{"id":547555,"date":"2021-10-11T05:18:21","date_gmt":"2021-10-11T05:18:21","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=547555"},"modified":"2021-10-11T08:51:19","modified_gmt":"2021-10-11T08:51:19","slug":"rd-sharma-solutions-for-class-6-maths-chapter-20-mensuration","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-6-maths-chapter-20-mensuration\/","title":{"rendered":"RD Sharma Solutions for Class 6 Maths Chapter 20\u2013Mensuration"},"content":{"rendered":"\n

Class 6: Maths Chapter 20 solutions. Complete Class 6 Maths Chapter 20 Notes.<\/p>\n\n\n\n

RD Sharma Solutions for Class 6 Maths Chapter 20\u2013Mensuration<\/h2>\n\n\n\n

RD Sharma 6th Maths Chapter 20, Class 6 Maths Chapter 20 solutions<\/p>\n\n\n\n

Exercise 20.1 page: 20.5<\/h4>\n\n\n\n

1. Which of the following are closed curves? Which of them are simple?<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

The figures (ii), (iii), (iv), (vi) and (vii) are closed curves and the figures (ii), (iii), (iv) and (vi) are simple closed curves.<\/p>\n\n\n\n

2. Define perimeter of a closed figure.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

The length of the boundary of a closed figure is known as its perimeter.<\/p>\n\n\n\n

3. Find the perimeter of each of the following shapes:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We know that perimeter is the sum of lengths of all the sides of a closed figure.<\/p>\n\n\n\n

(i) Perimeter of the given figure = 4 + 2 + 1 + 5 = 12 cm<\/p>\n\n\n\n

(ii) Perimeter of the given figure = 23 + 35 + 40 + 35 = 133 cm<\/p>\n\n\n\n

(iii) Perimeter of the given figure = 15 + 15 + 15 + 15 = 60 cm<\/p>\n\n\n\n

(iv) Perimeter of the given figure = 3 + 3 + 3 + 3 + 3 = 15 cm<\/p>\n\n\n\n

\n\n\n\n

Exercise 20.2 page: 20.10<\/h4>\n\n\n\n

1. Find the perimeters of the rectangles whose lengths and breadths are given below:<\/strong><\/p>\n\n\n\n

(i) 7 cm, 5 cm<\/strong><\/p>\n\n\n\n

(ii) 5 cm, 4 cm<\/strong><\/p>\n\n\n\n

(iii) 7.5 cm, 4.5 cm<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) We know that the perimeter of a rectangle = 2 (L + B)<\/p>\n\n\n\n

It is given that L = 7 cm and B = 5 cm<\/p>\n\n\n\n

So the perimeter of a rectangle = 2 (7 + 5) = 2 \u00d7 12 = 24 cm<\/p>\n\n\n\n

(ii) We know that the perimeter of a rectangle = 2 (L + B)<\/p>\n\n\n\n

It is given that L = 5 cm and B = 4 cm<\/p>\n\n\n\n

So the perimeter of a rectangle = 2 (5 + 4) = 2 \u00d7 9 = 18 cm<\/p>\n\n\n\n

(iii) We know that the perimeter of a rectangle = 2 (L + B)<\/p>\n\n\n\n

It is given that L = 7.5 cm and B = 4.5 cm<\/p>\n\n\n\n

So the perimeter of a rectangle = 2 (7.5 + 4.5) = 2 \u00d7 12 = 24 cm<\/p>\n\n\n\n

2. Find the perimeters of the squares whose sides are given below:<\/strong><\/p>\n\n\n\n

(i) 10 cm<\/strong><\/p>\n\n\n\n

(ii) 5 m<\/strong><\/p>\n\n\n\n

(iii) 115.5 cm<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) We know that the perimeter of a square = 4 \u00d7 Length of one side<\/p>\n\n\n\n

It is given that L = 10 cm<\/p>\n\n\n\n

So the perimeter of a square = 4 \u00d7 10 = 40 cm<\/p>\n\n\n\n

(ii) We know that the perimeter of a square = 4 \u00d7 Length of one side<\/p>\n\n\n\n

It is given that L = 5 m<\/p>\n\n\n\n

So the perimeter of a square = 4 \u00d7 5 = 20 m<\/p>\n\n\n\n

(iii) We know that the perimeter of a square = 4 \u00d7 Length of one side<\/p>\n\n\n\n

It is given that L = 115.5 cm<\/p>\n\n\n\n

So the perimeter of a square = 4 \u00d7 115.5 = 462 cm<\/p>\n\n\n\n

3. Find the side of the square whose perimeter is:<\/strong><\/p>\n\n\n\n

(i) 16 m<\/strong><\/p>\n\n\n\n

(ii) 40 cm<\/strong><\/p>\n\n\n\n

(iii) 22 cm<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) We know that side of a square = perimeter\/ 4<\/p>\n\n\n\n

It is given that perimeter = 16 m<\/p>\n\n\n\n

So the side of the square = 16\/4 = 4 m<\/p>\n\n\n\n

(ii) We know that side of a square = perimeter\/ 4<\/p>\n\n\n\n

It is given that perimeter = 40 cm<\/p>\n\n\n\n

So the side of the square = 40\/4 = 10 cm<\/p>\n\n\n\n

(iii) We know that side of a square = perimeter\/ 4<\/p>\n\n\n\n

It is given that perimeter = 22 cm<\/p>\n\n\n\n

So the side of the square = 22\/4 = 5.5 cm<\/p>\n\n\n\n

4. Find the breadth of the rectangle whose perimeter is 360 cm and whose length is<\/strong><\/p>\n\n\n\n

(i) 116 cm<\/strong><\/p>\n\n\n\n

(ii) 140 cm<\/strong><\/p>\n\n\n\n

(iii) 102 cm<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We know that the perimeter of a rectangle = 2 (L + B)<\/p>\n\n\n\n

So the breadth of the rectangle = perimeter\/2 \u2013 length<\/p>\n\n\n\n

(i) It is given that perimeter = 360 cm and length = 116 cm<\/p>\n\n\n\n

So the breadth of the rectangle = 360\/2 \u2013 116 = 180 \u2013 116 = 64 cm<\/p>\n\n\n\n

(ii) It is given that perimeter = 360 cm and length = 140 cm<\/p>\n\n\n\n

So the breadth of the rectangle = 360\/2 \u2013 140 = 180 \u2013 140 = 40 cm<\/p>\n\n\n\n

(iii) It is given that perimeter = 360 cm and length = 102 cm<\/p>\n\n\n\n

So the breadth of the rectangle = 360\/2 \u2013 102 = 180 \u2013 102 = 78 cm<\/p>\n\n\n\n

