Class 8 Exam > Class 8 Notes > Mathematics (Maths) Class 8 > Mensuration - I (Area of a Trapezium and a Polygon) - EXERCISE 20.1
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Page 1
RD Sharma Solutions for Class 8 Maths Chapter 20 –
Mensuration – I (Area of a Trapezium and a Polygon)
EXERCISE 20.1 PAGE NO: 20.13
1. A flooring tile has the shape of a parallelogram whose base is 24 cm and the
corresponding height is 10 cm. How many such tiles are required to cover a floor of
area 1080 m
2
?
Solution:
Given that,
Base of parallelogram = 24cm
Height of parallelogram = 10cm
Area of floor = 1080m
2
We know that,
Area of parallelogram = Base × Height
Area of 1 tile = 24 × 10 = 240cm
2
We know that, 1m = 100cm
So for 1080m
2
= 1080 × 100 × 100 cm
2
To calculate the Number of tiles required = Area of floor/Area of 1 tile
i.e., Number of tiles required = (1080 × 100 × 100) / (24 × 10) = 45000
? Number of tiles required = 45000
2. A plot is in the form of a rectangle ABCD having semi-circle on BC as shown in
Fig. 20.23. If AB = 60 m and BC = 28 m, Find the area of the plot.
Solution:
Area of the plot = Area of the rectangle + Area of semi-circle
Radius of semi-circle = BC/2 = 28/2 = 14m
Area of the Rectangular plot = Length × Breadth = 60 × 28 = 1680 m
2
Area of the Semi-circular portion = pr
2
/2
= 1/2 × 22/7 × 14 × 14
= 308 m
2
? The total area of the plot = 1680 + 308 = 1988 m
2
Page 2
RD Sharma Solutions for Class 8 Maths Chapter 20 –
Mensuration – I (Area of a Trapezium and a Polygon)
EXERCISE 20.1 PAGE NO: 20.13
1. A flooring tile has the shape of a parallelogram whose base is 24 cm and the
corresponding height is 10 cm. How many such tiles are required to cover a floor of
area 1080 m
2
?
Solution:
Given that,
Base of parallelogram = 24cm
Height of parallelogram = 10cm
Area of floor = 1080m
2
We know that,
Area of parallelogram = Base × Height
Area of 1 tile = 24 × 10 = 240cm
2
We know that, 1m = 100cm
So for 1080m
2
= 1080 × 100 × 100 cm
2
To calculate the Number of tiles required = Area of floor/Area of 1 tile
i.e., Number of tiles required = (1080 × 100 × 100) / (24 × 10) = 45000
? Number of tiles required = 45000
2. A plot is in the form of a rectangle ABCD having semi-circle on BC as shown in
Fig. 20.23. If AB = 60 m and BC = 28 m, Find the area of the plot.
Solution:
Area of the plot = Area of the rectangle + Area of semi-circle
Radius of semi-circle = BC/2 = 28/2 = 14m
Area of the Rectangular plot = Length × Breadth = 60 × 28 = 1680 m
2
Area of the Semi-circular portion = pr
2
/2
= 1/2 × 22/7 × 14 × 14
= 308 m
2
? The total area of the plot = 1680 + 308 = 1988 m
2
RD Sharma Solutions for Class 8 Maths Chapter 20 –
Mensuration – I (Area of a Trapezium and a Polygon)
3. A playground has the shape of a rectangle, with two semi-circles on its smaller
sides as diameters, added to its outside. If the sides of the rectangle are 36 m and
24.5 m, find the area of the playground. (Take p= 22/7.)
Solution:
Area of the plot = Area of the Rectangle + 2 × area of one semi-circle
Radius of semi-circle = BC/2 = 24.5/2 = 12.25m
Area of the Rectangular plot = Length × Breadth = 36 × 24.5 = 882 m
2
Area of the Semi-circular portions = 2 × pr
2
/2
= 2 × 1/2 × 22/7 × 12.25 × 12.25 = 471.625 m
2
Area of the plot = 882 + 471.625 = 1353.625 m
2
4. A rectangular piece is 20 m long and 15 m wide. From its four corners, quadrants
of radii 3.5 m have been cut. Find the area of the remaining part.
