Gujarat Board GSEB Textbook Solutions Class 6 Maths Chapter 10 Mensuration Ex 10.1 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 6 Maths Chapter 10 Mensuration Ex 10.1

Question 1.

Find the perimeter of each of the following figures:

Solution:

(a) Perimeter of the given figure = sum of the sides

= 4 cm + 5 cm + 1 cm + 2 cm = 12 cm

(b) Perimeter of the given figure = sum of the sides

= 23 cm + 35 cm + 40 cm + 35 cm = 133 cm

(c) Perimeter of the given figure = sum of the sides

= 15 cm + 15 cm + 15 cm + 15 cm = 60 cm

(d) Perimeter of the given figure = sum of the sides

= 4 cm + 4 cm + 4 cm + 4 cm + 4 cm = 20 cm

(e) Perimeter of the given figure = sum of the sides

= 1 cm + 4 cm + 0.5 cm + 2.5 cm + 2.5 cm + 0.5 cm + 4 cm = 15 cm

(f) Perimeter of the given figure = sum of the sides

= 4 cm + 3 cm + 2 cm + 3 cm + 1 cm

+ 4 cm + 3 cm + 2 cm + 3 cm + 1 cm

+ 4 cm + 3 cm + 2 cm + 3 cm + 1 cm

+ 4 cm = 52 cm + 3 cm + 2 cm + 3 cm + 1 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:

The length of the tape required

= perimeter of the lid of a rectangular box

= 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 = 225 cm

Breadth of the table top = 1 m 50 cm = 150 cm

∴ Perimeter of the table top

= 2(length + breadth) = 2(225 + 150) cm

= 2(375 cm)

= 750 cm or 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 frame = 32 cm

Breadth of the frame = 21 cm

∴ Perimeter of the frame = 2 [length + breadth] = 2(32 cm + 21 cm) = 2(53 cm) = 106 cm

∴ Length of the wooden strip required = 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:

0.7 km by 0.5 km means.

Length = 0.7 km and breadth = 0.5 km

∴ Perimeter of the rectangular piece of land = 2 (length + breadth) = 2 (0.7 km + 0.5 km) = 2 (1.2 km) = 2.4 km

∴ Length of wire fencing for 1 row = 2.4 km Length of wire fencing for 4 rows = 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) Sides of the triangle are: 3 cm, 4 cm, 5 cm

∴ Perimeter = sum of the sides = 3 cm + 4 cm + 5 cm = 12 cm

(b) Length of each side = 9 cm

∵ The perimeter of an equilateral triangle

= 3 × length of one side

∴ The perimeter of the given equilateral triangle

= 3 × 9 cm = 27 cm

(c) Sides of the given triangle are: 8 cm, 6 cm and 8 cm

∴ Perimeter of the triangle = sum of the sides

= 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:

Sides of the triangle are: 10 cm, 14 cm and 15 cm

∴ Perimeter = sum of the sides

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

Question 8.

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

Solution:

∵ The perimeter of a regular hexagon = 6 × one side of the hexagon

∴ Perimeter of the given regular hexagon = 6 × 8 m = 48 m

Question 9.

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

Solution:

∵ The perimeter of the square = 20 m.

∵ Perimeter of square = 4 × length of one side

∴ 4 x Length of a side = 20 m

or Length of a side = \(\frac { 20 m }{ 2 }\) = 5 m

Question 10.

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

Solution:

Perimeter of a regular pentagon = 100 cm

∵ Perimeter = 5 × Length of one side

∴ 5 × Length of one side = 100 cm

∴ Length of one side = \(\frac{100 \mathrm{~cm}}{5}\) = 20 cm

Question 11.

A piece of string is 30 cm long. What will be the length of each side of the string is used to form:

(a) a square?

(b) an equilateral triangle?

(c) a regular hexagon?

Solution:

Length of the given string = 30 cm

(a) ∵ The string is in the form of a square and perimeter of a square = 4 × Side

∴ side = \(\frac{30 \mathrm{~cm}}{4}\) = 7.5 cm

∴ length of each side = 7.5 cm

(b) ∵ The string is in the form of an equilateral triangle

∴ perimeter = 3 × length of one side

or 3 × length of one side = 30 cm

or length of one side = \(\frac{30 \mathrm{~cm}}{3}\) = 10 cm

(c) ∵ String is in the form of a regular hexagon and perimeter of a regular hexagon

= 6 × length of one side

6 × Length of one side = 30 cm

or Length of one side =\(\frac{30 \mathrm{~cm}}{6}\)= 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:

Perimeter of the triangle = 90 sum of its sides

= 12 cm + 14 cm + third side

= 26 cm + third side

But perimeter = 36 cm

∴ 26 cm + third side = 36 cm

or third side = 36 – 26 = 10 cm

Hence, the length of the third side of the triangle = 10 cm

Question 13.

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

Solution:

Side of the park = 250 m

∵ The park is in the form of a square.

∴ Perimeter = 4 × length of one side = 4 × 250 m = 1000 m

∵ Rate of fencing = ₹ 20 per meter

∴ Cost of fencing = ₹ 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 meter.

Solution:

Length of the park = 175 m

The breadth of the park = 125 m

Park is in the form of a rectangle.

Perimeter of a rectangle

= 2 × [length + breadth]

∴ Perimeter of the given park = 2 × [175 m + 125 m]

= 2 × 300 m = 600 m

Rate of fencing = ₹ 12 per metre

∴ Cost of fencing = ₹ 12 × 600 = ₹ 7200

Question 15.

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

Solution:

Side of the square park = 75 m

∴ Perimeter of the square park = 4 × side = 4 × 75 m = 300 m

i.e. Distance covered by Sweety = 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

i.e. Distance covered by Bulbul = 210 m

Since 210 < 300,

Thus, 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) Length of a side of the square = 25 cm

∴ Perimeter = 4 × side = 4 × 25 = 100 cm

(b) Length of the rectangle = 40 cm

Breadth of the rectangle = 10 cm

∴ Perimeter = 2(length + breadth)

= 2(40 + 10) cm = 2 × (50 cm) = 100 cm

(c) Length of the rectangle = 30 cm

Breadth of the rectangle = 20 cm

∴ Perimeter = 2 (length + breadth)

= 2(30 cm + 20 cm) = 2 × 50 cm = 100 cm

(d) Sides of the given triangle are: 30 cm, 40 cm, and 30 cm

∴ Perimeter of the triangle = sum of the sides = 30 cm + 40 cm + 30 cm = 100 cm

Here, we observe that the perimeter of each figure is 100 cm or they have an equal perimeter.

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 [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 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) The arrangement is in the form of a square of side

∴ The perimeter of the square arrangement = 4 × side

(b) Perimeter of the cross-arrangement

(c) ∵ 10 m > 6 m

∴ Cross-arrangement has a greater perimeter.

(d) ∵ Total number of tiles = 9

∴ The following arrangement will also have a greater perimeter.

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

∴ Perimeter =