5. A rectangular piece of lawn is 55 m wide and 98 m long. Find the length of the fence around it.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

The dimensions of lawn are<\/p>\n\n\n\n

Breadth = 55 m<\/p>\n\n\n\n

Length = 98 m<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

Perimeter of lawn = 2 (L + B)<\/p>\n\n\n\n

By substituting the values<\/p>\n\n\n\n

Perimeter of lawn = 2 (98 +55)<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

Perimeter of lawn = 2 \u00d7 153 = 306 m<\/p>\n\n\n\n

Hence, the length of the fence around the lawn is 306 m.<\/p>\n\n\n\n

6. The side of a square field is 65 m. What is the length of the fence required all around it?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Side of a square field = 65 m<\/p>\n\n\n\n

So the perimeter of square field = 4 \u00d7 side of the square<\/p>\n\n\n\n

By substituting the values<\/p>\n\n\n\n

Perimeter of square field = 4 \u00d7 65 = 260 m<\/p>\n\n\n\n

Hence, the length of the fence required all around the square field is 260 m.<\/p>\n\n\n\n

7. Two sides of a triangle are 15 cm and 20 cm. The perimeter of the triangle is 50 cm. What is the third side?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

First side of triangle = 15 cm<\/p>\n\n\n\n

Second side of triangle = 20 cm<\/p>\n\n\n\n

In order to find the length of third side<\/p>\n\n\n\n

We know that perimeter of a triangle is the sum of all three sides of a triangle<\/p>\n\n\n\n

So the length of third side = perimeter of triangle \u2013 sum of length of other two sides<\/p>\n\n\n\n

By substituting the values<\/p>\n\n\n\n

Length of third side = 50 \u2013 (15 + 20) = 15 cm.<\/p>\n\n\n\n

Hence, the length of third side is 15 cm.<\/p>\n\n\n\n

8. 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?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

Length of wire 20 m is folded in the form of rectangle<\/p>\n\n\n\n

So the perimeter = 20 m<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

2 (L + B) = 20 m<\/p>\n\n\n\n

On further calculation<\/p>\n\n\n\n

L + B = 10 m<\/p>\n\n\n\n

If the sides are positive integers in metres the possible dimensions are (1m, 9m), (2m, 8m), (3m, 7m), (4m, 6m) and (5m, 5m)<\/p>\n\n\n\n

Hence, five rectangles can be formed using the given wire.<\/p>\n\n\n\n

9. A square piece of land has each side equal to 100 m. If 3 layers of metal wire has to be used to fence it, what is the length of the wire needed?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Each side of a square field = 100 m<\/p>\n\n\n\n

We can find the wire required to fence the square field by determining the perimeter = 4 \u00d7 each side of a square field<\/p>\n\n\n\n

By substituting the values<\/p>\n\n\n\n

Perimeter of the square field = 4 \u00d7 100 = 400 m<\/p>\n\n\n\n

So the length of wire which is required to fence three layers is = 3 \u00d7 400 = 1200 m<\/p>\n\n\n\n

Hence, the length of wire needed to fence 3 layers is 1200 m.<\/p>\n\n\n\n

10. Shikha runs around a square of side 75 m. Priya runs around a rectangle with length 60 m and breadth 45 m. Who covers the smaller distance?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Shikha runs around a square of side = 75 m<\/p>\n\n\n\n

So the perimeter = 4 \u00d7 75 = 300 m<\/p>\n\n\n\n

Priya runs around a rectangle having<\/p>\n\n\n\n

Length = 60 m<\/p>\n\n\n\n

Breadth = 45 m<\/p>\n\n\n\n

So the distance covered can be found from the perimeter = 2 (L + B)<\/p>\n\n\n\n

By substituting the values<\/p>\n\n\n\n

Perimeter = 2 (60 + 45) = 2 \u00d7 105 = 210 m<\/p>\n\n\n\n

Hence, Priya covers the smaller distance of 210 m.<\/p>\n\n\n\n

11. The dimensions of a photographs are 30 cm \u00d7 20 cm. What length of wooden frame is needed to frame the picture?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Dimensions of a photographs = 30 cm \u00d7 20 cm<\/p>\n\n\n\n

So the required length of the wooden frame can be determined from the perimeter of the photograph = 2 (L + B)<\/p>\n\n\n\n

By substituting the values = 2 (30 + 20) = 2 \u00d7 50 = 100 cm<\/p>\n\n\n\n

Hence, the length of the wooden frame required to frame the picture is 100 cm.<\/p>\n\n\n\n

12. The length of a rectangular field is 100 m. If the perimeter is 300 m, what is its breadth?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

The dimensions of rectangular field are<\/p>\n\n\n\n

Length = 100 m<\/p>\n\n\n\n

Perimeter = 300 m<\/p>\n\n\n\n

We know that perimeter = 2 (L + B)<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

Breadth = perimeter\/2 \u2013 length<\/p>\n\n\n\n

By substituting the values<\/p>\n\n\n\n

Breadth = (300-200)\/2 = 100\/2 = 50 m<\/p>\n\n\n\n

Hence, the breadth of the rectangular field is 50 m.<\/p>\n\n\n\n

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 the total length of the pipes he bought for the posts?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

The dimensions of garden are<\/p>\n\n\n\n

Length = 70 m<\/p>\n\n\n\n

Breadth = 50 m<\/p>\n\n\n\n

So the perimeter = 2 (L + B)<\/p>\n\n\n\n

By substituting the values<\/p>\n\n\n\n

Perimeter = 2 (70 + 50) = 2 \u00d7 120 = 240 m<\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

Arvind fixes a post every 5 metres apart<\/p>\n\n\n\n

No. of posts required = 240\/5 = 48<\/p>\n\n\n\n

The length of each post = 2 m<\/p>\n\n\n\n

So the total length of the pipe required = 48 \u00d7 2 = 96 m<\/p>\n\n\n\n

Hence, the total length of the pipes he bought for the posts is 96 m.<\/p>\n\n\n\n

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.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

The dimensions of the rectangular park are<\/p>\n\n\n\n

Length = 175 m<\/p>\n\n\n\n

Breadth = 125 m<\/p>\n\n\n\n

So the perimeter = 2 (L + B)<\/p>\n\n\n\n

By substituting the values<\/p>\n\n\n\n

Perimeter = 2 (175 + 125) = 2 \u00d7 300 = 600 m<\/p>\n\n\n\n

It is given that the cost of fencing = Rs 12 per meter<\/p>\n\n\n\n

So the total cost of fencing = 12 \u00d7 600 = Rs 7200<\/p>\n\n\n\n

Hence, the cost of fencing a rectangular park is Rs 7200.<\/p>\n\n\n\n

15. The perimeter of a regular pentagon is 100 cm. How long is each side?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We know that a regular pentagon is a closed polygon having 5 sides of same length.<\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Perimeter of a regular pentagon = 100 cm<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

