Class 6 Exam > Class 6 Notes > Mathematics (Maths) Class 6 > RD Sharma Solutions: Mensuration (Exercise 20.4)
Mensuration (Exercise 20.4) RD Sharma Solutions | Mathematics (Maths) Class 6 PDF Download
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Page 1
1. Find the area of a rectangle, whose
(i) Length = 6 cm, breadth = 3 cm
(ii) Length = 8 cm, breadth = 3 cm
(iii) Length = 4.5 cm, breadth = 2 cm.
Solution:
(i) We know that area of a rectangle = L × B
It is given that Length = 6 cm, breadth = 3 cm
By substituting the values
Area of a rectangle = 6 × 3 = 18 cm
2
(ii) We know that area of a rectangle = L × B
It is given that Length = 8 cm, breadth = 3 cm
By substituting the values
Area of a rectangle = 8 × 3 = 24 cm
2
(iii) We know that area of a rectangle = L × B
It is given that Length = 4.5 cm, breadth = 2 cm
By substituting the values
Area of a rectangle = 4.5 × 2 = 9 cm
2
2. Find the area of a square whose side is:
(i) 5 cm
(ii) 4.1 cm
(iii) 5.5 cm
(iv) 2.6 cm
Solution:
(i) We know that area of a square = side × side
It is given that side of a square = 5 cm
So the area of the square = 5 × 5 = 25 cm
2
(ii) We know that area of a square = side × side
It is given that side of a square = 4.1 cm
So the area of the square = 4.1 × 4.1 = 16.81 cm
2
(iii) We know that area of a square = side × side
It is given that side of a square = 5.5 cm
So the area of the square = 5.5 × 5.5 = 30.25 cm
2
(iv) We know that area of a square = side × side
It is given that side of a square = 2.6 cm
So the area of the square = 2.6 × 2.6 = 6.76 cm
2
3. The area of a rectangle is 49 cm
2
and its breadth is 2.8 cm. Find the length of the rectangle.
Solution:
It is given that area of a rectangle = 49 cm
2
Page 2
1. Find the area of a rectangle, whose
(i) Length = 6 cm, breadth = 3 cm
(ii) Length = 8 cm, breadth = 3 cm
(iii) Length = 4.5 cm, breadth = 2 cm.
Solution:
(i) We know that area of a rectangle = L × B
It is given that Length = 6 cm, breadth = 3 cm
By substituting the values
Area of a rectangle = 6 × 3 = 18 cm
2
(ii) We know that area of a rectangle = L × B
It is given that Length = 8 cm, breadth = 3 cm
By substituting the values
Area of a rectangle = 8 × 3 = 24 cm
2
(iii) We know that area of a rectangle = L × B
It is given that Length = 4.5 cm, breadth = 2 cm
By substituting the values
Area of a rectangle = 4.5 × 2 = 9 cm
2
2. Find the area of a square whose side is:
(i) 5 cm
(ii) 4.1 cm
(iii) 5.5 cm
(iv) 2.6 cm
Solution:
(i) We know that area of a square = side × side
It is given that side of a square = 5 cm
So the area of the square = 5 × 5 = 25 cm
2
(ii) We know that area of a square = side × side
It is given that side of a square = 4.1 cm
So the area of the square = 4.1 × 4.1 = 16.81 cm
2
(iii) We know that area of a square = side × side
It is given that side of a square = 5.5 cm
So the area of the square = 5.5 × 5.5 = 30.25 cm
2
(iv) We know that area of a square = side × side
It is given that side of a square = 2.6 cm
So the area of the square = 2.6 × 2.6 = 6.76 cm
2
3. The area of a rectangle is 49 cm
2
and its breadth is 2.8 cm. Find the length of the rectangle.
Solution:
It is given that area of a rectangle = 49 cm
2
Breadth of a rectangle = 2.8 cm
We know that
Area of a rectangle = L × B
It can be written as
L = Area/B = 49/2.8 = 17.5 cm
Hence, the length of the rectangle is 17.5 cm.