Solution:
Area of the plot = Area of the rectangle - 4 × area of one quadrant
Radius of semi-circle = 3.5 m
Area of four quadrants = area of one circle
Area of the plot = Length × Breadth - pr
2
Page 3
RD Sharma Solutions for Class 8 Maths Chapter 20 –
Mensuration – I (Area of a Trapezium and a Polygon)
EXERCISE 20.1 PAGE NO: 20.13
1. A flooring tile has the shape of a parallelogram whose base is 24 cm and the
corresponding height is 10 cm. How many such tiles are required to cover a floor of
area 1080 m
2
?
Solution:
Given that,
Base of parallelogram = 24cm
Height of parallelogram = 10cm
Area of floor = 1080m
2
We know that,
Area of parallelogram = Base × Height
Area of 1 tile = 24 × 10 = 240cm
2
We know that, 1m = 100cm
So for 1080m
2
= 1080 × 100 × 100 cm
2
To calculate the Number of tiles required = Area of floor/Area of 1 tile
i.e., Number of tiles required = (1080 × 100 × 100) / (24 × 10) = 45000
? Number of tiles required = 45000
2. A plot is in the form of a rectangle ABCD having semi-circle on BC as shown in
Fig. 20.23. If AB = 60 m and BC = 28 m, Find the area of the plot.
Solution:
Area of the plot = Area of the rectangle + Area of semi-circle
Radius of semi-circle = BC/2 = 28/2 = 14m
Area of the Rectangular plot = Length × Breadth = 60 × 28 = 1680 m
2
Area of the Semi-circular portion = pr
2
/2
= 1/2 × 22/7 × 14 × 14
= 308 m
2
? The total area of the plot = 1680 + 308 = 1988 m
2
RD Sharma Solutions for Class 8 Maths Chapter 20 –
Mensuration – I (Area of a Trapezium and a Polygon)
3. A playground has the shape of a rectangle, with two semi-circles on its smaller
sides as diameters, added to its outside. If the sides of the rectangle are 36 m and
24.5 m, find the area of the playground. (Take p= 22/7.)
Solution:
Area of the plot = Area of the Rectangle + 2 × area of one semi-circle
Radius of semi-circle = BC/2 = 24.5/2 = 12.25m
Area of the Rectangular plot = Length × Breadth = 36 × 24.5 = 882 m
2
Area of the Semi-circular portions = 2 × pr
2
/2
= 2 × 1/2 × 22/7 × 12.25 × 12.25 = 471.625 m
2
Area of the plot = 882 + 471.625 = 1353.625 m
2
4. A rectangular piece is 20 m long and 15 m wide. From its four corners, quadrants
of radii 3.5 m have been cut. Find the area of the remaining part.
Solution:
Area of the plot = Area of the rectangle - 4 × area of one quadrant
Radius of semi-circle = 3.5 m
Area of four quadrants = area of one circle
Area of the plot = Length × Breadth - pr
2
RD Sharma Solutions for Class 8 Maths Chapter 20 –
Mensuration – I (Area of a Trapezium and a Polygon)
Area of the plot = 20 × 15 – (22/7 × 3.5 × 3.5)
Area of the plot = 300 – 38.5 = 261.5 m
2
5. The inside perimeter of a running track (shown in Fig. 20.24) is 400 m. The length
of each of the straight portion is 90 m and the ends are semi-circles. If track is
everywhere 14 m wide, find the area of the track. Also, find the length of the outer
running track.
Solution:
Perimeter of the inner track = 2 × Length of rectangle + perimeter of two semi-circular
ends
Perimeter of the inner track = Length + Length + 2pr
400 = 90 + 90 + (2 × 22/7 × r)
(2 × 22/7 × r) = 400 – 180
(2 × 22/7 × r) = 220
44r = 220 × 7
44r = 1540
r = 1540/44 = 35
r = 35m
So, the radius of inner circle = 35 m
Now, let’s calculate the radius of outer track
Radius of outer track = Radius of inner track + width of the track
Radius of outer track = 35 + 14 = 49m
Length of outer track = 2× Length of rectangle + perimeter of two outer semi-circular
ends
Length of outer track = 2× 90 + 2pr
Length of outer track = 2× 90 + (2 × 22/7 × 49)
Length of outer track = 180 + 308 = 488
So, Length of outer track = 488m
Area of inner track = Area of inner rectangle + Area of two inner semi-circles
Page 4
RD Sharma Solutions for Class 8 Maths Chapter 20 –
Mensuration – I (Area of a Trapezium and a Polygon)
EXERCISE 20.1 PAGE NO: 20.13
1. A flooring tile has the shape of a parallelogram whose base is 24 cm and the
corresponding height is 10 cm. How many such tiles are required to cover a floor of
area 1080 m
2
?