Perimeter = 5 \u00d7 side of the regular pentagon<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

Side of the regular pentagon = Perimeter\/5<\/p>\n\n\n\n

By substituting the values<\/p>\n\n\n\n

Side of the regular pentagon = 100\/5 = 20 cm<\/p>\n\n\n\n

Hence, the side of the regular pentagon measures 20 cm.<\/p>\n\n\n\n

16. Find the perimeter of a regular hexagon with each side measuring 8 m.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We know that a regular hexagon is a closed polygon which has six sides of same length.<\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Side of the regular hexagon = 8 m<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

Perimeter = 6 \u00d7 side of the regular hexagon<\/p>\n\n\n\n

By substituting the values<\/p>\n\n\n\n

Perimeter = 6 \u00d7 8 = 48 m<\/p>\n\n\n\n

Hence, the perimeter of a regular hexagon is 48 m.<\/p>\n\n\n\n

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?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Measure of rectangular piece of land = 0.7 km \u00d7 0.5 km<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

Perimeter = 2 (L + B)<\/p>\n\n\n\n

By substituting the values<\/p>\n\n\n\n

Perimeter = 2 (0.7 + 0.5) = 2 \u00d7 1.2 = 2.4 km<\/p>\n\n\n\n

The above obtained perimeter = one row of wire needed to fence the rectangular piece of land<\/p>\n\n\n\n

So the length of wire needed to fence the land with 4 rows of wire = 4 \u00d7 2.4 = 9.6 km<\/p>\n\n\n\n

Hence, the length of wire needed is 9.6 km.<\/p>\n\n\n\n

18. Avneet buys 9 square paving slabs, each with a side of \u00bd m. He lays them in the form of a square.<\/strong><\/p>\n\n\n\n

(i) What is the perimeter of his arrangement?<\/strong><\/p>\n\n\n\n

(ii) Shari does not like his arrangement. She gets him to lay them out like a cross. What is the perimeter of her arrangement?<\/strong><\/p>\n\n\n\n

(iii) Which has greater perimeter?<\/strong><\/p>\n\n\n\n

(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)<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) It is given that length of each side of the slab = \u00bd m<\/p>\n\n\n\n

One side of the square is formed by three slabs in a square arrangement<\/p>\n\n\n\n

Length of side = 3 \u00d7 \u00bd = 3\/2 m<\/p>\n\n\n\n

So the perimeter of the square arrangement = 4 \u00d7 3\/2 = 6 m<\/p>\n\n\n\n

(ii) From the figure, cross arrangement has 8 sides which form periphery of the arrangement and measure 1 m each.<\/p>\n\n\n\n

It also has 4 sides which measure \u00bd m each<\/p>\n\n\n\n

Perimeter of the cross arrangement = 1 + \u00bd + 1 + 1 + \u00bd + 1 + 1 + \u00bd + 1 + 1 + \u00bd + 1 = 8 + 2 = 10 m<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(iii) We know that<\/p>\n\n\n\n

Perimeter of cross arrangement = 10 m<\/p>\n\n\n\n

Perimeter of square arrangement = 6 m<\/p>\n\n\n\n

Hence, the perimeter of cross arrangement is greater than the perimeter of square arrangement.<\/p>\n\n\n\n

(iv) No, Avneet cannot arrange the slabs having perimeter more than 10 m.<\/p>\n\n\n\n

\n\n\n\n

Exercise 20.3 page: 20.14<\/h4>\n\n\n\n

1. The following figures are drawn on a squared paper. Count the number of squares enclosed by each figure and find its area, taking the area of each square as 1 cm2<\/sup>. (Fig. 20.25).<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) The given shape has 16 complete squares.<\/p>\n\n\n\n

It is given that area of one square = 1 cm2<\/sup><\/p>\n\n\n\n

So the area of the given shape = 16 \u00d7 1 = 16 cm2<\/sup><\/p>\n\n\n\n

(ii) The given shape has 36 complete squares.<\/p>\n\n\n\n

It is given that area of one square = 1 cm2<\/sup><\/p>\n\n\n\n

So the area of the given shape = 36 \u00d7 1 = 36 cm2<\/sup><\/p>\n\n\n\n

(iii) The given shape has 15 complete and 6 half squares.<\/p>\n\n\n\n

It is given that area of one square = 1 cm2<\/sup><\/p>\n\n\n\n

So the area of the given shape = 15 + 6 \u00d7 12 = 18 cm2<\/sup><\/p>\n\n\n\n

(iv) The given shape has 20 complete and 8 half squares.<\/p>\n\n\n\n

It is given that area of one square = 1 cm2<\/sup><\/p>\n\n\n\n

So the area of the given shape = 20 + 8 \u00d7 12 = 24 cm2<\/sup><\/p>\n\n\n\n

(v) The given shape has 13 complete, 8 more than half and 7 less than half squares.<\/p>\n\n\n\n

It is given that area of one square = 1 cm2<\/sup><\/p>\n\n\n\n

So the area of the given shape = 13 + 8 \u00d7 1 = 21 cm2<\/sup><\/p>\n\n\n\n

(vi) The given shape has 8 complete, 6 more than half and 4 less than half squares.<\/p>\n\n\n\n

It is given that area of one square = 1 cm2<\/sup><\/p>\n\n\n\n

So the area of the given shape = 8 + 6 \u00d7 1 = 14 cm2<\/sup><\/p>\n\n\n\n

2. On a squared paper, draw (i) a rectangle, (ii) a triangle (iii) any irregular closed figure. Find the approximate area of each by counting the number of squares complete, more than half and exactly half.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) A rectangle<\/p>\n\n\n\n

The given shape has 18 complete squares<\/p>\n\n\n\n

Assume that area of one square = 1 cm2<\/sup><\/p>\n\n\n\n

So the area of the rectangle = 18 \u00d7 1 = 18 cm2<\/sup><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(ii) A triangle<\/p>\n\n\n\n

The given shape has 4 complete, 6 more than half and 6 less than half squares.<\/p>\n\n\n\n

Assume that area of one square = 1 cm2<\/sup><\/p>\n\n\n\n

So the area of the square = 4 + 6 \u00d7 1 = 10 cm2<\/sup><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(iii) Any irregular figure<\/p>\n\n\n\n

The given shape has 10 complete, 1 exactly half, 7 more than half and 6 less than half squares.<\/p>\n\n\n\n

