RD Sharma Solutions Class 6 Mensuration Exercise 20.2

RD Sharma Solutions Class 6 Chapter 20 Exercise 20.2

RD Sharma Class 6 Solutions Chapter 20 Ex 20.2 PDF Free Download

Exercise 20.2

Q 1. Find the perimeter of the rectangle whose lengths and breadths are given below:

(i) 7 cm, 5 cm

Solution: Length = 7 cm, Breadth = 5 cm

Perimeter of a rectangle = 2 × (Length + Breadth)

Perimeter of a rectangle = 2 × (7 + 5) = 2 × (12) = 24 cm

(ii) 5 cm, 4 cm

Solution: Length = 5 cm, Breadth = 4 cm

Perimeter of a rectangle = 2 × (Length + Breadth)

Perimeter of a rectangle = 2 × (5 + 4) = 2 × (9) = 18 cm

(iii) 7.5 cm, 4.5 cm

Solution: Length = 7.5 cm, Breadth = 4.5 cm

Perimeter of a rectangle = 2 × (Length + Breadth)

Perimeter of a rectangle = 2 × (7.5 + 4.5) = 2 × (12) = 24 cm

Q 2. Find the perimeter of the squares whose sides are given below:

(i) 10 cm

Solution: Side = 10 cm

Perimeter of a square = 4 x Side

Perimeter of a square = 4 x 10 = 40 cm

(ii) 5 m

Solution: Side = 5 m

Perimeter of a square = 4 x Side

Perimeter of a square = 4 x 5 = 20 m

(iii) 115.5 cm

Solution: Side = 115.5 cm

Perimeter of a square = 4 x Side

Perimeter of a square = 4 x 115.5 = 462 cm

Q 3. Find the side of the square whose perimeter is:

(i) 16 m

Solution: Perimeter = 16 m

Side of a square = \(\frac{ 1 }{ 4 }\) × (Perimeter of the square)

Side of this square = \(\frac{ 16 }{ 4 }\) = 4 m

(ii) 40 cm

Solution: Perimeter = 40 cm

Side of a square = \(\frac{ 1 }{ 4 }\) × (Perimeter of the square)

Side of this square = \(\frac{ 40 }{ 4 }\) = 10 cm

(iii) 22 cm

Solution: Perimeter = 22 cm

Side of a square = \(\frac{ 1 }{ 4 }\) × (Perimeter of the square)

Side of this square = \(\frac{ 22 }{ 4 }\) = 5.5 cm

Q 4. Find the breadth of the rectangle whose perimeter is 360 cm and whose length is

(i) 116 cm

Solution: Perimeter = 360 cm, Length = 116 cm

Perimeter of a rectangle = 2 × (Length + Breadth)

360 = 2 × (116 + Breadth)

360 = 232 + (2×Breadth)

(2×Breadth) = 360 – 232

(2×Breadth) = 128

Breadth = 64 cm

(ii) 140 cm

Solution: Perimeter = 360 cm, Length = 140 cm

Perimeter of a rectangle = 2 × (Length + Breadth)

360 = 2 × (140 + Breadth)

360 = 280 + (2×Breadth)

(2×Breadth) = 360 – 280

(2×Breadth) = 80

Breadth = 40 cm

(iii) 102 cm

Solution: Perimeter = 360 cm, Length = 102 cm

Perimeter of a rectangle = 2 × (Length + Breadth)

360 = 2 × (102 + Breadth)

360 = 204 + (2×Breadth)

(2×Breadth) = 360 – 204

(2×Breadth) = 156

Breadth = 78 cm

Q 5. A rectangular piece of lawn is 55 m wide and 98 m long. Find the length of the fence around it.

Solution: Length of the lawn = 98 m

Breadth of the lawn = 55 m

Length of the fence around the lawn = Perimeter of the lawn

Length of the fence around the lawn = 2 × (Length + Breadth)

Length of the fence around the law = 2 × (98 + 55) = 2 × (153) = 306 m

Q6. The side of a square field is 65 m. What is the length of the fence required all around it?

Solution: Side of the square field = 65 m

Length of the fence around the square field = Perimeter of the square field

Length of the fence around the square field = 4 × (Side of the square)

Length of the fence around the square field = 4 × 65 = 260 m

Q7. Two sides of a triangle are 15 cm and 20 cm. The perimeter of the triangles is 50 cm. What is the third side?

Solution: Let S1, S2, Sbe the side of a triangle.

S1 = 15 cm

S2 = 20 cm

Perimeter of the triangle = 50 cm

We know that;

Perimeter of a triangle = Sum of all three sides of the triangle

Perimeter of a triangle = S1 + S2 + S3

50 = 15 + 20 + S3

S3 = 15 cm

Q8. A wire of length 20 m is to be folded in the form of a rectangle. How many rectangles can be formed by folding the wire if the sides are positive integers in metres?

Solution: Length of the wire = 20 m

The wire is folded in the form of a rectangle;

Therefore, Perimeter of the rectangle = Length of the wire

2 (Length + Breadth) = 20

Length + Breadth = 20/2

Length + Breadth = 10 m

The length and breadth are positive integers in metres.

Therefore, the possible dimensions are: (1m, 9m), (2m, 8m), (3m, 7m), (4m, 6m) and (5m, 5m)

Thus, five rectangles can be formed with the given wire.

Q 9. A square piece of land has each side equal to 100 m. If 3 layers of metal wire have to be used to fence it, what is the length of the wire?