4. The side of a square is 70 cm. Find its area and perimeter.
Solution:
It is given that side of a square = 70 cm
We know that area of a square = side × side
By substituting the values
Area of a square = 70 × 70 = 4900 cm
2
We know that perimeter of a square = 4 × side
By substituting the values
Perimeter of a square = 4 × 70 = 280 cm
Hence, the area of square is 4900 cm
2
and the perimeter of square is 280 cm.
5. The area of a rectangle is 225 cm
2
and its one side is 25 cm, find its other side.
Solution:
It is given that
Area of a rectangle = 225 cm
2
Length of one side = 25 cm
We know that area of a rectangle = Product of length of two sides
So the other side = area/side
By substituting the values
Other side = 225/25 = 9 cm
Hence, the other side of the rectangle is 9 cm.
6. What will happen to the area of rectangle if its
(i) Length and breadth are trebled
(ii) Length is doubled and breadth is same
(iii) Length is doubled and breadth is halved.
Solution:
(i) Length and breadth are trebled
Consider l as the initial length and b as the initial breadth
So the original area = l × b
If the length and breadth are trebled it becomes three times more than the original value
New length = 3l
New breadth = 3b
New area of the rectangle = 3l × 3b = 9lb
Hence, the area of the rectangle becomes 9 times more than its original area.
Page 3
1. Find the area of a rectangle, whose
(i) Length = 6 cm, breadth = 3 cm
(ii) Length = 8 cm, breadth = 3 cm
(iii) Length = 4.5 cm, breadth = 2 cm.
Solution:
(i) We know that area of a rectangle = L × B
It is given that Length = 6 cm, breadth = 3 cm
By substituting the values
Area of a rectangle = 6 × 3 = 18 cm
2
(ii) We know that area of a rectangle = L × B
It is given that Length = 8 cm, breadth = 3 cm
By substituting the values
Area of a rectangle = 8 × 3 = 24 cm
2
(iii) We know that area of a rectangle = L × B
It is given that Length = 4.5 cm, breadth = 2 cm
By substituting the values
Area of a rectangle = 4.5 × 2 = 9 cm
2
2. Find the area of a square whose side is:
(i) 5 cm
(ii) 4.1 cm
(iii) 5.5 cm
(iv) 2.6 cm
Solution:
(i) We know that area of a square = side × side
It is given that side of a square = 5 cm
So the area of the square = 5 × 5 = 25 cm
2
(ii) We know that area of a square = side × side
It is given that side of a square = 4.1 cm
So the area of the square = 4.1 × 4.1 = 16.81 cm
2
(iii) We know that area of a square = side × side
It is given that side of a square = 5.5 cm
So the area of the square = 5.5 × 5.5 = 30.25 cm
2
(iv) We know that area of a square = side × side
It is given that side of a square = 2.6 cm
So the area of the square = 2.6 × 2.6 = 6.76 cm
2
3. The area of a rectangle is 49 cm
2
and its breadth is 2.8 cm. Find the length of the rectangle.
Solution:
It is given that area of a rectangle = 49 cm
2
Breadth of a rectangle = 2.8 cm
We know that
Area of a rectangle = L × B
It can be written as
L = Area/B = 49/2.8 = 17.5 cm
Hence, the length of the rectangle is 17.5 cm.
4. The side of a square is 70 cm. Find its area and perimeter.
Solution:
It is given that side of a square = 70 cm
We know that area of a square = side × side
By substituting the values
Area of a square = 70 × 70 = 4900 cm
2
We know that perimeter of a square = 4 × side
By substituting the values
Perimeter of a square = 4 × 70 = 280 cm
Hence, the area of square is 4900 cm
2
and the perimeter of square is 280 cm.
5. The area of a rectangle is 225 cm
2
and its one side is 25 cm, find its other side.
Solution:
It is given that
Area of a rectangle = 225 cm
2
Length of one side = 25 cm
We know that area of a rectangle = Product of length of two sides
So the other side = area/side
By substituting the values
Other side = 225/25 = 9 cm
Hence, the other side of the rectangle is 9 cm.
6. What will happen to the area of rectangle if its
(i) Length and breadth are trebled
(ii) Length is doubled and breadth is same
(iii) Length is doubled and breadth is halved.