Assume that area of one square = 1 cm2<\/sup><\/p>\n\n\n\n

So the area of the shape = 10 + 1 \u00d7 12 + 7 \u00d7 1 = 17.5 cm2<\/sup><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

3. Draw any circle on the graph paper. Count the squares and use them to estimate the area of the circular region.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

The given circles has 21 complete, 15 more than half and 8 less than half squares.<\/p>\n\n\n\n

Assume that area of one square = 1 cm2<\/sup><\/p>\n\n\n\n

By neglecting less than half squares, we get<\/p>\n\n\n\n

Area of the circle = 21 + 15 = 36 cm2<\/sup><\/p>\n\n\n\n

4. Use tracing paper and centimetre graph paper to compare the areas of the following pairs of figures:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

With the help of tracing paper trace both the figures on a graph<\/p>\n\n\n\n

Figure (i) has 4 complete, 9 more than half and 9 less than half squares.<\/p>\n\n\n\n

Assume that area of one square = 1 cm2<\/sup><\/p>\n\n\n\n

By neglecting less than half squares, we get<\/p>\n\n\n\n

Area of the shape = 4 + 9 = 13 cm2<\/sup><\/p>\n\n\n\n

Figure (ii) has 8 complete, 11 more than half and 10 less than half squares.<\/p>\n\n\n\n

Assume that area of one square = 1 cm2<\/sup><\/p>\n\n\n\n

By neglecting less than half squares, we get<\/p>\n\n\n\n

Area of the shape = 8 + 11 = 19 cm2<\/sup><\/p>\n\n\n\n

By comparing the areas of both the shapes, we know that the figure (ii) has area greater than that of figure (i).<\/p>\n\n\n\n

\n\n\n\n

Exercise 20.4 page: 20.21<\/h4>\n\n\n\n

1. Find the area of a rectangle, whose<\/strong><\/p>\n\n\n\n

(i) Length = 6 cm, breadth = 3 cm<\/strong><\/p>\n\n\n\n

(ii) Length = 8 cm, breadth = 3 cm<\/strong><\/p>\n\n\n\n

(iii) Length = 4.5 cm, breadth = 2 cm.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) We know that area of a rectangle = L \u00d7 B<\/p>\n\n\n\n

It is given that Length = 6 cm, breadth = 3 cm<\/p>\n\n\n\n

By substituting the values<\/p>\n\n\n\n

Area of a rectangle = 6 \u00d7 3 = 18 cm2<\/sup><\/p>\n\n\n\n

(ii) We know that area of a rectangle = L \u00d7 B<\/p>\n\n\n\n

It is given that Length = 8 cm, breadth = 3 cm<\/p>\n\n\n\n

By substituting the values<\/p>\n\n\n\n

Area of a rectangle = 8 \u00d7 3 = 24 cm2<\/sup><\/p>\n\n\n\n

(iii) We know that area of a rectangle = L \u00d7 B<\/p>\n\n\n\n

It is given that Length = 4.5 cm, breadth = 2 cm<\/p>\n\n\n\n

By substituting the values<\/p>\n\n\n\n

Area of a rectangle = 4.5 \u00d7 2 = 9 cm2<\/sup><\/p>\n\n\n\n

2. Find the area of a square whose side is:<\/strong><\/p>\n\n\n\n

(i) 5 cm<\/strong><\/p>\n\n\n\n

(ii) 4.1 cm<\/strong><\/p>\n\n\n\n

(iii) 5.5 cm<\/strong><\/p>\n\n\n\n

(iv) 2.6 cm<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) We know that area of a square = side \u00d7 side<\/p>\n\n\n\n

It is given that side of a square = 5 cm<\/p>\n\n\n\n

So the area of the square = 5 \u00d7 5 = 25 cm2<\/sup><\/p>\n\n\n\n

(ii) We know that area of a square = side \u00d7 side<\/p>\n\n\n\n

It is given that side of a square = 4.1 cm<\/p>\n\n\n\n

So the area of the square = 4.1 \u00d7 4.1 = 16.81 cm2<\/sup><\/p>\n\n\n\n

(iii) We know that area of a square = side \u00d7 side<\/p>\n\n\n\n

It is given that side of a square = 5.5 cm<\/p>\n\n\n\n

So the area of the square = 5.5 \u00d7 5.5 = 30.25 cm2<\/sup><\/p>\n\n\n\n

(iv) We know that area of a square = side \u00d7 side<\/p>\n\n\n\n

It is given that side of a square = 2.6 cm<\/p>\n\n\n\n

So the area of the square = 2.6 \u00d7 2.6 = 6.76 cm2<\/sup><\/p>\n\n\n\n

3. The area of a rectangle is 49 cm2<\/sup> and its breadth is 2.8 cm. Find the length of the rectangle.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that area of a rectangle = 49 cm2<\/sup><\/p>\n\n\n\n

Breadth of a rectangle = 2.8 cm<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

Area of a rectangle = L \u00d7 B<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

L = Area\/B = 49\/2.8 = 17.5 cm<\/p>\n\n\n\n

Hence, the length of the rectangle is 17.5 cm.<\/p>\n\n\n\n

4. The side of a square is 70 cm. Find its area and perimeter.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that side of a square = 70 cm<\/p>\n\n\n\n

We know that area of a square = side \u00d7 side<\/p>\n\n\n\n

By substituting the values<\/p>\n\n\n\n

Area of a square = 70 \u00d7 70 = 4900 cm2<\/sup><\/p>\n\n\n\n

We know that perimeter of a square = 4 \u00d7 side<\/p>\n\n\n\n

By substituting the values<\/p>\n\n\n\n

Perimeter of a square = 4 \u00d7 70 = 280 cm<\/p>\n\n\n\n

Hence, the area of square is 4900 cm2<\/sup> and the perimeter of square is 280 cm.<\/p>\n\n\n\n

5. The area of a rectangle is 225 cm2 <\/sup>and its one side is 25 cm, find its other side.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Area of a rectangle = 225 cm2<\/sup><\/p>\n\n\n\n

Length of one side = 25 cm<\/p>\n\n\n\n

We know that area of a rectangle = Product of length of two sides<\/p>\n\n\n\n

So the other side = area\/side<\/p>\n\n\n\n

By substituting the values<\/p>\n\n\n\n

Other side = 225\/25 = 9 cm<\/p>\n\n\n\n

Hence, the other side of the rectangle is 9 cm.<\/p>\n\n\n\n

6. What will happen to the area of rectangle if its<\/strong><\/p>\n\n\n\n

(i) Length and breadth are trebled<\/strong><\/p>\n\n\n\n

(ii) Length is doubled and breadth is same<\/strong><\/p>\n\n\n\n

(iii) Length is doubled and breadth is halved.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) Length and breadth are trebled<\/p>\n\n\n\n