Solution:
Given that,
Base of parallelogram = 24cm
Height of parallelogram = 10cm
Area of floor = 1080m
2
We know that,
Area of parallelogram = Base × Height
Area of 1 tile = 24 × 10 = 240cm
2
We know that, 1m = 100cm
So for 1080m
2
= 1080 × 100 × 100 cm
2
To calculate the Number of tiles required = Area of floor/Area of 1 tile
i.e., Number of tiles required = (1080 × 100 × 100) / (24 × 10) = 45000
? Number of tiles required = 45000
2. A plot is in the form of a rectangle ABCD having semi-circle on BC as shown in
Fig. 20.23. If AB = 60 m and BC = 28 m, Find the area of the plot.
Solution:
Area of the plot = Area of the rectangle + Area of semi-circle
Radius of semi-circle = BC/2 = 28/2 = 14m
Area of the Rectangular plot = Length × Breadth = 60 × 28 = 1680 m
2
Area of the Semi-circular portion = pr
2
/2
= 1/2 × 22/7 × 14 × 14
= 308 m
2
? The total area of the plot = 1680 + 308 = 1988 m
2
RD Sharma Solutions for Class 8 Maths Chapter 20 –
Mensuration – I (Area of a Trapezium and a Polygon)
3. A playground has the shape of a rectangle, with two semi-circles on its smaller
sides as diameters, added to its outside. If the sides of the rectangle are 36 m and
24.5 m, find the area of the playground. (Take p= 22/7.)
Solution:
Area of the plot = Area of the Rectangle + 2 × area of one semi-circle
Radius of semi-circle = BC/2 = 24.5/2 = 12.25m
Area of the Rectangular plot = Length × Breadth = 36 × 24.5 = 882 m
2
Area of the Semi-circular portions = 2 × pr
2
/2
= 2 × 1/2 × 22/7 × 12.25 × 12.25 = 471.625 m
2
Area of the plot = 882 + 471.625 = 1353.625 m
2
4. A rectangular piece is 20 m long and 15 m wide. From its four corners, quadrants
of radii 3.5 m have been cut. Find the area of the remaining part.
Solution:
Area of the plot = Area of the rectangle - 4 × area of one quadrant
Radius of semi-circle = 3.5 m
Area of four quadrants = area of one circle
Area of the plot = Length × Breadth - pr
2
RD Sharma Solutions for Class 8 Maths Chapter 20 –
Mensuration – I (Area of a Trapezium and a Polygon)
Area of the plot = 20 × 15 – (22/7 × 3.5 × 3.5)
Area of the plot = 300 – 38.5 = 261.5 m
2
5. The inside perimeter of a running track (shown in Fig. 20.24) is 400 m. The length
of each of the straight portion is 90 m and the ends are semi-circles. If track is
everywhere 14 m wide, find the area of the track. Also, find the length of the outer
running track.
Solution:
Perimeter of the inner track = 2 × Length of rectangle + perimeter of two semi-circular
ends
Perimeter of the inner track = Length + Length + 2pr
400 = 90 + 90 + (2 × 22/7 × r)
(2 × 22/7 × r) = 400 – 180
(2 × 22/7 × r) = 220
44r = 220 × 7
44r = 1540
r = 1540/44 = 35
r = 35m
So, the radius of inner circle = 35 m
Now, let’s calculate the radius of outer track
Radius of outer track = Radius of inner track + width of the track
Radius of outer track = 35 + 14 = 49m
Length of outer track = 2× Length of rectangle + perimeter of two outer semi-circular
ends
Length of outer track = 2× 90 + 2pr
Length of outer track = 2× 90 + (2 × 22/7 × 49)
Length of outer track = 180 + 308 = 488
So, Length of outer track = 488m
Area of inner track = Area of inner rectangle + Area of two inner semi-circles
RD Sharma Solutions for Class 8 Maths Chapter 20 –
Mensuration – I (Area of a Trapezium and a Polygon)
Area of inner track = Length × Breadth + pr
2
Area of inner track = 90 × 70 + (22/7 × 35 × 35)
Area of inner track = 6300 + 3850
So, Area of inner track = 10150 m
2
Area of outer track = Area of outer rectangle + Area of two outer semi-circles
Breadth of outer track = 35 + 35 +14 + 14 = 98 m
Area of outer track = length× breadth + pr
2
Area of outer track = 90 × 98 + (22/7 × 49 × 49)
Area of outer track = 8820 + 7546
So, Area of outer track = 16366 m
2
Now, let’s calculate area of path
Area of path = Area of outer track – Area of inner track
Area of path = 16366 – 10150 = 6216
So, Area of path = 6216 m
2
6. Find the area of Fig. 20.25, in square cm, correct to one place of decimal. (Take p
=22/7)
Solution:
Area of the Figure = Area of square + Area of semi-circle – Area of right angled triangle
Area of the Figure = side × side + pr
2
/2 – (1/2 × base × height)
Area of the Figure = 10 × 10 + (1/2 × 22/7 × 5 × 5) – (1/2 × 8 × 6)
Area of the Figure = 100 + 39.28 – 24
Area of the Figure = 115.3
So, Area of the Figure = 115.3 cm
2
7. The diameter of a wheel of a bus is 90 cm which makes 315 revolutions per
minute. Determine its speed in kilometres per hour. (Take p=22/7)
Solution:
Page 5
RD Sharma Solutions for Class 8 Maths Chapter 20 –
Mensuration – I (Area of a Trapezium and a Polygon)
EXERCISE 20.1 PAGE NO: 20.13
1. A flooring tile has the shape of a parallelogram whose base is 24 cm and the
corresponding height is 10 cm. How many such tiles are required to cover a floor of
area 1080 m
2
?
