These NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.3 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Exercise 10.3

Question 1.

Find the areas of the rectangles whose sides are:

(a) 3 cm and 4 cm

(b) 12 m and 21 m

(c) 2 km and 3 km

(d) 2 m and 70 cm

Answer:

(a) Area of rectangle

= length × breadth

= 3 cm × 4 cm = 12 cm^{2}

(b) Area of rectangle

= length × breadth

= 12 m × 21 m = 252 m^{2}

(c) Area of rectangle

= length × breadth

= 2 km × 3 km = 6 km^{2}

(d) Area of rectangle

= lenght × breadth

= 2m ( lm = 100cm)

= 200 cm × 70 cm = 14000sq.cm

Question 2.

Find the areas of the squares whose sides are:

(a) 10 cm

(b) 14 cm

(c) 5 m

Answer:

(a) Area of square = side × side

= 10 cm × 10 cm = 100 cm^{2}

(b) Area of square = side × side

= 14 cm × 14 cm = 196 cm^{2}

(c) Area of square = side × side

= 5 m × 5 m = 25 m^{2}

Question 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

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

Answer:

(a) Area of rectangle

= length × breadth

= 9 m × 6 m = 54 m^{2}

(b) Area of rectangle

= length × breadth

= 3 m × 17 m = 51 m^{2}

(c) Area of rectangle

= length × breadth

= 4 m × 14 m = 56 m^{2}

Thus, the rectangle (c) has largest area, and rectangle (b) has smallest area.

Question 4.

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

Answer:

Length of rectangle

= 50 m and Area of rectangle = 300 m^{2 }Since, Area of rectangle

= length × breadth

Therefore, breadth

\(\frac{\text { Area of rectangle }}{\text { Length }}\) = \(\frac{300}{50}\) = 6m

Thus, the breadth of the garden is 6 m.

Question 5.

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?

Answer:

Length of land

= 500 m and Breadth of land = 200 m

Area of land = length × breadth

= 500 m × 200 m = 1,00,000 m^{2}

Cost of tilling 100 sq. m of land = 8

∴ Cost of tilling 1,00,000 sq. m of land

\(\frac{8 \times 100000}{100}\) = 8000

Question 6.

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

Answer:

Length of table = 2 m

Breadth of table = 1 m 50 cm = 1.50 m

Area of table = length × breadth

= 2 m × 1.50 m = 3 m^{2}

Question 7.

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?

Answer:

Length of room = 4 m

Breadth of room = 3 m 50 cm = 3.50 m

Area of carpet = length × breadth

= 4 × 3.50= 14 m^{2}

Question 8.

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

Answer:

Length of floor

= 5 m and breadth of floor = 4 m

Area of floor = length × breadth

= 5m × 4m = 20m^{2}

Now, Side of square carpet = 3 m

Area of square carpet = side × side

= 3 × 3 = 9 m^{2}

Area of floor that is not carpeted = 20m^{2} – 9m^{2} = 11m^{2}

Question 9.

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?

Answer:

Side of square bed = 1 m

Area of square bed = side × side

= 1m × 1m = 1m^{2}

∴ Area of 5 square beds = 1 × 5 = 5 m^{2}

Now, Length of land = 5 m

Breadth of land = 4 m

∴ Area of land = length × breadth

= 5 m × 4 m = 20 m^{2}

Area of remaining part

= Area of land – Area of 5 flower beds = 20 m^{2} – 5 m^{2} = 15 m^{2}

Question 10.

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

Answer:

Area of HKLM = 3 × 3 = 9 cm^{2}

Area of IJGH = 1 × 2 = 2 cm^{2}

Area of FEDG = 3 × 3 = 9 cm^{2}

Area of ABCD = 2 × 4 = 8 cm^{2}

Total area of the figure

= 9 + 2 +9 + 8

= 28 cm^{2}

Area of ABCD = 3 × 1 = 3 cm^{2}

Area of BDEF = 3 × 1 = 3 cm^{2}

Area of FGHI = 3 × 1 = 3 cm^{2}

Total area of the figure = 3 + 3 + 3 = 9 cm^{2}

Question 11.

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

Answer:

(a) Area of rectangle ABCD

= 2 × 10 = 20 cm^{2}

Area of rectangle DEFG

= 10 × 2 = 20 cm^{2}

Total area of the figure

= 20 + 20 = 40 cm^{2}

(b) There are 5 squares each of side 7 cm.

Area of one square

= 7 × 7 = 49 cm^{2}

Area of 5 square

= 49 × 5 = 245 cm2

(c) Area of rectangle ABCD

= 5 × 1 = 5 cm^{2}

Area of rectangle EFGH

= 4 × 1 = 4 cm^{2}

Total area of the figure

= 5 + 4 cm^{2}

Question 12.

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

(b) 70 cm and 36 cm

Answer:

(a) Area of region

= 100 cm × 144 cm = 14400 cm^{2}

Area of one tile

= 5 cm × 12 cm = 60 cm^{2}

Number of tiles = \(\frac{\text { Area of region }}{\text { Area of one tile }}\) = \(\frac{14400}{60}\) = 240

Thus, 240 tiles are required.

(b) Area of region

= 70 cm × 36 cm = 2520 cm^{2}

Area of one tile

= 5 cm × 12 cm = 60 cm^{2}

Number of tiles

= \(\frac{\text { Area of region }}{\text { Area of one tile }}\) = \(\frac{2520}{60}\) = 42

Thus, 42 tiles are required.