Consider l as the initial length and b as the initial breadth<\/p>\n\n\n\n

So the original area = l \u00d7 b<\/p>\n\n\n\n

If the length and breadth are trebled it becomes three times more than the original value<\/p>\n\n\n\n

New length = 3l<\/p>\n\n\n\n

New breadth = 3b<\/p>\n\n\n\n

New area of the rectangle = 3l \u00d7 3b = 9lb<\/p>\n\n\n\n

Hence, the area of the rectangle becomes 9 times more than its original area.<\/p>\n\n\n\n

(ii) Length is doubled and breadth is same<\/p>\n\n\n\n

Consider l as the initial length and b as the initial breadth<\/p>\n\n\n\n

So the original area = l \u00d7 b<\/p>\n\n\n\n

If the length is doubled and breadth is same we get<\/p>\n\n\n\n

New length = 2l<\/p>\n\n\n\n

New breadth = b<\/p>\n\n\n\n

New area of the rectangle = 2l \u00d7 b = 2lb<\/p>\n\n\n\n

Hence, the area of the rectangle becomes 2 times more than the original area.<\/p>\n\n\n\n

(iii) Length is doubled and breadth is halved<\/p>\n\n\n\n

Consider l as the initial length and b as the initial breadth<\/p>\n\n\n\n

So the original area = l \u00d7 b<\/p>\n\n\n\n

If the length is doubled and breadth is halved we get<\/p>\n\n\n\n

New length = 2l<\/p>\n\n\n\n

New breadth = b\/2<\/p>\n\n\n\n

New area of the rectangle = 2l \u00d7 b\/2 = lb<\/p>\n\n\n\n

Hence, the area of the rectangle does not change.<\/p>\n\n\n\n

7. What will happen to the area of a square if its side is:<\/strong><\/p>\n\n\n\n

(i) Tripled<\/strong><\/p>\n\n\n\n

(ii) Increased by half of it.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) Tripled<\/p>\n\n\n\n

Consider s as the original side of the square<\/p>\n\n\n\n

We know that original area = s \u00d7 s = s2<\/sup><\/p>\n\n\n\n

If the side of the square is tripled we get<\/p>\n\n\n\n

New side = 3s<\/p>\n\n\n\n

So the new area of the square = 3s \u00d7 3s = 9s2<\/sup><\/p>\n\n\n\n

Hence, the area becomes 9 times more than that of the original area.<\/p>\n\n\n\n

(ii) Increased by half of it<\/p>\n\n\n\n

Consider s as the original side of the square<\/p>\n\n\n\n

We know that original area = s \u00d7 s = s2<\/sup><\/p>\n\n\n\n

If the side of the square is increased by half of it we get<\/p>\n\n\n\n

New side = s + s\/2 = 3s\/2<\/p>\n\n\n\n

So the new area of the square = 3s\/2 \u00d7 3s\/2 = 9s2<\/sup>\/4<\/p>\n\n\n\n

Hence, the area becomes 9\/4 times more than that of the original area.<\/p>\n\n\n\n

8. Find the perimeter of a rectangle whose area is 500 cm2<\/sup> and breadth is 20 cm.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Area of the rectangle = 500 cm2<\/sup><\/p>\n\n\n\n

Breadth of the rectangle = 20 cm<\/p>\n\n\n\n

We know that area = L \u00d7 B<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

L = Area\/B<\/p>\n\n\n\n

By substituting the values<\/p>\n\n\n\n

L = 500\/20 = 25 cm<\/p>\n\n\n\n

We know that perimeter = 2 (L + B)<\/p>\n\n\n\n

By substituting the values<\/p>\n\n\n\n

Perimeter = 2 (25 + 20) = 2 \u00d7 45 = 90 cm<\/p>\n\n\n\n

Hence, the perimeter of the rectangle is 90 cm.<\/p>\n\n\n\n

9. A rectangle has the area equal to that of a square of side 80 cm. If the breadth of the rectangle is 20 cm, find its length.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Side of a square = 80 cm<\/p>\n\n\n\n

So the area of the square = side \u00d7 side<\/p>\n\n\n\n

By substituting the values<\/p>\n\n\n\n

Area of square = 80 \u00d7 80 = 6400 cm2<\/sup><\/p>\n\n\n\n

We know that area of rectangle = area of square = 6400 cm2<\/sup><\/p>\n\n\n\n

Breadth = 20 cm<\/p>\n\n\n\n

Area of rectangle = L \u00d7 B<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

L = Area\/B = 6400\/20 = 320 cm<\/p>\n\n\n\n

Hence, the length of the rectangle is 320 cm.<\/p>\n\n\n\n

10. Area of a rectangle of breadth 17 cm is 340 cm2<\/sup>. Find the perimeter of the rectangle.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

The dimensions of rectangle are<\/p>\n\n\n\n

Breadth = 17 cm<\/p>\n\n\n\n

Area = 340 cm2<\/sup><\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

Area of rectangle = L \u00d7 B<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

L = Area\/B = 340\/17 = 20 cm<\/p>\n\n\n\n

So the perimeter = 2 (L + B)<\/p>\n\n\n\n

By substituting the values<\/p>\n\n\n\n

Perimeter = 2 (20 + 17) = 2 \u00d7 37 = 74 cm<\/p>\n\n\n\n

Hence, the perimeter of the rectangle is 74 cm.<\/p>\n\n\n\n

11. A marble tile measures 15 cm \u00d7 20 cm. How many tiles will be required to cover a wall of size 4 m \u00d7 6 m?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Measure of marble tile = 15 cm \u00d7 20 cm<\/p>\n\n\n\n

Size of wall = 4 m \u00d7 6 m = 400 cm \u00d7 600 cm<\/p>\n\n\n\n

So we get area of tile = 15 cm \u00d7 20 cm = 300 cm2<\/sup><\/p>\n\n\n\n

Area of wall = 400 cm \u00d7 600 cm = 240000 cm2<\/sup><\/p>\n\n\n\n

No. of tiles required to cover the wall = Area of wall\/ Area of one tile<\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

No. of tiles required to cover the wall = 240000\/300 = 800 tiles<\/p>\n\n\n\n

Hence, 800 tiles are required to cover a wall of size 4 m \u00d7 6 m.<\/p>\n\n\n\n

12. A marble tile measures 10 cm \u00d7 12 cm. How many tiles will be required to cover a wall of size 3 m \u00d7 4 m? Also, find the total cost of the tiles at the rate of Rs 2 per tile.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Measure of marble tile = 10 cm \u00d7 12 cm<\/p>\n\n\n\n

