greater perimeter.

Since, this arrangement is in the form of rectangle with length and breadth as \(\frac { 9 }{ 2 }\) m and \(\frac { 1 }{ 2 }\) respectively.

∴ Perimeter = 2 \(\left(\frac{9}{2}+\frac{1}{2}\right)\)

= 2 × = 10m \(\left[\frac{10}{2}\right]\) = 10m