# NCERT Solutions for Class 6 Maths Chapter 10 Exercise 10.3

NCERT Solutions for Class 6 Maths Chapter 10 Exercise 10.3 (Ex. 10.3) Mensuration in PDF file format Hindi and English Medium free to use or download. All the contents are updated for academic session 2021-2022 CBSE and other state board’s students.

Class 6 math exercise 10.3 explains the way to find area of square and rectangles using formulae. Here some practical questions are also given based on daily life word problems.## Class 6 Maths Chapter 10 Exercise 10.3 Solution

### CBSE NCERT Class 6 Maths Chapter 10 Exercise 10.3 Solution in Hindi and English Medium

Class: 6 | Mathematics |

Chapter: 10 | Mensuration |

Exercise: 10.3 | English and Hindi Medium Solution |

### Class 6 Maths Chapter 10 Exercise 10.3 Solution in Videos

##### Area by Counting Squares on a Squared Paper

We can find the area of some regular and irregular figures by using squared paper. To find area some steps are given below:

Step-1: Count the number of complete squares enclosed by the figure.

Step-2: Count the number of those squares in which more than half part is enclosed by the figure and treat them as one complete square.

Step-3: Count the number of those squares in which half part is enclosed by the figure and divide it by 2 to get the number of complete squares.

Step-4: Leave those squares in which less than half part is enclosed by the figure.

##### Area of a Rectangle and a Square

So far, we have learnt how to find the area of any given figure by counting the number of squares enclosed by the figure. This method is time consuming and a bit difficult for calculating the area of given figures. There is an easier method for calculating the areas of rectangles and squares.

We observe that the area of a rectangle (A) = length × breath

or A = l × b

Also, we have: Length = area/breadth

And breadth = area/length

In the case of a square, we know that l = b

Hence, the area of a square (A) = side × side

or A = l × l

### Class 6 Maths Exercise 10.3 Extra Questions with Answer

##### Find the area of a rectangle if its length and breadth are 24 cm and 13 cm respectively.

Here, length = 24 cm and breadth = 13 cm

So, Area of the rectangle = length × breadth

= (24 × 13) sq.cm = 312 sq.cm

##### What is the area of the floor of a square room of side 8.3 m? Also, find the cost of flooring the room at Rs. 10 per sq. m.

Side of the square = 8.3 m

So, Area of the square = side × side

= (8.3 × 8.3) sq. m

= 68.89 sq. m

So, cost of flooring the room = Rs. (10 × 68.89) = Rs. 688.90.

##### Summary:

Perimeter of rectangle = 2 (length + breadth)

Area of rectangle = length x breadth

Perimeter of square = 4 x side

Area of square = side²

### Class 6 Maths Exercise 10.3 Important Questions

##### Does area increase as perimeter increases?

The area increases much faster than the length of the perimeter (you could graph this and discuss linear versus exponential growth). Sides increase by one, perimeters increase by four, and area increases by three, five, seven, then nine!

##### The area of a rectangle is 90 sq.cm. Find the breadth of the rectangle if its length is 20 cm.

Area of the rectangle = 90 sq.cm.

Length of the rectangle = 20 cm

So, Breadth of the rectangle = Area/length

= 90/20 cm

= 4.5 cm

##### The perimeter of a rectangular field is 220 m and its breadth (width) is 48 m. Find the area and total cost of cultivating the field at Rs. 3.50 per sq. m.

Perimeter of the field = 220 m Breadth of the field = 48 m

So, Length of the field = ½ x Perimeter – Breadth

= (1/2 × 220 – 48) m

= (110 – 48) m = 62 m

Now, area of the field = length × breadth

= (62 × 48) sq. m

= 2976 sq. m

So, Total cost of cultivating the field = Rs (3.50 × 2976)

= Rs 10416