Size of the wall = 3 m \u00d7 4 m = 300 cm \u00d7 400 cm<\/p>\n\n\n\n

So the area of marble tile = 10 cm \u00d7 12 cm = 120 cm2<\/sup><\/p>\n\n\n\n

Area of wall = 300 cm \u00d7 400 cm = 120000 cm2<\/sup><\/p>\n\n\n\n

No. of tiles required to cover the wall = Area of wall\/ Area of one tile<\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

No. of tiles required to cover the wall = 120000\/120 = 1000 tiles<\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Cost of one tile = Rs 2<\/p>\n\n\n\n

So the cost of 1000 tiles = 1000 \u00d7 2 = Rs 2000<\/p>\n\n\n\n

Hence, 1000 number of tiles are required to cover the wall and the cost is Rs 2000.<\/p>\n\n\n\n

13. One side of a square plot is 250 m, find the cost of levelling it at the rate of Rs 2 per square metre.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

It is given that<\/p>\n\n\n\n

Side of one tile of a square plot = 250 m<\/p>\n\n\n\n

So the area = side \u00d7 side = 250 \u00d7 250 = 62500 m2<\/sup><\/p>\n\n\n\n

Cost of levelling = Rs 2 per square meter<\/p>\n\n\n\n

So the cost of levelling 62500 m2<\/sup> = 62500 \u00d7 2 = Rs 125000<\/p>\n\n\n\n

Hence, the cost of levelling is Rs 125000.<\/p>\n\n\n\n

14. The following figures have been split into rectangles. Find their areas. (The measures are given in centimetres)<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) The given figure has two rectangles II and IV and two squares I and III.<\/p>\n\n\n\n

So the area of square I = side \u00d7 side = 3 \u00d7 3 = 9 cm2<\/sup><\/p>\n\n\n\n

The same way area of rectangle II = L \u00d7 B = 2 \u00d7 1 = 2 cm2<\/sup><\/p>\n\n\n\n

Area of square III = side \u00d7 side = 3 \u00d7 3 = 9 cm2<\/sup><\/p>\n\n\n\n

Similarly area of rectangle IV = L \u00d7 B = 2 \u00d7 4 = 8 cm2<\/sup><\/p>\n\n\n\n

So the total area of the figure = Area of square I + Area of rectangle II + Area of square III + Area of rectangle IV<\/p>\n\n\n\n

By substituting the values<\/p>\n\n\n\n

Total area of the figure = 9 + 2 + 9 + 8 = 28 cm2<\/sup><\/p>\n\n\n\n\n\n

(ii) The given figure has three rectangles I, II and III.<\/p>\n\n\n\n

So the area of rectangle I = L \u00d7 B = 3 \u00d7 1 = 3 cm2<\/sup><\/p>\n\n\n\n

Area of rectangle II = L \u00d7 B = 3 \u00d7 1 = 3 cm2<\/sup><\/p>\n\n\n\n

Area of rectangle III = L \u00d7 B = 3 \u00d7 1 = 3 cm2<\/sup><\/p>\n\n\n\n

So the total area of the figure = Area of rectangle I + Area of rectangle II + Area of rectangle III<\/p>\n\n\n\n

By substituting the values<\/p>\n\n\n\n

Total area of the figure = 3 + 3 + 3 = 9 cm2<\/sup><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

15. Split the following shapes into rectangles and find the area of each. (The measures are given in centimetres)<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) The given figure has two rectangles I and II.<\/p>\n\n\n\n

So the area of rectangle I = L \u00d7 B = 10 \u00d7 2 = 20 cm2<\/sup><\/p>\n\n\n\n

In the same way area of rectangle II = L \u00d7 B = 10 \u00d7 3\/2 = 15 cm2<\/sup><\/p>\n\n\n\n

So the total area of the figure = Area of rectangle I + Area of rectangle II<\/p>\n\n\n\n

By substituting the values<\/p>\n\n\n\n

Total area of the figure = 20 + 15 = 35 cm2<\/sup><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(ii) The given figure has two squares I and III and one rectangle II.<\/p>\n\n\n\n

So the area of square I = Area of square III = side \u00d7 side = 7 \u00d7 7 = 49 cm2<\/sup><\/p>\n\n\n\n

The area of rectangle II = 21 \u00d7 7 = 147 cm2<\/sup><\/p>\n\n\n\n

So the total area of the figure = Area of square I + Area of rectangle II + Area of square III<\/p>\n\n\n\n

By substituting the values<\/p>\n\n\n\n

Total area of the figure = 49 + 49 + 147 = 245 cm2<\/sup><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

(iii) The given figure has two rectangles I and II.<\/p>\n\n\n\n

So the area of rectangle I = L \u00d7 B = 5 \u00d7 1 = 5 cm2<\/sup><\/p>\n\n\n\n

The same way, area of rectangle II = L \u00d7 B = 4 \u00d7 1 = 4 cm2<\/sup><\/p>\n\n\n\n

So the total area of the figure = Area of rectangle I + Area of rectangle II<\/p>\n\n\n\n

By substituting the values<\/p>\n\n\n\n

Total area of the figure = 5 + 4 = 9 cm2<\/sup><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

16. How many tiles with dimensions 5 cm and 12 cm will be needed to fit a region whose length and breadth are respectively:<\/strong><\/p>\n\n\n\n

(i) 100 cm and 144 cm<\/strong><\/p>\n\n\n\n

(ii) 70 cm and 36 cm<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) Tile dimensions = 5 cm \u00d7 12 cm<\/p>\n\n\n\n

Region dimensions = 100 cm \u00d7 144 cm<\/p>\n\n\n\n

So the area of tile = 5 cm \u00d7 12 cm = 60 cm2<\/sup><\/p>\n\n\n\n

Similarly area of region = 100 cm \u00d7 144 cm = 14400 cm2<\/sup><\/p>\n\n\n\n

No. of tiles which is required to cover the region = Area of region\/ Area of one tile<\/p>\n\n\n\n

By substituting the values<\/p>\n\n\n\n

No. of tiles which is required to cover the region = 14400\/60 = 240 tiles<\/p>\n\n\n\n

(ii) Tile dimensions = 5 cm \u00d7 12 cm<\/p>\n\n\n\n

Region dimensions = 70 cm \u00d7 36 cm<\/p>\n\n\n\n

So the area of tile = 5 cm \u00d7 12 cm = 60 cm2<\/sup><\/p>\n\n\n\n

Similarly area of region = 70 cm \u00d7 36 cm = 2520 cm2<\/sup><\/p>\n\n\n\n

No. of tiles which is required to cover the region = Area of region\/ Area of one tile<\/p>\n\n\n\n

