Rd Sharma Solutions Class 6 Chapter 20 Mensuration
Welcome to NCTB Solution. Here with this post we are going to help 6th class students for the Solutions of Rd Sharma Class 6 Mathematics, Chapter 20, Mensuration. Here students can easily find Exercise wise solution for chapter 20, Mensuration. Students will find proper solutions for Exercise 20.1, 20.2, 20.3 and 20.4. Our teachers solved every problem with easily understandable methods so that every students can understand easily. Here all solutions are based on the CBSE latest curriculum.
Mensuration Exercise 20.1 Solution
Question no – (1)
Solution :
The length of the boundary of a closed figure is known as its Perimeter.
Question no – (2)
Solution :
(i) Perimeter of the given figure,
= 4 + 2 + 1 – 5
= 12 cm
(ii) Perimeter of the given figure,
= (23 + 40 + 35 + 35) cm
= 133 cm
(iii) Perimeter of the given figure,
= (15 + 15 + 15 + 15) cm
= 60 cm
(iv) Perimeter of the given figure,
= (3 + 3 + 3 + 3 + 3) cm
= 15 cm
Mensuration Exercise 20.2 Solution
Question no – (1)
Solution :
(i) Given, 7 cm, 5 cm
∴ Perimeter of rectangle,
= 2 (7 + 5) cm
= 2 × 12
= 24 cm
Hence, the perimeter will be 24 cm.
(ii) Given, 5 cm, 4 cm
∴ Perimeter of rectangle,
= 2 ( 5 + 4)
= (2 × 9) cm
= 18 cm
Thus, the perimeter will be 18 cm.
(iii) Given, 7.5 cm, 4.5 cm
∴ Perimeter of rectangle,
= 2 (7.5 + 4.5) cm
= 2 × 12
= 24 cm
Therefore, the perimeter will be 24 cm.
Question no – (2)
Solution :
As we know,
Perimeter of a square = 4 × (side length)
(i) Given, 10 cm
∴ Perimeter of square,
= (4 × 10) cm
= 40 cm
Thus, the perimeter will be 40 cm.
(ii) Given, 5 m
∴ Perimeter of square,
= (4 × 5) m
= 20 cm
Hence, the perimeter will be 20 cm.
(iii) Given, 115.5 cm
∴ Perimeter of square,
= (4 × 115.5) cm
= 462 cm
Therefore, the perimeter will be 462 cm.
Question no – (3)
Solution :
As we know that,
Side of a square = perimeter /4
(i) Given perimeter = 16 cm
∴ Side of a square,
= 16/4
= 4 cm
Thus, side of a square is 4 cm.
(ii) Given, perimeter = 40 cm
∴ Side of a square,
= 40/4 cm
= 10 cm
Hence, side of a square is 10 cm.
(iii) Given, perimeter = 22 cm
∴ Side of a square,
= 22/4
= 5.5 cm
Therefore, side of a square is 5.5 cm.
Question no – (4)
Solution :
(i) Given length is = 116 cm
∴ Breadth of rectangle = 360/2 – 116
= 180 – 116
= 64 cm
Therefore, breadth of rectangle is 64 cm.
(ii) Given, Perimeter is = 360 cm
length is = 140 cm
∴ Breadth of rectangle,
= 360/2 – 140
= 180 – 140
= 40 cm
Thus, breadth of rectangle is 40 cm.
(iii) Given, Perimeter is = 360 cm
Length is = 102
∴ Breadth of rectangle,
= 360/2 – 102
= 180 – 102
= 78 cm.
Therefore, breadth of rectangle is 78 cm.
Question no – (5)
Solution :
Given, Length is = 98 m
Breadth is = 55 m
∴ Perimeter of the lawn,
= 2 (l + b)
= 2 (98 + 55) m
= 2 × 153
= 306 m
Therefore, the length of the fence will be 306 m.
Question no – (6)
Solution :
As per the question,
side of a square field is 65m
the length of the required = ?
∴ Perimeter of square field,
= (4 × 65) m
= 260 m
Therefore, the length of the fence required is 260 m.
Question no – (7)
Solution :
As per the question,
Two sides of a rectangle = 15 cm and 20 cm
Perimeter of the triangle = 50 cm
Third side = ?
∴ Length of third side,
= 50 – (15 + 20)
= 50 – 35
= 15 cm
Therefore, the third side will be 15 cm.