Solution:
(i) Length and breadth are trebled
Consider l as the initial length and b as the initial breadth
So the original area = l × b
If the length and breadth are trebled it becomes three times more than the original value
New length = 3l
New breadth = 3b
New area of the rectangle = 3l × 3b = 9lb
Hence, the area of the rectangle becomes 9 times more than its original area.
(ii) Length is doubled and breadth is same
Consider l as the initial length and b as the initial breadth
So the original area = l × b
If the length is doubled and breadth is same we get
New length = 2l
New breadth = b
New area of the rectangle = 2l × b = 2lb
Hence, the area of the rectangle becomes 2 times more than the original area.
(iii) Length is doubled and breadth is halved
Consider l as the initial length and b as the initial breadth
So the original area = l × b
If the length is doubled and breadth is halved we get
New length = 2l
New breadth = b/2
New area of the rectangle = 2l × b/2 = lb
Hence, the area of the rectangle does not change.
7. What will happen to the area of a square if its side is:
(i) Tripled
(ii) Increased by half of it.
Solution:
(i) Tripled
Consider s as the original side of the square
We know that original area = s × s = s
2
If the side of the square is tripled we get
New side = 3s
So the new area of the square = 3s × 3s = 9s
2
Hence, the area becomes 9 times more than that of the original area.
(ii) Increased by half of it
Consider s as the original side of the square
We know that original area = s × s = s
2
If the side of the square is increased by half of it we get
New side = s + s/2 = 3s/2
So the new area of the square = 3s/2 × 3s/2 = 9s
2
/4
Hence, the area becomes 9/4 times more than that of the original area.
8. Find the perimeter of a rectangle whose area is 500 cm
2
and breadth is 20 cm.
Solution:
It is given that
Area of the rectangle = 500 cm
2
Breadth of the rectangle = 20 cm
We know that area = L × B
Page 4
1. Find the area of a rectangle, whose
(i) Length = 6 cm, breadth = 3 cm
(ii) Length = 8 cm, breadth = 3 cm
(iii) Length = 4.5 cm, breadth = 2 cm.
Solution:
(i) We know that area of a rectangle = L × B
It is given that Length = 6 cm, breadth = 3 cm
By substituting the values
Area of a rectangle = 6 × 3 = 18 cm
2
(ii) We know that area of a rectangle = L × B
It is given that Length = 8 cm, breadth = 3 cm
By substituting the values
Area of a rectangle = 8 × 3 = 24 cm
2
(iii) We know that area of a rectangle = L × B
It is given that Length = 4.5 cm, breadth = 2 cm
By substituting the values
Area of a rectangle = 4.5 × 2 = 9 cm
2
2. Find the area of a square whose side is:
(i) 5 cm
(ii) 4.1 cm
(iii) 5.5 cm
(iv) 2.6 cm
Solution:
(i) We know that area of a square = side × side
It is given that side of a square = 5 cm
So the area of the square = 5 × 5 = 25 cm
2
(ii) We know that area of a square = side × side
It is given that side of a square = 4.1 cm
So the area of the square = 4.1 × 4.1 = 16.81 cm
2
(iii) We know that area of a square = side × side
It is given that side of a square = 5.5 cm
So the area of the square = 5.5 × 5.5 = 30.25 cm
2
(iv) We know that area of a square = side × side
It is given that side of a square = 2.6 cm
So the area of the square = 2.6 × 2.6 = 6.76 cm
2
3. The area of a rectangle is 49 cm
2
and its breadth is 2.8 cm. Find the length of the rectangle.
Solution:
It is given that area of a rectangle = 49 cm
2
Breadth of a rectangle = 2.8 cm
We know that
Area of a rectangle = L × B
It can be written as
L = Area/B = 49/2.8 = 17.5 cm
Hence, the length of the rectangle is 17.5 cm.
4. The side of a square is 70 cm. Find its area and perimeter.
Solution:
It is given that side of a square = 70 cm
We know that area of a square = side × side
By substituting the values
Area of a square = 70 × 70 = 4900 cm
2
We know that perimeter of a square = 4 × side
By substituting the values
Perimeter of a square = 4 × 70 = 280 cm
Hence, the area of square is 4900 cm
2
and the perimeter of square is 280 cm.
5. The area of a rectangle is 225 cm
2
and its one side is 25 cm, find its other side.