Solution:
Given that,
Base of parallelogram = 24cm
Height of parallelogram = 10cm
Area of floor = 1080m
2
We know that,
Area of parallelogram = Base × Height
Area of 1 tile = 24 × 10 = 240cm
2
We know that, 1m = 100cm
So for 1080m
2
= 1080 × 100 × 100 cm
2
To calculate the Number of tiles required = Area of floor/Area of 1 tile
i.e., Number of tiles required = (1080 × 100 × 100) / (24 × 10) = 45000
? Number of tiles required = 45000
2. A plot is in the form of a rectangle ABCD having semi-circle on BC as shown in
Fig. 20.23. If AB = 60 m and BC = 28 m, Find the area of the plot.
Solution:
Area of the plot = Area of the rectangle + Area of semi-circle
Radius of semi-circle = BC/2 = 28/2 = 14m
Area of the Rectangular plot = Length × Breadth = 60 × 28 = 1680 m
2
Area of the Semi-circular portion = pr
2
/2
= 1/2 × 22/7 × 14 × 14
= 308 m
2
? The total area of the plot = 1680 + 308 = 1988 m
2
RD Sharma Solutions for Class 8 Maths Chapter 20 –
Mensuration – I (Area of a Trapezium and a Polygon)
3. A playground has the shape of a rectangle, with two semi-circles on its smaller
sides as diameters, added to its outside. If the sides of the rectangle are 36 m and
24.5 m, find the area of the playground. (Take p= 22/7.)
Solution:
Area of the plot = Area of the Rectangle + 2 × area of one semi-circle
Radius of semi-circle = BC/2 = 24.5/2 = 12.25m
Area of the Rectangular plot = Length × Breadth = 36 × 24.5 = 882 m
2
Area of the Semi-circular portions = 2 × pr
2
/2
= 2 × 1/2 × 22/7 × 12.25 × 12.25 = 471.625 m
2
Area of the plot = 882 + 471.625 = 1353.625 m
2
4. A rectangular piece is 20 m long and 15 m wide. From its four corners, quadrants
of radii 3.5 m have been cut. Find the area of the remaining part.
Solution:
Area of the plot = Area of the rectangle - 4 × area of one quadrant
Radius of semi-circle = 3.5 m
Area of four quadrants = area of one circle
Area of the plot = Length × Breadth - pr
2
RD Sharma Solutions for Class 8 Maths Chapter 20 –
Mensuration – I (Area of a Trapezium and a Polygon)
Area of the plot = 20 × 15 – (22/7 × 3.5 × 3.5)
Area of the plot = 300 – 38.5 = 261.5 m
2
5. The inside perimeter of a running track (shown in Fig. 20.24) is 400 m. The length
of each of the straight portion is 90 m and the ends are semi-circles. If track is
everywhere 14 m wide, find the area of the track. Also, find the length of the outer
running track.
