Class 8 Exam > Class 8 Notes > Mathematics (Maths) Class 8 > Mensuration - I (Area of a Trapezium and a Polygon) - EXERCISE 20.2
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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 20.2 PAGE NO: 20.22
1. Find the area, in square metres, of the trapezium whose bases and altitudes are as
under:
(i) bases = 12 dm and 20 dm, altitude = 10 dm
(ii) bases = 28 cm and 3 dm, altitude = 25 cm
(iii) bases = 8 m and 60 dm, altitude = 40 dm
(iv) bases = 150 cm and 30 dm, altitude = 9 dm
Solution:
(i) Given that,
Length of bases of trapezium = 12 dm and 20 dm
Length of altitude = 10 dm
We know that, 10 dm = 1 m
? Length of bases in m = 1.2 m and 2 m
Similarly, length of altitude in m = 1 m
Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude
Area of trapezium = 1/2 (1.2 + 2.0) × 1
Area of trapezium = 1/2 × 3.2 = 1.6
So, Area of trapezium = 1.6m
2
(ii) Given that,
Length of bases of trapezium = 28 cm and 3 dm
Length of altitude = 25 cm
We know that, 10 dm = 1 m
? Length of bases in m = 0.28 m and 0.3 m
Similarly, length of altitude in m = 0.25 m
Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude
Area of trapezium = 1/2 (0.28 + 0.3) × 0.25
Area of trapezium = 1/2 × 0.58× 0.25 = 0.0725
So, Area of trapezium = 0.0725m
2
(iii) Given that,
Length of bases of trapezium = 8 m and 60 dm
Length of altitude = 40 dm
We know that, 10 dm = 1 m
? Length of bases in m = 8 m and 6 m
Similarly, length of altitude in m = 4 m
Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude
Area of trapezium = 1/2 (8 + 6) × 4
Page 2
RD Sharma Solutions for Class 8 Maths Chapter 20 –
Mensuration – I (Area of a Trapezium and a Polygon)
EXERCISE 20.2 PAGE NO: 20.22
1. Find the area, in square metres, of the trapezium whose bases and altitudes are as
under:
(i) bases = 12 dm and 20 dm, altitude = 10 dm
(ii) bases = 28 cm and 3 dm, altitude = 25 cm
(iii) bases = 8 m and 60 dm, altitude = 40 dm
(iv) bases = 150 cm and 30 dm, altitude = 9 dm
Solution:
(i) Given that,
Length of bases of trapezium = 12 dm and 20 dm
Length of altitude = 10 dm
We know that, 10 dm = 1 m
? Length of bases in m = 1.2 m and 2 m
Similarly, length of altitude in m = 1 m
Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude
Area of trapezium = 1/2 (1.2 + 2.0) × 1
Area of trapezium = 1/2 × 3.2 = 1.6
So, Area of trapezium = 1.6m
2
(ii) Given that,
Length of bases of trapezium = 28 cm and 3 dm
Length of altitude = 25 cm
We know that, 10 dm = 1 m
? Length of bases in m = 0.28 m and 0.3 m
Similarly, length of altitude in m = 0.25 m
Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude
Area of trapezium = 1/2 (0.28 + 0.3) × 0.25
Area of trapezium = 1/2 × 0.58× 0.25 = 0.0725
So, Area of trapezium = 0.0725m
2
(iii) Given that,
Length of bases of trapezium = 8 m and 60 dm
Length of altitude = 40 dm
We know that, 10 dm = 1 m
? Length of bases in m = 8 m and 6 m
Similarly, length of altitude in m = 4 m
Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude
Area of trapezium = 1/2 (8 + 6) × 4
RD Sharma Solutions for Class 8 Maths Chapter 20 –
Mensuration – I (Area of a Trapezium and a Polygon)
Area of trapezium = 1/2 × 56 = 28
So, Area of trapezium = 28m
2
(iv) Given that,
Length of bases of trapezium = 150 cm and 30 dm
Length of altitude = 9 dm
We know that, 10 dm = 1 m
? Length of bases in m = 1.5 m and 3 m
Similarly, length of altitude in m = 0.9 m
Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude
Area of trapezium = 1/2 (1.5 + 3) × 0.9
Area of trapezium = 1/2 × 4.5 × 0.9 = 2.025
So, Area of trapezium = 2.025m
2
2. Find the area of trapezium with base 15 cm and height 8 cm, if the side parallel to
the given base is 9 cm long.
Solution:
Given that,
Length of bases of trapezium = 15 cm and 9 cm
Length of altitude = 8 cm
We know that,
Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude
Area of trapezium = 1/2 (15 + 9) × 8
Area of trapezium = 1/2 × 192 = 96
So, Area of trapezium = 96m
2
3. Find the area of a trapezium whose parallel sides are of length 16 dm and 22 dm
and whose height is 12 dm.