Solution:
It is given that
Area of a rectangle = 225 cm
2
Length of one side = 25 cm
We know that area of a rectangle = Product of length of two sides
So the other side = area/side
By substituting the values
Other side = 225/25 = 9 cm
Hence, the other side of the rectangle is 9 cm.
6. What will happen to the area of rectangle if its
(i) Length and breadth are trebled
(ii) Length is doubled and breadth is same
(iii) Length is doubled and breadth is halved.
Solution:
(i) Length and breadth are trebled
Consider l as the initial length and b as the initial breadth
So the original area = l × b
If the length and breadth are trebled it becomes three times more than the original value
New length = 3l
New breadth = 3b
New area of the rectangle = 3l × 3b = 9lb
Hence, the area of the rectangle becomes 9 times more than its original area.
(ii) Length is doubled and breadth is same
Consider l as the initial length and b as the initial breadth
So the original area = l × b
If the length is doubled and breadth is same we get
New length = 2l
New breadth = b
New area of the rectangle = 2l × b = 2lb
Hence, the area of the rectangle becomes 2 times more than the original area.
(iii) Length is doubled and breadth is halved
Consider l as the initial length and b as the initial breadth
So the original area = l × b
If the length is doubled and breadth is halved we get
New length = 2l
New breadth = b/2
New area of the rectangle = 2l × b/2 = lb
Hence, the area of the rectangle does not change.
7. What will happen to the area of a square if its side is:
(i) Tripled
(ii) Increased by half of it.
Solution:
(i) Tripled
Consider s as the original side of the square
We know that original area = s × s = s
2
If the side of the square is tripled we get
New side = 3s
So the new area of the square = 3s × 3s = 9s
2
Hence, the area becomes 9 times more than that of the original area.
(ii) Increased by half of it
Consider s as the original side of the square
We know that original area = s × s = s
2
If the side of the square is increased by half of it we get
New side = s + s/2 = 3s/2
So the new area of the square = 3s/2 × 3s/2 = 9s
2
/4
Hence, the area becomes 9/4 times more than that of the original area.
8. Find the perimeter of a rectangle whose area is 500 cm
2
and breadth is 20 cm.
Solution:
It is given that
Area of the rectangle = 500 cm
2
Breadth of the rectangle = 20 cm
We know that area = L × B
It can be written as
L = Area/B
By substituting the values
L = 500/20 = 25 cm
We know that perimeter = 2 (L + B)
By substituting the values
Perimeter = 2 (25 + 20) = 2 × 45 = 90 cm
Hence, the perimeter of the rectangle is 90 cm.
9. A rectangle has the area equal to that of a square of side 80 cm. If the breadth of the rectangle is 20 cm,
find its length.
Solution:
It is given that
Side of a square = 80 cm
So the area of the square = side × side
By substituting the values
Area of square = 80 × 80 = 6400 cm
2
We know that area of rectangle = area of square = 6400 cm
2
Breadth = 20 cm
Area of rectangle = L × B
It can be written as
L = Area/B = 6400/20 = 320 cm
Hence, the length of the rectangle is 320 cm.
10. Area of a rectangle of breadth 17 cm is 340 cm
2
. Find the perimeter of the rectangle.
Solution:
The dimensions of rectangle are
Breadth = 17 cm
Area = 340 cm
2
We know that
Area of rectangle = L × B
It can be written as
L = Area/B = 340/17 = 20 cm
So the perimeter = 2 (L + B)
By substituting the values
Perimeter = 2 (20 + 17) = 2 × 37 = 74 cm
Hence, the perimeter of the rectangle is 74 cm.
11. A marble tile measures 15 cm × 20 cm. How many tiles will be required to cover a wall of size 4 m × 6
m?
Solution:
Page 5
1. Find the area of a rectangle, whose
(i) Length = 6 cm, breadth = 3 cm
(ii) Length = 8 cm, breadth = 3 cm
(iii) Length = 4.5 cm, breadth = 2 cm.