Solution:
Perimeter of the inner track = 2 × Length of rectangle + perimeter of two semi-circular
ends
Perimeter of the inner track = Length + Length + 2pr
400 = 90 + 90 + (2 × 22/7 × r)
(2 × 22/7 × r) = 400 – 180
(2 × 22/7 × r) = 220
44r = 220 × 7
44r = 1540
r = 1540/44 = 35
r = 35m
So, the radius of inner circle = 35 m
Now, let’s calculate the radius of outer track
Radius of outer track = Radius of inner track + width of the track
Radius of outer track = 35 + 14 = 49m
Length of outer track = 2× Length of rectangle + perimeter of two outer semi-circular
ends
Length of outer track = 2× 90 + 2pr
Length of outer track = 2× 90 + (2 × 22/7 × 49)
Length of outer track = 180 + 308 = 488
So, Length of outer track = 488m
Area of inner track = Area of inner rectangle + Area of two inner semi-circles
RD Sharma Solutions for Class 8 Maths Chapter 20 –
Mensuration – I (Area of a Trapezium and a Polygon)
Area of inner track = Length × Breadth + pr
2
Area of inner track = 90 × 70 + (22/7 × 35 × 35)
Area of inner track = 6300 + 3850
So, Area of inner track = 10150 m
2
Area of outer track = Area of outer rectangle + Area of two outer semi-circles
Breadth of outer track = 35 + 35 +14 + 14 = 98 m
Area of outer track = length× breadth + pr
2
Area of outer track = 90 × 98 + (22/7 × 49 × 49)
Area of outer track = 8820 + 7546
So, Area of outer track = 16366 m
2
Now, let’s calculate area of path
Area of path = Area of outer track – Area of inner track
Area of path = 16366 – 10150 = 6216
So, Area of path = 6216 m
2
6. Find the area of Fig. 20.25, in square cm, correct to one place of decimal. (Take p
=22/7)
Solution:
Area of the Figure = Area of square + Area of semi-circle – Area of right angled triangle
Area of the Figure = side × side + pr
2
/2 – (1/2 × base × height)
Area of the Figure = 10 × 10 + (1/2 × 22/7 × 5 × 5) – (1/2 × 8 × 6)
Area of the Figure = 100 + 39.28 – 24
Area of the Figure = 115.3
So, Area of the Figure = 115.3 cm
2
7. The diameter of a wheel of a bus is 90 cm which makes 315 revolutions per
minute. Determine its speed in kilometres per hour. (Take p=22/7)
Solution:
RD Sharma Solutions for Class 8 Maths Chapter 20 –
Mensuration – I (Area of a Trapezium and a Polygon)
Given that, Diameter of a wheel = 90 cm
We know that, Perimeter of wheel = pd
Perimeter of wheel = 22/7 × 90 = 282.857
So, Perimeter of a wheel = 282.857 cm
Distance covered in 315 revolutions = 282.857× 315 = 89099.955 cm
One km = 100000 cm
Therefore, Distance covered = 89099.955/100000 = 0.89 km
Speed in km per hour = 0.89 × 60 = 53.4 km per hour
8. The area of a rhombus is 240 cm
2
and one of the diagonal is 16 cm. Find another
diagonal.
Solution:
Area of rhombus = 1/2 × d
1
× d
2
240 = 1/2 × 16 × d
2
240 = 8 × d
2
d
2
= 240/8 = 30
So, the other diagonal is 30 cm
9. The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.
Solution:
Area of rhombus = 1/2 × d
1
× d
2
Area of rhombus = 1/2 × 7.5 × 12
Area of rhombus = 6 × 7.5 = 45
So, Area of rhombus = 45 cm
2
10. The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars
dropped on it from the remaining opposite vertices are 8 m and 13 m. Find the area
of the field.
Solution:
Area of quadrilateral = 1/2 × d
1
× (p
1
+ p
2
)
Area of quadrilateral = 1/2 × 24 × (8 + 13)
Area of quadrilateral = 12 × 21 = 252
So, Area of quadrilateral is 252 cm
2
11. Find the area of a rhombus whose side is 6 cm and whose altitude is 4 cm. If one
of its diagonals is 8 cm long, find the length of the other diagonal.
Solution:
Given that,
Side of rhombus = 6 cm
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Study Mensuration - I (Area of a Trapezium and a Polygon) - EXERCISE 20.1 on the App
Students of Class 8 can study Mensuration - I (Area of a Trapezium and a Polygon) - EXERCISE 20.1 alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the Mensuration - I (Area of a Trapezium and a Polygon) - EXERCISE 20.1,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of Mensuration - I (Area of a Trapezium and a Polygon) - EXERCISE 20.1 is prepared as per the latest Class 8 syllabus.
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