Question no – (8)
Solution :
According to the question,
Here, perimeter = 20 m
∴ 2 (l + b) = 20 m
and, (l + b) = 10 m
Now, if the ides are positive integers in meters.
∴ The possible dimensions are,
= (1 m, 9 m), (2 m, 8 m), (3 m 7 m), (4 m, 6 m), (5 m, 5 m)
Question no – (9)
Solution :
As per the question,
Square piece of land has each side equal to = 100 m
∴ Perimeter of square field,
= (4 × 100) m
= 1200 m
Therefore, 1200 m of the wire will be needed.
Question no – (11)
Solution :
According to the question,
Dimensions of a photographs = 30 cm × 20 cm
∴ Length of wooden frame needed,
= 2 (l + b)
= 2 (30 + 20)
= 2 × 50 cm
= 100 cm
Hence, 100 cm wooden frame is needed to frame the picture.
Question no – (12)
Solution :
As per the question,
Length of a rectangular field = 100 m.
Perimeter = 300 m
∴ Breadth of rectangle field,
= (300/2 – 100)
= 150 – 100) m
= 50 cm
Thus, the its breadth is 50 cm.
Question no – (13)
Solution :
First we need to find the perimeter of garden,
= 2 (70 + 50) m
= 2 × 120 m
= 240 m
Now, number of posts required,
= 240/5
= 48
length of each post = 2m
∴ Total length of pipe required,
= (48 × 2)
= 96 m
Thus, total length of the pipes he bought is 96 m.
Question no – (15)
Solution :
As per question,
Perimeter of a rectangular pentagon = 100 cm
∴ Side of pentagon,
= 100/5
= 20 cm
Therefore, each side of regular pentagon is 20 cm.
Question no – (16)
Solution :
According to question,
Side of regular hexagon = 8 m
∴ Perimeter,
= (6 × 8)
= 48 m
Therefore, the perimeter of a regular hexagon is 48 m.
Question no – (17)
Solution :
1st, we need to find the perimeter of land,
= 2 (0.7 + 0.5)
= (2 × 1,2) km
= 2.4 km
Now, length of wire needed,
= (4 × 2.4) km
= 9.6 km
Therefore, 9.6 km of the wire is needed.
Mensuration Exercise 20.3 Solution
Question no – (1)
Solution :
(i) Area of the given figure
= (16 × 1)
= 16 cm2
(ii) Area of the given figure
= (36 × 1)
= 36 cm2
(iii) Area of the given figure
= (15 + 6 × ½)
= (15 + 3)
= 18 cm2
(iv) Area of the given figure
= (20 + 8 × 1/2) cm2
= (20 + 4)
= 24 cm3
(v) Area of the given figure
= (13 + 8 × 1)
= (13 + 8)
= 21cm2
Question no – (2)
Solution :
Here, has 18 completed squares.
∴ Area of rectangle,
= (18 × 1) cm²
= 18 cm²
Therefore, the approximate area will be 18 cm².
Mensuration Exercise 20.4 Solution
Question no – (1)
Solution :
We know that, Area = length × breadth
(i) Given, Length = 6 cm,
Breadth = 3 cm
∴ Area of rectangle,
= (6 × 3)
= 18 cm2
Therefore, the area of the rectangle will be 18 cm2
(ii) Given, Length = 8 cm,
Breadth = 3 cm
∴ Area of rectangle
= (8 × 3)
= 24 cm2
Hence, the area of rectangle will be 24 cm2
(iii) Given, Length = 4.5 cm, Breadth = 2 cm
∴ Area of rectangle,
= (4.5 × 2)
= 9 cm2
Therefore, the area of rectangle is 9 cm2
Question no – (2)
Solution :
As we know that, Area of square = Side × side
(i) Given side = 5 cm
∴ Area of square,
= (5 × 5) cm2
= 25 cm2
Thus, the area of square will be 25 cm2
(ii) As we know that,
Area of square = Side x side
Given side = 4.1 cm
∴ Area of square,
= (4.1 × 4.1) cm²
= 16.81 cm²
Therefore, the area of square is 16.81 cm²
(iii) Given side = 5.5 cm
∴ Area of square,
= (5.5 × 5.5) cm2
= 30.25 cm2
Therefore, the area of square will be 30.25 cm2
(iv) Given side = 2.6 cm
∴ Area of square,
= (2.6 × 2.6) cm2
= 6.76 cm2
Hence, the area of square will be 6.76 cm2
Question no – (3)
Solution :
As per the question, Area of a rectangle is 49 cm²
and, breadth is 2.8 cm
(As we know that,
Length = Area of a rectangle / breadth)
∴ Length of rectangle,
= 49/2.8 cm
= 17.5 cm
Therefore, the length of rectangle will be 17.5 cm.
