AP 6th Class Maths 10th Chapter Mensuration Exercise 10.3 Solutions

Well-designed AP 6th Class Maths Guide Chapter 10 Mensuration Exercise 10.3 offers step-by-step explanations to help students understand problem-solving strategies.

Mensuration Class 6 Exercise 10.3 Solutions – 6th Class Maths 10.3 Exercise Solutions

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) Length of the rectangle = 3 cm
Breadth of the rectangle = 4 cm
Area of the rectangle
= length × breadth
= 3 cm × 4 cm
= 12sq.cm

b) Length of the rectangle = 12 m
Breadth of the rectangle = 21 m
Area of the rectangle
= length × breadth
= 12 m × 21 m
= 252 sq.m

c) Length of the rectangle = 2 km
Breadth of the rectangle = 3 km
Area of the rectangle
= length × breadth
= 2 km × 3 km
= 6 sq. km.

d) Length of the rectangle = 2 m
Breadth of the rectangle = 70 cm
= 0.70 m
Area of the rectangle
= length × breadth
= 2 m × 0.70 m
= 1.40 sq.m

AP 6th Class Maths 10th Chapter Mensuration Exercise 10.3 Solutions

Question 2.
Find the areas of the squares whose sides are:
a) 10 cm
b) 14 cm
c) 5 m
Solution:
a) Side of the square = 10 cm
∴ Area of the square = side × side
= 10 cm × 10 cm
= 100 sq.cm

b) Side of the square = 14 cm
∴ Area of the square = side × side
= 14 cm × 14 cm
= 196 sq.cm

c) Side of the square = 5 m
∴ Area of the square
= side × side
= 5 m × 5 m
= 25 sq.m

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) Length of the rectangle = 9 m
Breadth of the rectangle = 6 m
∴ Area of the rectangle
= length × breadth
= 9 m × 6 m
= 54 sq.m.

b) Length of the rectangle = 17 m
Breadth of the rectangle = 3 m
∴ Area of the rectangle = length × breadth
= 17 m × 3 m
= 51 sq.m.

c) Length of the rectangle = 4 m
Breadth of the rectangle = 14 m
∴ Area of the rectangle = length × breadth
= 4 m × 14 m
= 56 sq.m.
From the above information
Rectangle (c) has the largest area i.e., 56 sq.m and rectangle
(b) has smallest area i.e. 51 sq.m

Question 4.
The area of a rectangular garden 50 m long is 300 sq m . Find the width of the garden.
Solution:
Length of the rectangular garden = 50 m
Area of the rectangular garden = 300 sq.m.
∴ Width of the rectangular garden = Area + Length
= 300 ÷ 50
= 6 m
∴ Width of the rectangular garden 1 m.

AP 6th Class Maths 10th Chapter Mensuration Exercise 10.3 Solutions

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:
Length of the rectangular plot = 500 m
and breadth = 200 m
Area of rectangular park
= length × breadth
= 500 m × 200 m
= 1,00,000 sq.m
Rate of the tiling the plot = ₹8 per 100 sq.m.
∴ Cost of tiling the garden = 100000 × \(\frac { 8 }{ 106 }\) =₹ 8000
Hence the cost of tiling the rectangular plot = ₹ 8000

Question 6.
A table-top measures 2 m by 1 m 50 cm . What is its area in square metres ?
Solution:
Length of the table – top = 2 m
and its breadth = 1 m 50 cm = 1.50 m
Area of the table top = length × breadth
= 2 m × 1.50 m = 3 m2
Hence, the area of the table top = 3m2 (or) 3 sq.m.

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 of the room = 4 m
Breadth of the room =3.5 m
Area of the room = length × breadth
= 4 m × 3.5 m
= 14 m2 (or) 14 sq.m
Hence, the area of the carpet needed = 14sq.m

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:
Length of the floor = 5 m
Breadth of the floor = 4m
AP 6th Class Maths 10th Chapter Mensuration Exercise 10.3 Solutions Img 1
∴ Area of the floor = length × breadth
= 5 m × 4 m
= 20 sq.m
Side of the square 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
= Area of floor – Area of square carpet
= 20 m2 – 9m2
= 11 sq.m.