By substituting the values<\/p>\n\n\n\n

No. of tiles which is required to cover the region = 2520\/60 = 42 tiles<\/p>\n\n\n\n

\n\n\n\n

Objective Type Questions page: 20.23<\/h4>\n\n\n\n

Mark the correct alternative in each of the following:<\/strong><\/p>\n\n\n\n

1. The sides of a rectangle are in the ratio 5: 4. If its perimeter is 72 cm, then its length is
(a) 40 cm
(b) 20 cm
(c) 30 cm
(d) 60 cm<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

The option (b) is the correct answer.<\/p>\n\n\n\n

Consider the sides of the rectangle as 5x <\/em>and 4x. <\/p>\n\n\n\n

We know that, perimeter of rectangle = 2 (Length + Breadth)

By substituting the values<\/p>\n\n\n\n

72 = 2 (5x <\/em>+ 4x)<\/p>\n\n\n\n

On further calculation<\/p>\n\n\n\n

72 = 2 \u00d7 9x<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

72 = 18x<\/p>\n\n\n\n

By division<\/p>\n\n\n\n

x <\/em>= 72\/18 = 4<\/p>\n\n\n\n

Hence, the length of the rectangle = 5x <\/em>= 5 \u00d7 4 = 20 cm<\/p>\n\n\n\n

2. The cost of fencing a rectangular field 34 m long and 18 m wide at Rs 2.25 per meter is
(a) Rs 243
(b) Rs 234
(c) Rs 240
(d) Rs 334<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

The option (b) is the correct answer.<\/p>\n\n\n\n

We must find the perimeter of the rectangle for fencing the field.

The dimensions of the rectangle are<\/p>\n\n\n\n

Length = 34m

Breadth = 18m<\/p>\n\n\n\n

We know that Perimeter = 2 (Length + Breadth)<\/p>\n\n\n\n

By substituting the values<\/p>\n\n\n\n

Perimeter of the rectangle = 2 (34 + 18) = 2 \u00d7 52 = 104 m<\/p>\n\n\n\n

So the cost of fencing the field at the rate of Rs. 2.25 per meter = 104 \u00d7 2.25 = Rs. 234<\/p>\n\n\n\n

3. If the cost of fencing a rectangular field at Rs. 7.50 per meter is Rs. 600, and the length of the field is 24 m, then the breadth of the field is
(a) 8 m
(b) 18 m
(c) 24 m
(d) 16 m<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

The option (d) is the correct answer.<\/p>\n\n\n\n

It is given that cost of fencing the rectangular field = Rs. 600<\/p>\n\n\n\n

So the rate of fencing the field = Rs. 7.50 per m<\/p>\n\n\n\n

We know that perimeter of the field = Cost of fencing\/Rate of fencing <\/p>\n\n\n\n

By substituting the values<\/p>\n\n\n\n

Perimeter of the field = 600\/7.50 = 80 m

Length of the field = 24 m<\/p>\n\n\n\n

So we get breadth of the field = Perimeter\/2- Length = 80\/2- 24 = 16 m<\/p>\n\n\n\n

4. The cost of putting a fence around a square field at Rs 2.50 per meter is Rs 200. The length of each side of the field is
(a) 80 m
(b) 40 m
(c) 20 m
(d) None of these<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

The option (c) is the correct answer.<\/p>\n\n\n\n

It is given that cost of fencing the square field = Rs. 200<\/p>\n\n\n\n

So the rate of fencing the field = Rs. 2.50<\/p>\n\n\n\n

We know that, perimeter of the square field = Cost of fencing\/Rate of fencing <\/p>\n\n\n\n

By substituting the values<\/p>\n\n\n\n

Perimeter of the square field = 200\/2.50 = 80 m<\/p>\n\n\n\n

Perimeter of square = 4 \u00d7 Side of the square

It can be written as

Side of the square = Perimeter\/4 = 80\/4 = 20 m<\/p>\n\n\n\n

5. The length of a rectangle is three times of its width. If the length of the diagonal is 8\u221a10 m, then the perimeter of the rectangle is
(a) 15\u221a10 m
(b) 16\u221a10 m
(c) 24\u221a10 m
(d) 64 m<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

The option (d) is the correct answer.<\/p>\n\n\n\n

Consider ABCD as a rectangle.

Assume that the width of the rectangle BC = x m<\/p>\n\n\n\n

We know that the length is three times width of the rectangle.

So, length of the rectangle AB = 3x m<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

AC is the diagonal of rectangle

Consider ABC as a right angled triangle.<\/p>\n\n\n\n

AC2<\/sup> = AB2<\/sup> + BC2<\/sup>                                <\/p>\n\n\n\n

By substituting the values<\/p>\n\n\n\n

640 = 9x2<\/sup> + x2<\/sup><\/p>\n\n\n\n

We get<\/p>\n\n\n\n

640 = 10x2<\/sup><\/p>\n\n\n\n

On further calculation<\/p>\n\n\n\n

x2<\/sup> = 640\/10 = 64<\/p>\n\n\n\n

x = \u221a64 = 8 m<\/p>\n\n\n\n

So the breadth of the rectangle x <\/em>= 8 m<\/p>\n\n\n\n

Length of the rectangle 3x = 3 \u00d7 8 = 24 m

Perimeter = 2 (Length + Breadth)<\/p>\n\n\n\n

By substituting the values<\/p>\n\n\n\n

Perimeter = 2 (24 + 8) = 2 \u00d7 32 = 64 m<\/p>\n\n\n\n

6. If a diagonal of a rectangle is thrice its smaller side, then its length and breadth are in the ratio
(a) 3: 1
(b) \u221a3: 1
(c) \u221a2 : 1
(d) 2\u221a2: 1<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

The option (d) is the correct answer.<\/p>\n\n\n\n

Assume that the length of the smaller side of the rectangle BC = x<\/p>\n\n\n\n

Length of the larger side AB = y<\/p>\n\n\n\n

It is given that the length of the diagonal is three times that of the smaller side.

