**NCERT
Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.3**

NCERT
Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.3 are the part of NCERT Solutions for Class
6 Maths. Here you can find the NCERT Solutions for Class 6 Maths Chapter 10 Mensuration
Ex 10.3.

**NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.1****NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.2****NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.3**

**Ex 10.3 Class 6 Maths 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

**Solution:
**(a) Given: Length of the rectangle = 3 cm

Breadth of the rectangle = 4 cm

∴ Area of the rectangle = length × breadth = 3 cm × 4 cm

= 12 cm

^{2}

(b) Given: Length of the rectangle = 12 m

Breadth of the rectangle = 21 m

∴ Area of the rectangle = length × breadth = 12 m × 21 m

= 252 m^{2}

(c) Given: Length of the rectangle = 2 km

Breadth of the rectangle = 3 km

∴ Area of the rectangle = length × breadth = 2 km × 3 km

= 6 km^{2}

(d) Given: Length of the rectangle = 2 m

Breadth of the rectangle = 70 cm = 70/100 m = 0.7 m

∴ Area of the rectangle = length × breadth = 2 m × 0.7 m

= 1.4 m^{2}

**Ex 10.3 Class 6 Maths Question 2.**

Find the areas of the squares whose sides are:(a) 10 cm

(b) 14 cm

(c) 5 m

**Solution:
**(a) Given: Side of the square = 10 cm

∴ Area of the square = Side × Side = 10 cm × 10 cm = 100 cm

^{2}

(b) Given: Side of the square = 14 cm

∴ Area of the square = Side × Side = 14 cm × 14 cm = 196 cm

^{2}

(c) Given: Side of the square = 5 m

∴ Area of the square = Side × Side = 5 m × 5 m = 25 m

^{2}

**Ex 10.3 Class 6 Maths 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?

**Solution:****
**(a) Given: Length of the rectangle
= 9 m

Breadth of the rectangle = 6 m

∴ Area of the rectangle = length × breadth

= 9 m × 6 m

= 54 m

^{2}

(b) Given: Length of the rectangle = 17 m

Breadth of the rectangle = 3 m

∴ Area of the rectangle

= length × breadth = 17 m × 3 m = 51 m^{2}

(c) Given: Length of the rectangle = 4 m

Breadth of the rectangle = 14 m

∴ Area of the rectangle = length × breadth

= 4 m × 14 m

= 56 m^{2}

Rectangle given in (c) has the largest area, i.e., 56 sq. m and the Rectangle
given in (b) has the smallest area, i.e., 51 sq. m.

**Ex 10.3 Class 6 Maths Question 4.**

The area of a rectangular garden 50 m long is 300 sq.
m. Find the width of the garden.**Solution:
**Given: Length of the rectangular garden = 50 m

Area of the rectangular garden = 300 sq. m

∴ Width of the garden = Area ÷ Length

= 300 sq. m ÷ 50 m = 6 m

Hence, width of the garden is 6 m.

**Ex
10.3 Class 6 Maths 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?**Solution:
**Given: Length of the rectangular plot = 500 m

Breadth of the rectangular plot = 200 m

∴ Area of the plot = length × breadth = 500 m × 200 m = 1,00,000 sq. m

Now, rate of tiling the plot = ₹ 8 per 100 sq. m

Cost of tiling the garden = ₹ (8/100 × 1,00,000) = ₹ 8000

Hence, the required cost is ₹ 8000.

**Ex 10.3 Class 6 Maths Question 6.**

A table-top measures 2 m by 1 m 50 cm. What is its
area in square metres?**Solution:
**Given: Length of the table-top = 2 m

Breadth of the table-top = 1 m 50 cm = 1.5 m

∴ Area of the table-top = length × breadth

= 2 m × 1.5 m

= 3 m

^{2}

Hence, the area of table-top is 3 sq. m.

**Ex 10.3 Class 6 Maths 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?**Solution:
**Given: Length of the room = 4 m

Breadth of the room = 3 m 50 cm = 3.5 m

Area of the room = length × breadth

= 4 m × 3.5 m = 14 sq. m

Hence, the area of the carpet needed to cover the floor is 14 sq. m.

