Students can download Chapter 10 Mensuration Ex 10.3 KSEEB Solutions for Class 6 Maths helps you to revise the complete syllabus.

## Karnataka State Syllabus Class 6 Maths Chapter 10 Mensuration Ex 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) 2m and 70cm.

Solution:

It is known that

Area of rectangle = length × Breadth

a) 1 = 3 cm, b = 4 cm Area = l × b = 3 × 4 = 12 cm^{2}

b) 1 = 12 m, b = 21 m

Area = l × b = 12 × 21 = 252 m^{2}

c) l = 2 km, b = 3 km

Area = l × b = 2 × 3 = 6km^{2}

d) l = 2 m , b = 70 cm = 0.70 m

Area = l × b = 2 × 0.70 = 1.40 m^{2}

Question 2.

Find the areas of the squares whose sides are:-

a) 10 cm

b) 14 cm

c) 5 m

Solution:

It’s known that, Area o f square = side × side

a) 10cm

side = 10 cm

Area of the square =10 × 10= 100 cm^{2}

b) 14 cm

side = 14 cm

Area of the square = 14 × 14 = 196 cm^{2}

c) 5m

Side = 5 m

Area of the square = 5 × 5 = 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

d) 4 m and 14 m

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

Solution:

It is known that

Area of rectangle = length × Breadth

a) l = 9 m b = 6 m

Area = l × b = 9 × 6 = 54 m^{2}

b) l = 17 m b = 3 m

Area = l × b = 17 × 3 = 51 m^{2}

c) l = 4 m b = 14 m

Area = l × b = 4 × 14 = 56 m^{2}

It can be see that rectangle (c) has the largest area and rectangle (b) has the smallest area.

Question 4.

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

Solution:

Given:-

Length of the garden = 50 m = l

and Areas of the garden = 300 sq m = A

Width or breadth = ? = b

A = l × b

300 = 50 × b

b = 6 m

The breadth or width of the garden = 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 Rs 8 per hundred sq m.?

Solution:

Area of rectangular plot = 500 × 200 = 100000 m^{2}

cost of tiling per 100 m^{2} = Rs. 8

Question 6.

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

Solution:

Areas = 1 × 6 = 2 × 1.5 = 3m^{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?

Solution:

Length (l) = 4 m,

Breadth (b) = 3 m 50 cm = 3.5 m

Area = l × b = 4 × 3.5 = 14 m^{2}

Question 8.

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.

Solution:

Length (l) = 5 m,

Breadth (b) = 4 m

Area = l × b = 5 × 4 = 20 m^{2}

Area covered by the carpet = (side)^{2} = (3)^{2} = 9 m^{2}

Area not covered by the carpet = 20 – 9 = 11 m^{2}

Question 9.

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

Solution:

Area of the land = 5 × 4 = 20 m^{2}

Area occupied by 5 flower beds = 5 × (side)^{2} = 5 × (1 )^{2} = 5m^{2}

Areas of the remaining part = 20 – 5 = 15 m^{2}

Question 10.

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

Solution :

(a) The given figure can be broken into rectangles as follows.

Area of 1^{st} rectangle = 4 × 2 = 8 cm^{2}

Area of 2^{nd} rectangle = 6 × 1 = 6 cm^{2}

Area of 3^{rd} rectangle = 3 × 2 = 6 cm^{2}

Area of 4^{th} rectangle = 4 × 2 = 8 cm^{2}

Total area of the complete figure = 8 + 6 + 6 + 8 = 28 cm^{2}

(b) The given figure can be broken into rectangles as follows.

Area of 1^{st} rectangle = 3 × 1 = 3 cm^{2}

Area of 2^{nd} rectangle = 3 × 1 = 3 cm^{2}

Area of 3^{rd} rectangle = 3 × 1 = 3 cm^{2}

Total area of the complete 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)

Solution:

(a) The given figure can be broken into rectangles as follows.

Area of 1^{st} rectangle = 12 × 2 = 24 cm

Area of 2^{nd} rectangle = 8 × 2 = 16 cm

Total area of the complete figure = 24 + 16 = 40 cm

(b) The given figure can be broken into rectangles as follows.

Area of 1^{st} rectangle = 21 × 7 = 147 cm

Area of 2^{nd} square = 7 × 7 = 49 cm

Area of 3^{rd} square = 7 × 7 = 49 cm

Total area of the complete figure = 147 + 49 + 49 = 245 cm

(c) The given figure can be broken into rectangles as follows:

Area of 1^{st} rectangle = 5 × 1 = 5 cm

Area of 2^{nd} rectangle = 4 × 1 = 4 cm

Total area of the complete figure = 5 + 4 = 9 cm

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:

(a) Total area of the region = 100 × 144 = 14400 cm^{2}

Area of one tile = 12 × 5 = 60 cm^{2}

img class=”alignnone size-full wp-image-45590″ src=”https://ktbssolutions.com/wp-content/uploads/2020/10/KSEEB-Solutions-for-Class-6-Maths-Chapter-10-Mensuration-Ex-10.3-60.png” alt=”KSEEB Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.3 60″ width=”433″ height=”57″ />

Therefore, 240 tiles are required.

(b) Total area of the region = 70 × 36 = 2520 cm^{2}

Area of one tile = 60 cm^{2}

Therefore, 42 tiles are required.