The DAV Class 6 Maths Book Solutions and **DAV Class 6 Maths Chapter 15 Worksheet 1 **Solutions of Perimeter and Area offer comprehensive answers to textbook questions.

## DAV Class 6 Maths Ch 15 WS 1 Solutions

Question 1.

A field is in the shape of a rectangle whose length is 130 m and breadth is 95 m respectively. Find the cost of fencing the field at ₹ 20 per metre.

Answer:

The perimeter of the rectangle

= 2[l + 6]

= 2[130 m + 95 m]

= 2 × [225 m]

= 450 m

Cost of fencing = ₹ 20 × 450

= ₹ 9000

Hence the required cost = ₹ 9000.

Question 2.

Akshay covers a rectangular field three times whose length and breadth are 25 m and 19 m respectively. How much distance does he cover?

Answer:

Perimeter of the field = 2[l + b]

= 2[25 m + 19 m]

= 2 × 44 m

= 88 m

Total Distance covered by Akshay = 3 × 88 m = 264 m

Hence the required distance = 264 m.

Question 3.

The length of one side of a square field is 25 m. Find the cost of fencing at the rate of ? 18 per metre.

Answer:

The perimeter of a square = 4 × side = 4 × 25 = 100 m

Cost of fencing = ₹ 100 × 18 = ₹ 1800

Hence the required cost = ₹ 1800

Question 4.

Raju can round a rectangular garden of length 30 m and breadth 15 m. Manu ran around a square field of side 35 m. Who covered more distance and by how much?

Answer:

Perimeter of a rectangle = 2 [l + b]

= 2[30 m + 15 m]

= 2 × 45 m = 90 m.

∴ The distance covered by Raju in one round = 90 m.

The distance covered by Manu in round = 4 × 35 = 140 m

∴ Manu covered more distance by 140 m – 90 m = 50 m.

Question 5.

Vineet wants to polish the floor of a room which is 140 cm long and 120 cm wide. The charges for polishing the floor is ₹ 52 per square metre. What will be the cost of polishing the floor of the room?

Answer:

Area of the rectangle

= length × breadth = 140 cm × 120 cm

= 16800 cm^{2} = 16800 m^{2} + 10000

= 1.68 m^{2}

Charge for polishing the floor = ₹ (1.68 × 52) = ₹ 87.36

Hence the required cost = ₹ 87.36

Question 6.

A carpet is 6 m long and 4.5 m wide. Find the area of the carpet. Also find the cost of the carpet if it costs ? 45 per square metre.

Answer:

Area of the carpet = l × b

= 6 m × 4.5 m = 27 m^{2}

Cost of the carpet = ₹(27 × 45)

= ₹ 1215

Hence the required cost = ₹ 1215.

Question 7.

A square park is to be watered. If one side of the park is 4.2 m, find the area to be watered.

Answer:

Side of the square park = 4.2 m

∴ Area of the park = (4.2 × 4.2) m^{2} = 17.64 m^{2}

Hence the required area = 17.64 m^{2}

Question 8.

Which playground has bigger area – one that measures -50 m by 40 m or the other that measures 75 m by 20 m?

Answer:

Area of the first playground = l × b

= 50 m × 40 m

= 2000 m^{2}

Area of the second playground = l × b

= 75 m × 20 m

= 1500 m^{2}

Hence the first playground has bigger area.

### DAV Class 6 Maths Chapter 15 Worksheet 1 Notes

1. Perimeter of rectangle = 2 × [length + breadth]

2. Perimeter of square = 4 × side

3. Area of rectangle = length × breadth.

4. Area of square = (Side)^{2}

Example 1:

Find the perimeter of the following figures:

Solution:

Perimeter of rectangle = 2 [l + b]

= 2[2.4 + 1.8]

= 2 × 4.2

= 8.4 cm

Solution:

Perimeter of the quadrilateral

= 3.6 cm + 2.4 cm + 1.8 cm + 3.00 cm

= 10.8 cm

Solution:

Perimeter of the pentagon

= 3.2 cm + 3.4 cm + 1.8 cm + 2.2 cm + 2.00 cm

= 12.6 cm

Solution:

Perimeter of the pentagon

= 4.2 cm + 2.4 cm + 2.2 cm + 4.2 cm + 2.8 cm

= 15.8 cm

Example 2:

If the perimeter of square is 32 cm, find its side.

Solution:

Perimeter of square = 4 × side

∴ Side = \(\frac{1}{4}\) × Perimeter

= \(\frac{1}{4}\) × 32 = 8 cm

Hence the side = 8 cm.

Example 3:

Find the perimeter of a rectangle if its length is 16 m and breadth 12 m.

Solution:

Length = 16 m,Breadth = 12 m

. .Perimeter of the rectangle= 2[l + b]

= 2[16 m + 12 m]

= 2 × 28 m = 56 m

Hence the perimeter of the rectangle is 56 m.

Example 4:

A lawn is 200 m long and 150 m wide. The gardener wants a fencing of barbbed wire all around it. Find the cost of the wire at the rate of ₹ 20 per metre.

Solution:

Perimeter of the lawn = 2 [l + b]

= 2 [200 m + 150 m]

= 2 × 350 m

= 700 m

Cost of barbbed wire = ₹ 700 × 20

= ₹ 14000

Hence the total cost = ₹ 14000

Example 5:

Find the area of a rectangular field having 20 m length and 15 m 75 cm breadth.

Solution:

Area of rectangle = length × breadth

= (20 × 15.75) m^{2}

= 315 m^{2}

Hence the required area = 315 m^{2}