**Ex 10.3 Class 6 Maths 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.**Solution:
**Given: Length of the floor = 5 m

Breadth of the floor = 4 m

∴ Area of the floor = length × breadth

= 5 m × 4 m = 20 sq. m

Side of the carpet = 3 m

∴ Area of the square carpet = side × side = 3 m × 3 m = 9 sq. m

∴ Area of the floor which is not carpeted = 20 sq. m – 9 sq. m

= 11 sq. m

**Ex 10.3 Class 6 Maths Question 9.**

Five square flower beds each of side 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?**Solution:
**Given: Side of the square flower bed = 1 m

∴ Area of 1 square flower bed = 1 m × 1 m = 1 sq. m

∴ Area of 5 square flower beds = 1 sq. m × 5 = 5 sq. m

Now, length of the land = 5 m

Breadth of the land = 4 m

∴ Area of the land = length × breadth = 5 m × 4 m = 20 sq. m

∴ Area of the remaining part of the land = 20 sq. m – 5 sq. m

= 15 sq. m

**Ex 10.3 Class 6 Maths Question 10.**

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

Splitting the given figure into the rectangles I, II, III and IV, we have

Area of the rectangle I = length × breadth

= 4 cm × 3 cm = 12 sq. cm

Area of the rectangle II = length × breadth

= 3 cm × 2 cm = 6 sq. cm

Area of the rectangle III = length × breadth

= 4 cm × 1 cm = 4 sq. cm

Area of the rectangle IV = length × breadth

= 3 cm × 2 cm = 6 sq. cm

∴ Total area of the whole figure

= 12 sq. cm + 6 sq. cm + 4 sq. cm + 6 sq. cm

= 28 sq. cm

(b) Splitting the given figure into the rectangles
I, II and III, we have

= 3 cm × 1 cm = 3 sq. cm

Area of the rectangle II

= 3 cm × 1 cm = 3 sq. cm

Area of rectangle III

= 3 cm × 1 cm = 3 sq. cm

∴ Total area of the given figure = 3 sq. cm + 3 sq. cm + 3 sq. cm = 9 sq. cm

**Ex 10.3 Class 6 Maths Question 11.**

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

(a) Splitting the given figure into the rectangles I and II, we have

Area of the rectangle I

= 12 cm × 2 cm = 24 sq. cm

Area of the rectangle II

= 8 cm × 2 cm = 16 sq. cm

∴ Total area of the whole figure = 24 sq. cm + 16 sq. cm = 40 sq. cm

(b) Splitting the given figure into the rectangles
I, II and III, we have

= 7 cm × 7 cm = 49 sq. cm

Area of the rectangle II

= 21 cm × 7 cm = 147 sq. cm

Area of the rectangle III

= 7 cm × 7 cm = 49 sq. cm

∴ Total area of the whole figure

= 49 sq. cm + 147 sq. cm + 49 sq. cm

= 245 sq. cm

(c) Splitting the given figure into two rectangles,
we have

Area of first rectangle = 5 × 1 = 5 sq. cm

Area of second rectangle = 4 × 1 = 4 sq. cm

∴ Total area of the whole figure

= 5 sq. cm + 4 sq. cm = 9 sq. cm

**Ex 10.3 Class 6 Maths 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

**Solution:
**Given: Length of the tile = 12 cm

Breadth of the tile = 5 cm

∴ Area of a tile = length × breadth = 12 cm × 5 cm = 60 sq. cm

(a) Length of the rectangular region = 144 cm

Breadth of the region = 100 cm

∴ Area of the rectangular region = length × breadth = 144 cm ×
100 cm

= 14400 sq. cm

∴ Number of tiles needed to cover the whole rectangular region

= 14400 sq. cm ÷ 60 sq. cm

= 240 tiles

(b) Length of the rectangular region = 70 cm

Breadth of the rectangular region = 36 cm

∴ Area of the rectangular region = length × breadth = 70 cm × 36
cm = 2520 sq. cm

∴ Number of tiles needed to cover the whole rectangular region

= 2520 sq. cm ÷ 60 sq. cm

= 42 tiles

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