Solution:
(i) We know that area of a rectangle = L × B
It is given that Length = 6 cm, breadth = 3 cm
By substituting the values
Area of a rectangle = 6 × 3 = 18 cm
2
(ii) We know that area of a rectangle = L × B
It is given that Length = 8 cm, breadth = 3 cm
By substituting the values
Area of a rectangle = 8 × 3 = 24 cm
2
(iii) We know that area of a rectangle = L × B
It is given that Length = 4.5 cm, breadth = 2 cm
By substituting the values
Area of a rectangle = 4.5 × 2 = 9 cm
2
2. Find the area of a square whose side is:
(i) 5 cm
(ii) 4.1 cm
(iii) 5.5 cm
(iv) 2.6 cm
Solution:
(i) We know that area of a square = side × side
It is given that side of a square = 5 cm
So the area of the square = 5 × 5 = 25 cm
2
(ii) We know that area of a square = side × side
It is given that side of a square = 4.1 cm
So the area of the square = 4.1 × 4.1 = 16.81 cm
2
(iii) We know that area of a square = side × side
It is given that side of a square = 5.5 cm
So the area of the square = 5.5 × 5.5 = 30.25 cm
2
(iv) We know that area of a square = side × side
It is given that side of a square = 2.6 cm
So the area of the square = 2.6 × 2.6 = 6.76 cm
2
3. The area of a rectangle is 49 cm
2
and its breadth is 2.8 cm. Find the length of the rectangle.
Solution:
It is given that area of a rectangle = 49 cm
2
Breadth of a rectangle = 2.8 cm
We know that
Area of a rectangle = L × B
It can be written as
L = Area/B = 49/2.8 = 17.5 cm
Hence, the length of the rectangle is 17.5 cm.
4. The side of a square is 70 cm. Find its area and perimeter.
Solution:
It is given that side of a square = 70 cm
We know that area of a square = side × side
By substituting the values
Area of a square = 70 × 70 = 4900 cm
2
We know that perimeter of a square = 4 × side
By substituting the values
Perimeter of a square = 4 × 70 = 280 cm
Hence, the area of square is 4900 cm
2
and the perimeter of square is 280 cm.
5. The area of a rectangle is 225 cm
2
and its one side is 25 cm, find its other side.
Solution:
It is given that
Area of a rectangle = 225 cm
2
Length of one side = 25 cm
We know that area of a rectangle = Product of length of two sides
So the other side = area/side
By substituting the values
Other side = 225/25 = 9 cm
Hence, the other side of the rectangle is 9 cm.
6. What will happen to the area of rectangle if its
(i) Length and breadth are trebled
(ii) Length is doubled and breadth is same
(iii) Length is doubled and breadth is halved.
Solution:
(i) Length and breadth are trebled
Consider l as the initial length and b as the initial breadth
So the original area = l × b
If the length and breadth are trebled it becomes three times more than the original value
New length = 3l
New breadth = 3b
New area of the rectangle = 3l × 3b = 9lb
Hence, the area of the rectangle becomes 9 times more than its original area.
(ii) Length is doubled and breadth is same
Consider l as the initial length and b as the initial breadth
So the original area = l × b
If the length is doubled and breadth is same we get
New length = 2l
New breadth = b
New area of the rectangle = 2l × b = 2lb
Hence, the area of the rectangle becomes 2 times more than the original area.
(iii) Length is doubled and breadth is halved
Consider l as the initial length and b as the initial breadth
So the original area = l × b
If the length is doubled and breadth is halved we get
New length = 2l
New breadth = b/2
New area of the rectangle = 2l × b/2 = lb
Hence, the area of the rectangle does not change.
7. What will happen to the area of a square if its side is:
(i) Tripled
(ii) Increased by half of it.
Solution:
(i) Tripled
Consider s as the original side of the square
We know that original area = s × s = s
2
If the side of the square is tripled we get
New side = 3s
So the new area of the square = 3s × 3s = 9s
2
Hence, the area becomes 9 times more than that of the original area.
(ii) Increased by half of it
Consider s as the original side of the square
We know that original area = s × s = s
2
If the side of the square is increased by half of it we get
New side = s + s/2 = 3s/2
So the new area of the square = 3s/2 × 3s/2 = 9s
2
/4
Hence, the area becomes 9/4 times more than that of the original area.
8. Find the perimeter of a rectangle whose area is 500 cm
2
and breadth is 20 cm.