<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Diagonal of the rectangle 3x = AC<\/p>\n\n\n\n

By using Pythagoras theorem<\/p>\n\n\n\n

(AC) 2<\/sup> = (AB) 2   <\/sup>+ (BC) 2<\/sup><\/p>\n\n\n\n

By substituting the values<\/p>\n\n\n\n

(3x) 2<\/sup> = (x) 2<\/sup> + (y) 2<\/sup><\/p>\n\n\n\n

On further calculation<\/p>\n\n\n\n

9x2<\/sup> = x2<\/sup> + y2<\/sup><\/p>\n\n\n\n

We get<\/p>\n\n\n\n

8x2<\/sup> = y2<\/sup><\/p>\n\n\n\n

By taking square roots of both sides,<\/p>\n\n\n\n

2\u221a2 x <\/em>= y<\/p>\n\n\n\n

Hence, the ratio of the larger side to the smaller side is 2\u221a2: 1.<\/p>\n\n\n\n

7. The ratio of the areas of two squares, one having its diagonal double than the other, is
(a) 1: 2
(b) 2: 3
(c) 3: 1
(d) 4: 1<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

The option (d) is the correct answer.<\/p>\n\n\n\n

Consider ABCD and PQRS as the two squares. We know that, the diagonal of square PQRS is twice the diagonal of square ABCD.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

PR = 2 AC<\/p>\n\n\n\n

Area of the square = Diagonal2<\/sup>\/2

Area of PQRS = PR2<\/sup>\/2 

In the same way, area of ABCD = AC2<\/sup>\/2<\/p>\n\n\n\n

From the question, we know that: 

If AC = x units, we get PR = 2x units

Area of PQRS\/Area of ABCD = (PR2<\/sup>\u00d72)\/ (2\u00d7AC2<\/sup>) <\/p>\n\n\n\n

By substituting the value<\/p>\n\n\n\n

Area of PQRS\/Area of ABCD = [(2x) 2<\/sup>\u00d72]\/ [2\u00d7x2<\/sup>] = 4\/1<\/p>\n\n\n\n

Hence, the ratio of the areas of squares PQRS and ABCD is 4: 1.<\/p>\n\n\n\n

8. If the ratio of areas of two squares is 225 : 256, then the ratio of their perimeters is
(a) 225 : 256
(b) 256 : 225
(c) 15 : 16
(d) 16 : 15<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

The option (c) is the correct answer.<\/p>\n\n\n\n

Consider ABCD and PQRS as the two squares.

Let the lengths of each side of ABCD and PQRS be x and y.<\/p>\n\n\n\n

We know that

Area of sq. ABCD\/Area of sq. PQRS = x2<\/sup>\/y2<\/sup>

So we get x2<\/sup>\/y2<\/sup>= 225\/256

By taking square roots on both sides,

x\/y = 15\/16<\/p>\n\n\n\n

By taking the ratio of their perimeters, we get

Perimeter of sq. ABCD\/Primeter of sq. PQRS = (4 \u00d7 side of sq. ABCD)\/ (4 \u00d7 side of sq. PQRS) = 4x\/4y

By removing common terms in numerator and denominator<\/p>\n\n\n\n

Perimeter of sq. ABCD\/Perimeter of sq. PQRS = x\/ y<\/p>\n\n\n\n

So perimeter of sq. ABCD\/Perimeter of sq. PQRS = 15\/16<\/p>\n\n\n\n

Hence, the ratio of their perimeters = 15: 16<\/p>\n\n\n\n

9. If the sides of a square are halved, then its area
(a) remains same
(b) becomes half
(c) becomes one fourth
(d) becomes double<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

The option (c) is the correct answer.<\/p>\n\n\n\n

Consider x as the side of the square.

We know that area of a square = Side \u00d7 Side = x \u00d7 x = x2<\/sup>

If the sides are halved, we get new side = x\/2<\/p>\n\n\n\n

So the new area = (x\/2)2<\/sup>= x2<\/sup>\/4<\/p>\n\n\n\n

From this we know that the area has become one fourth of its previous value.<\/p>\n\n\n\n

10. A rectangular carpet has area 120 m2<\/sup> and perimeter 46 meters. The length of its diagonal is
(a) 15 m
(b) 16 m
(c) 17 m
(d) 20 m<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

The option (c) is the correct answer.<\/p>\n\n\n\n

It is given that area of the rectangle = 120 m2<\/sup>

Perimeter of the rectangle = 46 m

Consider l and b as the length and breadth.

Area of the rectangle = l \u00d7 b = 120 m2<\/sup>                 

Perimeter of the rectangle = 2 (l <\/em>+ b) = 46<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

(l <\/em>+ b) = 46\/2 = 23 m             

Length of the diagonal of the rectangle = l2<\/sup> + b2<\/sup>

It can be written as

(l2<\/sup> + b2<\/sup>) = (l <\/em>+ b) 2<\/sup> \u2013 2 (l \u00d7 b)            <\/p>\n\n\n\n

By substituting the values

(l2<\/sup> + b2<\/sup>) = (23)2<\/sup> \u2013 2 (120) = 529 \u2013 240 = 289<\/p>\n\n\n\n

By adding square roots on both sides

Length of the diagonal of the rectangle = \u221al2<\/sup> + b2<\/sup> = \u221a289 = 17 m<\/p>\n\n\n\n

Hence, the length of the diagonal of the rectangle is 17m.<\/p>\n\n\n\n

11. If the ratio between the length and the perimeter of a rectangular plot is 1: 3, then the ratio between the length and breadth of the plot is
(a) 1: 2
(b) 2: 1
(c) 3: 2
(d) 2: 3<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

The option (b) is the correct answer.<\/p>\n\n\n\n

Given that Length of rectangle\/Perimeter of rectangle = 1\/3<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

l\/ (2l + 2b) = 1\/3

By cross multiplication, we get:

3l = 2l + 2b

On further calculation<\/p>\n\n\n\n

l = 2b<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

l\/ b= 2\/1

Hence, the ratio of the length and the breadth is 2: 1.<\/p>\n\n\n\n

12. If the length of the diagonal of a square is 20 cm, then its perimeter is
(a) 10\u221a2cm
(b) 40 cm
(c) 40\u221a2cm
(d) 200 cm<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

The option (c) is the correct answer.<\/p>\n\n\n\n

It is given that length of the diagonal = 20 cm<\/p>\n\n\n\n

So the length of the side of a square = Length of Diagonal\/\u221a2 = 20\/\u221a2 = (2 \u00d7 10)\/\u221a2<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

Length of the side of a square = (\u221a2 \u00d7 \u221a2 \u00d7 10)\/\u221a2 = 10\u221a2 cm

Hence, perimeter of the square = 4 \u00d7 Side = 4 \u00d710\u221a2 = 40\u221a2 cm<\/p>\n\n\n\n

RD Sharma Solutions for Class 6 Maths Chapter 20: Download PDF<\/strong><\/h2>\n\n\n\n

RD Sharma Solutions for Class 6 Maths Chapter 20\u2013Mensuration<\/p>\n\n\n\n

Download PDF<\/strong>: RD Sharma Solutions for Class 6 Maths Chapter 20\u2013Mensuration PDF<\/a><\/p>\n\n\n\n

Chapterwise RD Sharma Solutions for Class 6 Maths :<\/strong><\/h2>\n\n\n\n