Solution: Side of the square field = 100 m

Wire required to fence 1 layer of the square field = Perimeter of the square field

Wire required to fence 1 layer of the square field = 4 × (Side of the square field)

Wire required to fence 1 layer of the square field = 4 × 100 = 400 m

The length of wire required to fence 3 layers = 3 × 400 m = 1200 m

Q 10. Shikha runs around a square of side 75m. Priya runs around a rectangle with length 60 m and breadth 45 m. Who covers the smaller distance?

Solution: Distance covered by Shikha = Perimeter of the square

Distance covered by Shikha = 4 × 75 m = 300 m

Similarly, distance covered by Priya = Perimeter of the rectangle

Distance covered by Priya = 2 × (60 + 45)

Distance covered by Priya = 2 × 105 = 210 m

As we can see that distance covered by Priya is less than Shikha. So, Priya covers the smallest distance.

Q 11. The dimensions of a photographs are 30 cm × 20 cm. What length of wooden frame is needed to frame the picture?

Solution: The length of the photograph = 30 cm

The breadth of the photograph = 20 cm

The required length of wooden frame = Perimeter of the photograph

The required length of wooden frame = 2 (Length + Breadth)

The required length of wooden frame = 2 × (30 + 20)

The required length of wooden frame = 2 × 50 cm = 100 cm

Q 12. The length of a rectangular field is 100 m. If its perimeter is 300 m, what is its breadth?

Solution: It is given that;

l = 100 m, P = 300 m

P = 2 (l + b)

300 = 2 (100 + b)

300 = 200 + 2b

2b = 100

b = 50 m

Thus, the breadth of the rectangular field is 50 m.

Q 13. To fix fence wires in a garden, 70 m long and 50 m wide, Arvind bought metal pipes for posts. He fixed a post every 5 metres apart. Each post was 2 m long. What is total length of the pipes he bought for the posts?

Solution: It is given that;

Length of the garden = 70 m, Breadth of the garden = 50 m

Perimeter of the garden = 2 × (Length + Breadth)

Perimeter of the garden = 2 × (70 + 50)

Perimeter of the garden = 2 × 120 = 240 m

It is given that Arvind fixes a post every 5 metres apart.

So, the number of posts required = 240/5 = 48

The length of each post = 2 m

Therefore, Total length of the pipe required = 48 × 2 = 96 m

Q 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.

Solution: It is given that;

Length of the park = 175 m, Breadth of the park = 125 m

Perimeter of the park = 2 × (Length + Breadth)

Perimeter of the park = 2 × (175 + 125)

Perimeter of the park = 2 × 300 = 600 m

Cost of fencing 1 meter = Rs. 12

Cost of fencing 600 m = 12 × 600 = Rs. 7,200

Q 15. The perimeter of a rectangular pentagon is 100 cm. How long is each side?

Solution: It is given that the perimeter of the regular pentagon = 100 cm

A regular pentagon has 5 sides of equal length.

So, Perimeter of the regular pentagon = 5 x Side of the pentagon

100 = 5 × Side of the pentagon

Side of the pentagon = 100 / 5 = 20 cm

Thus, each side of the pentagon is 20 cm.

Q 16. Find the perimeter of a regular hexagon with each side measuring 8 m.

Solution: The side of the hexagon = 8 m

A hexagon is a closed polygon having 6 sides of equal lengths.

So, Perimeter of the hexagon = 6 × Side of the hexagon

Perimeter of the hexagon = 6 × 8 = 48 m

Q 17. A rectangular piece of land measure 0.7 km by 0.5 km. Each side is to be fenced with four rows of wires. What length of the wire is needed?

Solution: Length of the rectangular land = 0.7 km

Breadth of the rectangular land = 0.5 km

Perimeter of the rectangular land = 2 (Length + Breadth)

Perimeter of the rectangular land = 2 (0.7 + 0.5) km

Perimeter of the rectangular land = 2 × 1.2 km = 2.4 km

One row of wire required to fence the land = 2.4 km

Therefore, length of wire required to fence the land with 4 rows of wire = 4 × 2.4 km = 9.6 km

Q 18. Avneet buys 9 square paving slabs, each with a side of ½ m. He lays them in the form of a square.

Class 6 Maths RD Sharma Solutions Chapter 20 Mensuration

(i) What is the perimeter of his arrangement?

Solution: Length of the side of one slab = 1/2 m

In the square arrangement, one side of the square is formed by three slabs.

So, length of the side of the square = 3 × ½ = 3/2 m

The perimeter of the square arrangement = 4 × (Side of the square)

The perimeter of the square arrangement = 4 × 3/2 = 6 m

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

Solution: The cross arrangement consists of 8 sides.

These sides form the periphery of the arrangement and measure 1 m each.

Also, this arrangement consists of other 4 sides that measure 1/2 m each.

So, the perimeter of the cross arrangement = (1 + 1/2 + 1 + 1 + 1/2 + 1 + 1 + 1/2 + 1 + 1 + 1/2 + 1)

= (8 + 2) = 10 m

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(iii) Which has greater perimeter?

Solution: Perimeter of the cross arrangement = 10 m

Perimeter of the square arrangement = 6 m

Thus, the perimeter of the cross arrangement is more than the square arrangement.

(iv) 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 they cannot be broken)

Solution: No, there is no way of arranging these slabs where the perimeter is more than 10 m.

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