Solution:
It is given that
Area of the rectangle = 500 cm
2
Breadth of the rectangle = 20 cm
We know that area = L × B
It can be written as
L = Area/B
By substituting the values
L = 500/20 = 25 cm
We know that perimeter = 2 (L + B)
By substituting the values
Perimeter = 2 (25 + 20) = 2 × 45 = 90 cm
Hence, the perimeter of the rectangle is 90 cm.
9. A rectangle has the area equal to that of a square of side 80 cm. If the breadth of the rectangle is 20 cm,
find its length.
Solution:
It is given that
Side of a square = 80 cm
So the area of the square = side × side
By substituting the values
Area of square = 80 × 80 = 6400 cm
2
We know that area of rectangle = area of square = 6400 cm
2
Breadth = 20 cm
Area of rectangle = L × B
It can be written as
L = Area/B = 6400/20 = 320 cm
Hence, the length of the rectangle is 320 cm.
10. Area of a rectangle of breadth 17 cm is 340 cm
2
. Find the perimeter of the rectangle.
Solution:
The dimensions of rectangle are
Breadth = 17 cm
Area = 340 cm
2
We know that
Area of rectangle = L × B
It can be written as
L = Area/B = 340/17 = 20 cm
So the perimeter = 2 (L + B)
By substituting the values
Perimeter = 2 (20 + 17) = 2 × 37 = 74 cm
Hence, the perimeter of the rectangle is 74 cm.
11. A marble tile measures 15 cm × 20 cm. How many tiles will be required to cover a wall of size 4 m × 6
m?
Solution:
Measure of marble tile = 15 cm × 20 cm
Size of wall = 4 m × 6 m = 400 cm × 600 cm
So we get area of tile = 15 cm × 20 cm = 300 cm
2
Area of wall = 400 cm × 600 cm = 240000 cm
2
No. of tiles required to cover the wall = Area of wall/ Area of one tile
Substituting the values
No. of tiles required to cover the wall = 240000/300 = 800 tiles
Hence, 800 tiles are required to cover a wall of size 4 m × 6 m.
12. A marble tile measures 10 cm × 12 cm. How many tiles will be required to cover a wall of size 3 m × 4
m? Also, find the total cost of the tiles at the rate of Rs 2 per tile.
Solution:
Measure of marble tile = 10 cm × 12 cm
Size of the wall = 3 m × 4 m = 300 cm × 400 cm
So the area of marble tile = 10 cm × 12 cm = 120 cm
2
Area of wall = 300 cm × 400 cm = 120000 cm
2
No. of tiles required to cover the wall = Area of wall/ Area of one tile
Substituting the values
No. of tiles required to cover the wall = 120000/120 = 1000 tiles
It is given that
Cost of one tile = Rs 2
So the cost of 1000 tiles = 1000 × 2 = Rs 2000
Hence, 1000 number of tiles are required to cover the wall and the cost is Rs 2000.
13. One side of a square plot is 250 m, find the cost of levelling it at the rate of Rs 2 per square metre.
Solution:
It is given that
Side of one tile of a square plot = 250 m
So the area = side × side = 250 × 250 = 62500 m
2
Cost of levelling = Rs 2 per square meter
So the cost of levelling 62500 m
2
= 62500 × 2 = Rs 125000
Hence, the cost of levelling is Rs 125000.
14. The following figures have been split into rectangles. Find their areas. (The measures are given in
centimetres)
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Document Description: RD Sharma Solutions: Mensuration (Exercise 20.4) for Class 6 2024 is part of Mathematics (Maths) Class 6 preparation.
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,Mensuration (Exercise 20.4) RD Sharma Solutions | Mathematics (Maths) Class 6
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,Mensuration (Exercise 20.4) RD Sharma Solutions | Mathematics (Maths) Class 6
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Additional Information about RD Sharma Solutions: Mensuration (Exercise 20.4) for Class 6 Preparation
RD Sharma Solutions: Mensuration (Exercise 20.4) Free PDF Download
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RD Sharma Solutions: Mensuration (Exercise 20.4) Notes
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RD Sharma Solutions: Mensuration (Exercise 20.4) Class 6 Questions
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The content of RD Sharma Solutions: Mensuration (Exercise 20.4) is prepared as per the latest Class 6 syllabus.
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