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 ?
Solution:
Side of the square flower bed = 1m
∴ Area of a square flower bed
= side × side
= 1 m × 1 m
= 1 sq.m
So, area of 5 square flower beds = 5 × 1 sq.m = 5
AP 6th Class Maths 10th Chapter Mensuration Exercise 10.3 Solutions Img 2
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 = Area of land – Area of 5 square flower beds.
= 20 sq.m – 5 sq.m = 15 sq.m

Question 10.
By splitting the following figures into rectangles, find their areas (The measures are given in centimetres).
AP 6th Class Maths 10th Chapter Mensuration Exercise 10.3 Solutions Img 3
Solution:
a) Splitting the given figure into four rectangles, they are I, II, II and IV
AP 6th Class Maths 10th Chapter Mensuration Exercise 10.3 Solutions Img 4
Area of the rectangle I
= length × breadth
= 4 cm × 3 cm
= 12 sq.m

Area of the rectangle II
= 3 cm × 2 cm
= 6 sq.m

Area of the rectangle III
= 4 cm × 1 cm
= 4 sq.cm

Area of the rectangle IV
= 3 cm × 2 cm
= 6 sq. cm
Total area of the given figure = 12 sq.cm + 6 sq.cm + 4 sq.cm + 6 sq.cm = 28 sq.cm

b) Splitting the given figure into three rectangles, they are I, II and III
AP 6th Class Maths 10th Chapter Mensuration Exercise 10.3 Solutions Img 5
Area of the Rectangle I = length × breadth
= 3 cm × 1 cm
= 3 sq.cm

Area of the rectangle II
= 3 cm × 1 cm
= 3 sq.m

Area of the 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

AP 6th Class Maths 10th Chapter Mensuration Exercise 10.3 Solutions

Question 11.
Split the following shapes into rectangles and find their areas. (The measures are given in centimetres)
AP 6th Class Maths 10th Chapter Mensuration Exercise 10.3 Solutions Img 6
Solution:
a) Splitting the given figure into two rectangles, they are I and II
AP 6th Class Maths 10th Chapter Mensuration Exercise 10.3 Solutions Img 7
Now, area of the rectangle I
= length × breadth
= 10 cm × 2 cm
= 20 sq.cm

Area of the rectangle II
= 10 cm × 2 cm
= 20 sq.cm
∴ Total area of the given figure
= 20 sq.cm + 20 sq.cm
= 40 sq.cm

b) Splitting the given figure into two squares, and one rectangle, they are I and III are squares and il is rectangle.
AP 6th Class Maths 10th Chapter Mensuration Exercise 10.3 Solutions Img 8
Now, Area of the square l = side × side = 7 cm × 7 cm = 49 sq.cm

Area of the rectangle II = length × breadth
= 21 cm × 7 cm
= 147 sq.cm

Area of the square III
= 7 cm × 7 cm
= 49 sq.cm
∴ Total area of the given figure = 49 sq.cm + 147 sq.cm + 49 sq.cm
= 245 sq.cm

c) Splitting the given figure into two rectangles, they are I and II
AP 6th Class Maths 10th Chapter Mensuration Exercise 10.3 Solutions Img 9
Area of the rectangle I = length × breadth
= 4 cm × 1 cm
= 4 sq.cm

Area of the rectangle II = 5 cm × 1 cm
= 5 sq.cm
∴ Total area of the given figure = 4 sq.cm + 5 sq.cm
= 9 sq.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:
Length of the tile = 12 cm.
Breadth-of the tile = 5 cm.
∴ Area of the tile = length × breadth = 12 cm × 5 cm = 60 sq.cm.

Now
a) Length of the rectangular region = 100 cm
Breadth of the region = 144 cm
∴ Area of the rectangular region = length × breadth
= 100 cm × 144 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.

Leave a Comment