# Ncert Solutions For Class 6 Maths Ex 10.3

## Ncert Solutions For Class 6 Maths Chapter 10 Ex 10.3

Q1. What is the area of the following rectangles whose sides’ measure:

(a) 2 cm and 5 cm

(b) 15 m and 20 m

(c) 4 km and 5 km

(d) 3 m and 60 cm

Ans: It is known that, Area of a rectangle = Length (l) x Breadth (b)

(a) l = 2 cm

b = 5 cm

Area = l x b = 2 x 5 = 12 cm2

(b) l = 15 m

b= 20 m

Area = l x b= 20 x 15 = 300 m2

(c) l = 4 km

b= 5 km

Area = l x b = 4 x 5 = 20 km2

(d) l = 3 m

b = 60 cm = 0.60 m

Area = l x b=2 x 0.60 = 1.20 m2

Q2. Find the areas of the squares whose sides are: (a) 8 cm (b) 11 cm (c) 6 m

Ans:

It is known that, Area of a square = (Side)2

(a) Side = 8 cm Area = (8)2 =64 cm2

(b) Side = 11 cm Area = (11)2 = 121 cm2

(c) Side = 6 m Area = (6)2 = 36 m2

Q3. Which one has the largest and smallest area among the 3 rectangles whose dimensions measure:

(a) 5 m by 6 m

(b) 12m by 4 m

(c) 2 m by 10 m

Ans: We know that, Area of rectangle = Length x Breadth

(a) l = 5m

b = 6m

Area = l x b = 5 x 6 = 30 m2

(b) l = 12m

b = 4m

Area = l x b = 12 x 4 = 48 m2

(c) l = 2m

b = 10m

Area = l x b = 2 x 10 = 20 m2

From the above results, we come to know that (b) has the largest area and rectangle (c) has the smallest area.

Q5: What is the cost of tiling a rectangular plot of land 400 m long and 300 m wide at the rate of Rs 5 per hundred sq in?

Ans:

Area of rectangular plot = 400 x 300 = 120000 m2

Cost of tiling per 100 m2 = Rs 5

Cost of tiling per 120000 m2 = $$\frac{5}{100}\times 120000 = Rs.6000$$

Q6: What is the area of table whose sides measure 1m 20 cm in length and 2m 50 cm in breadth?

Ans:

Length (l) = 1 m 20cm = $$(1+\frac{20}{100})\ m = 1.2\ m$$

Breadth (b) 2 m 50 cm = $$(2+\frac{50}{100})\ m = 2.5\ m$$

Area = l x b = 1.2 x 2.5 = 3 m2

Q7: How many sq meters of carpet is needed to cover the floor of a room of sides measuring 3 m in length and 4 m 20cm in breadth?

Ans:

Length (l) = 3 m

Breadth (b) = 4 m 20 cm = $$(4+\frac{20}{100})\ m = 4.2\ m$$

Area = l x b = 3 x 4.2 = 12.6 m2

Q8: A square carpet is laid on the floor of sides measuring 4 m in a floor of sides 4 m and 5 m. Find the remaining area of the floor that is not carpeted.

Ans:

Length (I) = 4 m

Area of floor = l x b = 4 x 5 = 20 m2

Area covered by the carpet = (Side) 2 = (4)2 = 16 m2

Area not covered by the carpet = 20 – 16 = 4 m2

Q9: From a piece of land measuring 4 m long and 6 m wide, 4 sq flower beds of side 2m each are dug out. Find out the remaining area of the land.

Ans:

Area of the land = 4 x 6 = 24 m2

Area occupied by 4 flower beds = 4 x (Side)2 = 4x (2)2 = 16m2

Area of the remaining part = 24 – 16 = 8 m2

Q10: Split the below figures into multiple rectangles and find out their area. All dimensions are given in centimeters.

Ans:

(a) The above figures can be split in to multiple rectangles as follows:

Area of 1st rectangle = 8 x 4 = 32 cm2

Area of 2nd rectangle = 12 x 2 = 24 cm2

Area of 3rd rectangle = 6 x 4 = 24 cm2

Area of 4th rectangle = 8 x 4 = 32 cm2

Total area of the whole figure = 32 + 24 + 24 + 32 = 112 cm2

(b) The above figures can be split in to multiple rectangles as follows:

Area of 1st rectangle = 6 x 2 = 12 cm2

Area of 2nd rectangle = 6 x 2 = 12 cm2

Area of 3rd rectangle = 6 x 2 = 12 cm2

Total area of the whole figure = 12 + 12 + 12 = 36 cm2

Q11. Split the below figures into multiple rectangles and find out their area. All dimensions are given in centimeters.

Ans:

(a) The above figures can be split in to multiple rectangles as follows:

Area of 1st rectangle = 24 x 4 = 96 cm2

Area of 2nd rectangle = 16 x 4 = 64 cm2

Total area of the whole figure = 96 + 64 = 160 cm2

(b) The above figures can be split in to multiple rectangles as follows:

Area of 1st rectangle = 42 x 14 = 588 cm2

Area of 2nd rectangle = 14 x 14 = 196 cm2

Area of 3rd rectangle = 14 x 14 = 196 cm2

Total area of the whole figure = 588 + 196 + 196 = 980 cm2

(c) The above figures can be split in to multiple rectangles as follows:

Area of 1st rectangle = 10 x 2 = 20 cm2

Area of 2nd rectangle = 8 x 2 = 16 cm2

Total area of the whole figure = 20 + 16 = 36 cm2

Q12. Tiles of sides measuring 10 cm by 4 cm each are to be fit in a rectangular region whose length and breadth are respectively:

(a) 144 cm and 100 cm

(b) 36 cm and 70 cm

Ans. (a) Total area of the region = 144 x 100 = 14400 cm2

Area of one tile = 10 x 4 = 40 cm2

Number of tiles required = $$\frac{14400}{40}$$ = 360

Therefore, 360 tiles are required.

(b) Total area of the region = 36 x 70 = 2520 cm2

Area of one tile = 40 cm2

Number of tiles required = $$\frac{2520}{40}$$ = 63

Therefore, 63 tiles are required.