NCERT Solutions for Class 6 Maths Chapter 10: Mensuration - Exercise 10.3

Exercise- 10.3

Refer to page 1 - 10 for exercise 10.3 in the PDF”

1. Find the areas of the rectangles whose sides are: 

a) 3 cm and 4 cm

Ans: We know that the area of a rectangle is given by

\[{\text{Area of rectangle  =  length }} \times {\text{ breadth}}\]

\[{\text{Area of rectangle  =  }}3 \times 4 = 12c{m^2}\]

b) 12 m and 21 m

Ans: We know that the area of the rectangle is given by

\[{\text{Area of rectangle  =  length }} \times {\text{ breadth}}\]

\[{\text{Area of rectangle  =  12}} \times 21 = 252{m^2}\]

c) 2 km and 3 km

Ans: We know that the area of the rectangle is given by

\[{\text{Area of rectangle  =  length }} \times {\text{ breadth}}\]

\[{\text{Area of rectangle  =  2}} \times 3 = 6k{m^2}\]

d) 2 m and 70 cm 

Ans: We know that the area of the rectangle is given by

\[{\text{Area of rectangle  =  length }} \times {\text{ breadth}}\]

Here the units should be the same. 

\[70cm = 70 \times {10^{ - 2}}m = 0.7m\]

\[{\text{Area of rectangle  =  2}} \times 0.7 = 1.4{m^2}\]

2. Find the areas of the squares whose sides are: 

a. 10 cm 

Ans: We know that the area of square is given by

\[{\text{Area of square  =  sid}}{{\text{e}}^2}\]

\[{\text{Area of square  =  }}{10^2} = 100c{m^2}\]

b. 14 cm 

Ans: We know that the area of the square is given by

\[{\text{Area of square  =  sid}}{{\text{e}}^2}\]

\[{\text{Area of square  =  1}}{{\text{4}}^2} = 196c{m^2}\]

c. 5 m 

Ans: We know that the area of the square is given by

\[{\text{Area of square  =  sid}}{{\text{e}}^2}\]

\[{\text{Area of square  =  }}{{\text{5}}^2} = 25{m^2}\]

3. The length and breadth of three rectangles are as given below: 

a) 9 m and 6 m

b) 17 m and 3 m 

c) 4 m and 14 m

4. Which one has the largest area and which one has the smallest?

Ans: We know that the area of the rectangle is a product of its length and breadth.

a) 9 m and 6 m

\[{\text{Area of rectangle  =  length }} \times {\text{ breadth}}\]

\[{\text{Area of rectangle  =  9 }} \times {\text{ }}6 = 54{m^2}\]

b) 17 m and 3 m

\[{\text{Area of rectangle  =  length }} \times {\text{ breadth}}\]

\[{\text{Area of rectangle  =  3 }} \times {\text{ 17}} = 51{m^2}\]

c) 4 m and 14 m

\[{\text{Area of rectangle  =  length }} \times {\text{ breadth}}\]

\[{\text{Area of rectangle  =  4 }} \times {\text{ 14}} = 56{m^2}\]

In the above, we can see that rectangle (c) has the largest area i.e., , and rectangle (b) has the smallest area i.e., \[{\text{51}}{{\text{m}}^2}\]

5. The area of a rectangular garden 50 m long is 300 sq m. Find the width of the garden.

Ans: We know the formula for the area of a rectangle.

\[{\text{Area of rectangle  =  length }} \times {\text{ breadth}}\] or \[{\text{Area of rectangle  =  length }} \times {\text{ width}}\]

Here the area is 300 sq m and length is 50 m

\[ \Rightarrow 300 = 50 \times {\text{ width}}\]

\[ \Rightarrow {\text{width}} = 6m\]

6. What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the rate of  8 per hundred sq m.? 

Ans:  First we need to find the area of land, here length, and width is given.

\[{\text{Area of land  =  length }} \times {\text{ width}}\]

\[{\text{Area of land  =  }}500 \times 200 = 100000{m^2}\]

Now,

The cost of tilling 1,00,000 sq. m of land is

\[ = \dfrac{{8 \times 100000}}{{100}}\]

\[ = Rs.8000\]

7. A table-top measures 2 m by 1 m 50 cm. What is its area in square metres?

Ans: First we convert 1 m 50 cm to meters.

We know that \[50cm = 0.5m\]

Then 1 m 50 cm is equal to 1.5 m

Now,

\[ \Rightarrow {\text{Area}} = 2m \times 1.5m = 3{m^2}\]

8. A room is 4 m long and 3 m 50 cm wide. How many square metres of carpet is needed to cover the floor of the room? 

Ans: First we convert 3 m 50 cm to meters.