Question no – (4)
Solution :
Given, Side of a square is = 70 cm
∴ Area of square,
= (70 × 70) cm²
= 4900 cm²
∴ Perimeter of square,
= (4 × 70) cm
= 280 cm
Therefore, the area will be 4900 cm² and the perimeter will be 280 cm.
Question no – (5)
Solution :
According to the question,
The area of a rectangle is 225 cm²
One side is = 25 cm,
Other side = ?
As we know that,
Length = Area of a rectangle / breadth
∴ Length of other side,
= 225/25 cm
= 9 cm
Therefore, its other side will be 9 cm.
Question no – (6)
Solution :
(i) According to the question,
Length of trebled = 3l
Breadth of trebled = 3b
∴ Area = (3l × 3b) = 9 lb
Original length = l
Breadth = b
∴ The area of rectangle becomes 9 times more than its original area.
(ii) Let, length = l; breadth = b
∴ Area = (l × b) = lb
New Length = l × 2 = 2l
Breadth = 6
∴ Area = (2l × b) 2 bl
∴ The area of rectangle becomes 2 times more than the original.
(iii) Let, Length = l; breadth = b
∴ Area = (l × b) = lb
New Length = (l × 2) = 2l
Breadth = b/2
∴ Area = (2l × b/2)
= lb
Therefore, Area of rectangle does not change.
Question no – (7)
Solution :
Let, side of original square = a
∴ Area = (a × a) = a²
Now, length of side,
= 2 + s/2
= 2s + 5/2
= 3s/2
∴ Area = (3s/2 × 3s/2)
= 9s²/4
Therefore, now Area is 9/4 times more than original Area.
Question no – (8)
Solution :
According to the question,
Area of rectangle = 500 cm²
and, its breadth = 20 cm.
∴ Length of rectangle,
= 500/20 cm
= 25 cm
Breadth = 20 cm
∴ Perimeter of rectangle,
= 2 (25 + 20) cm
= 2 × 35
= 90 cm
Therefore, the perimeter of a rectangle is 90 cm.
Question no – (9)
Solution :
Area of square,
= (80 × 80) cm2
= 6400 cm2
Here, area of rectangle = area of square
∴ Rectangle breadth = 200 cm
∴ Length = 6400/20
= 320 cm
Hence, its length will be 320 cm.
Question no – (10)
Solution :
According to the question,
Area of a rectangle of breadth 17 cm is = 340 cm2
∴ Length of rectangle,
= 340/17
= 20 cm
∴ Perimeter,
= 2 (l + b)
= 2 (20 + 17)
= 2 × 37
= 74 cm
Therefore, the perimeter of the rectangle is 74 cm.
Question no – (11)
Solution :
Length of tile,
= (15 × 20) cm2
= 300 cm2
Area of wall,
= (400 × 600) cm2
= 240000 cm2
∴ No of tiles required,
= 240000/300
= 800
Hence, 800 tiles will be required to cover the wall.
Question no – (12)
Solution :
1st, we need to find the Area of tile,
= (10 × 12) cm2
= 120 cm2
Now, we find the Area of wall,
= (300 × 400) cm2
= 120000 cm2
∴ No of tiles required to cover,
= 120000/120
= 1000 tiles
∴ Total cost,
= (2 × 1000) Rs
= 2000 Rs
Question no – (13)
Solution :
Area of the tile,
= (250 × 250)
= 62500 m2
∴ Cost of levelling,
= (62500 × 2) Rs
= 125000 Rs
Therefore, the cost of levelling will be 125000 Rs.
Question no – (16)
Solution :
First, we need find the Area of tile,
= (5 × 12) cm2
= 60 cm2
Now,
(i) 100 cm and 144 cm,
∴ Area of region,
= 14400/60
= 240 tiles
Therefore, total 240 tiles will be needed.
(ii) 70 cm and 36 cm,
∴ Area of tile,
= (5 × 12) cm2
= 60 cm2
∴ Area of region,
= (70 × 36) cm
= 2520 cm
Therefore, 2520 tiles will be required.
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