We know that \[50cm = 0.5m\]

Then 3 m 50 cm is equal to 3.5 m

Now,

\[ \Rightarrow {\text{Area}} = 4m \times 3.5m = 14{m^2}\]

9. A floor is 5 m long and 4 m wide. A square carpet of sides 3m is laid on the floor. Find the area of the floor that is not carpeted. 

Ans: First we need to find the area of floor and area of the square carpet.

\[{\text{Area of floor}} = 5m \times 4{\text{ }}m = 20{\text{ }}{m^2}\]

\[{\text{Area of square carpet}} = 3m \times 3m = 9{\text{ }}{m^2}\]

Now the area of the floor that is not carpeted is

\[ = 20 - 9\]

\[ = 11{m^2}\].

10. Five square flower beds each of sides 1 m are dug on a piece of land 5 m long and 4 m wide. What is the area of the remaining part of the land? 

Ans: The area of the square bed is \[ = 1 \times 1 = 1{m^2}\]

Area of 5 square beds \[ = 5 \times 1 = 5{m^2}\]

Now the area of land \[ = 5 \times 4 = 20{m^2}\]

The remaining area is the difference of area of land and the area of 5 flower beds.

\[{\text{Remaining area  =  Area of land }} - {\text{ Area of 5 flower beds}}{\text{.}}\]

\[{\text{Remaining area  =  }}20 - 5 = 15{m^2}\]

11. By splitting the following figures into rectangles, find their areas (The measures are given in centimetres).

(a) 

Ans: 

The area of orange region is

\[ \Rightarrow 3 \times 3 = 9\,c{m^2}\]

The area of blue region is

\[ \Rightarrow 1 \times 2 = 2\,c{m^2}\]

The area of Gold region is

\[ \Rightarrow 3 \times 3 = 9\,c{m^2}\]

The area of green region is

\[ \Rightarrow 2 \times 4 = 8\,c{m^2}\]

The total area \[ = 9 + 2 + 9 + 8 = 28\,c{m^2}\]

Hence, the total area is \[28\,c{m^2}\]

(b) 

Ans:

The area of green region is

\[ \Rightarrow 3 \times 1 = 3\,c{m^2}\]

The area of orange region is

\[ \Rightarrow 3 \times 1 = 3\,c{m^2}\]

The area of blue region is

\[ \Rightarrow 3 \times 1 = 3\,c{m^2}\]

The total area \[ = 3 + 3 + 3 = 9\,c{m^2}\]

Hence, the total area is \[9\,c{m^2}\]

12. Split the following shapes into rectangles and find their areas. (The measures are given in centimetres)

a) 

Ans: 

The total area of the figure is 

\[ \Rightarrow 12 \times 2 + 8 \times 2 = 40\,c{m^2}\]

b)

Ans:

As we can see that there are 5 squares. Each side is of the length 7cm.

The area of the figure is 

\[ \Rightarrow 5 \times {7^2} = 245\,c{m^2}\]

c) 

Ans:

 Area of the orange rectangle is

\[ \Rightarrow 2 \times 1 = 2\,c{m^2}\]

Area of the blue rectangle is

\[ \Rightarrow 2 \times 1 = 2\,c{m^2}\]

The area of the green rectangle is

\[ \Rightarrow 5 \times 1 = 5\;c{m^2}\]

Hence the total area is

\[ \Rightarrow 2 + 2 + 5 = 9\;c{m^2}\]

13. How many tiles whose length and breadth are 12 cm and 5 cm respectively will be needed to fit in a rectangular region whose length and breadth are respectively: 

(a) 100 cm and 144 cm 

Ans: Area of the rectangle is

\[ \Rightarrow 100 \times 144 = 14400\,c{m^2}\]

The area of one tile is

\[ \Rightarrow 5 \times 12 = 60\;c{m^2}\]

\[ \Rightarrow {\text{Number of tiles  = }}\dfrac{{{\text{area of rectangle}}}}{{{\text{area of one tile}}}}\]

\[ \Rightarrow {\text{Number of tiles  = }}\dfrac{{14400}}{{60}} = 240\]

Hence 240 tiles are needed.

(b) 70 cm and 36 cm.

The area of a rectangle is

\[ \Rightarrow 70 \times 36 = 2520\,c{m^2}\]

The area of one tile is

\[ \Rightarrow 5 \times 12 = 60\;c{m^2}\]

\[ \Rightarrow {\text{Number of tiles  = }}\dfrac{{{\text{area of rectangle}}}}{{{\text{area of one tile}}}}\]

\[ \Rightarrow {\text{Number of tiles  = }}\dfrac{{2520}}{{60}} = 42\]

Hence 42 tiles are needed.

NCERT Solutions For Class 6 Maths Chapter 10 Mensuration Exercise 10.3

Opting for the NCERT solutions for Ex 10.3 Class 6 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 10.3 Class 6 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 6 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 6 Maths Chapter 10 Exercise 10.3 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 6 Maths Chapter 10 Exercise 10.3, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

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