Solution:
Given that,
Length of bases of trapezium = 16 dm and 22 dm
Length of altitude = 12 dm
We know that, 10 dm = 1 m
? Length of bases in m = 1.6 m and 2.2 m
Similarly, length of altitude in m = 1.2 m
Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude
Area of trapezium = 1/2 (1.6 + 2.2) × 1.2
Area of trapezium = 1/2 × 3.8 × 1.2 = 2.28
So, Area of trapezium = 2.28m
2
Page 3
RD Sharma Solutions for Class 8 Maths Chapter 20 –
Mensuration – I (Area of a Trapezium and a Polygon)
EXERCISE 20.2 PAGE NO: 20.22
1. Find the area, in square metres, of the trapezium whose bases and altitudes are as
under:
(i) bases = 12 dm and 20 dm, altitude = 10 dm
(ii) bases = 28 cm and 3 dm, altitude = 25 cm
(iii) bases = 8 m and 60 dm, altitude = 40 dm
(iv) bases = 150 cm and 30 dm, altitude = 9 dm
Solution:
(i) Given that,
Length of bases of trapezium = 12 dm and 20 dm
Length of altitude = 10 dm
We know that, 10 dm = 1 m
? Length of bases in m = 1.2 m and 2 m
Similarly, length of altitude in m = 1 m
Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude
Area of trapezium = 1/2 (1.2 + 2.0) × 1
Area of trapezium = 1/2 × 3.2 = 1.6
So, Area of trapezium = 1.6m
2
(ii) Given that,
Length of bases of trapezium = 28 cm and 3 dm
Length of altitude = 25 cm
We know that, 10 dm = 1 m
? Length of bases in m = 0.28 m and 0.3 m
Similarly, length of altitude in m = 0.25 m
Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude
Area of trapezium = 1/2 (0.28 + 0.3) × 0.25
Area of trapezium = 1/2 × 0.58× 0.25 = 0.0725
So, Area of trapezium = 0.0725m
2
(iii) Given that,
Length of bases of trapezium = 8 m and 60 dm
Length of altitude = 40 dm
We know that, 10 dm = 1 m
? Length of bases in m = 8 m and 6 m
Similarly, length of altitude in m = 4 m
Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude
Area of trapezium = 1/2 (8 + 6) × 4
RD Sharma Solutions for Class 8 Maths Chapter 20 –
Mensuration – I (Area of a Trapezium and a Polygon)
Area of trapezium = 1/2 × 56 = 28
So, Area of trapezium = 28m
2
(iv) Given that,
Length of bases of trapezium = 150 cm and 30 dm
Length of altitude = 9 dm
We know that, 10 dm = 1 m
? Length of bases in m = 1.5 m and 3 m
Similarly, length of altitude in m = 0.9 m
Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude
Area of trapezium = 1/2 (1.5 + 3) × 0.9
Area of trapezium = 1/2 × 4.5 × 0.9 = 2.025
So, Area of trapezium = 2.025m
2
2. Find the area of trapezium with base 15 cm and height 8 cm, if the side parallel to
the given base is 9 cm long.
Solution:
Given that,
Length of bases of trapezium = 15 cm and 9 cm
Length of altitude = 8 cm
We know that,
Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude
Area of trapezium = 1/2 (15 + 9) × 8
Area of trapezium = 1/2 × 192 = 96
So, Area of trapezium = 96m
2
3. Find the area of a trapezium whose parallel sides are of length 16 dm and 22 dm
and whose height is 12 dm.
Solution:
Given that,
Length of bases of trapezium = 16 dm and 22 dm
Length of altitude = 12 dm
We know that, 10 dm = 1 m
? Length of bases in m = 1.6 m and 2.2 m
Similarly, length of altitude in m = 1.2 m
Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude
Area of trapezium = 1/2 (1.6 + 2.2) × 1.2
Area of trapezium = 1/2 × 3.8 × 1.2 = 2.28
So, Area of trapezium = 2.28m
2
RD Sharma Solutions for Class 8 Maths Chapter 20 –
Mensuration – I (Area of a Trapezium and a Polygon)
4. Find the height of a trapezium, the sum of the lengths of whose bases (parallel
sides) is 60 cm and whose area is 600 cm
2
.
Solution:
Given that,
Length of bases of trapezium = 60 cm
Area = 600 cm
2
We know that,
Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude
600 = 1/2 (60) × altitude
600 = 30 × altitude
Which implies, altitude = 600/30 = 20
? Length of altitude is 20 cm
5. Find the altitude of a trapezium whose area is 65 cm
2
and whose base are 13 cm
and 26 cm.
Solution:
Given that,
Length of bases of trapezium = 13 cm and 26 cm
Area = 65 cm
2
We know that,
Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude
65 = 1/2 (13 + 26) × altitude
65 = 39/2 × altitude
Which implies, altitude = (65×2) /39 = 130/39 = 10/3
? Length of altitude = 10/3 cm
6. Find the sum of the lengths of the bases of a trapezium whose area is 4.2 m
2
and
whose height is 280 cm.
Solution:
Given that,
Height of trapezium = 280 cm = 2.8m
Area = 4.2 m
2
We know that,
Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude
To calculate the length of parallel sides we can rewrite the above equation as,
Sum of lengths of parallel sides = (2 × Area) / altitude
Sum of lengths of parallel sides = (2 × 4.2) / 2.8 = 8.4/2.8 = 3
? Sum of lengths of parallel sides = 3 m
Page 4
RD Sharma Solutions for Class 8 Maths Chapter 20 –
Mensuration – I (Area of a Trapezium and a Polygon)
EXERCISE 20.2 PAGE NO: 20.22
1. Find the area, in square metres, of the trapezium whose bases and altitudes are as
under:
(i) bases = 12 dm and 20 dm, altitude = 10 dm
(ii) bases = 28 cm and 3 dm, altitude = 25 cm
(iii) bases = 8 m and 60 dm, altitude = 40 dm
(iv) bases = 150 cm and 30 dm, altitude = 9 dm
Solution:
(i) Given that,
Length of bases of trapezium = 12 dm and 20 dm
Length of altitude = 10 dm
We know that, 10 dm = 1 m
? Length of bases in m = 1.2 m and 2 m
Similarly, length of altitude in m = 1 m
Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude
Area of trapezium = 1/2 (1.2 + 2.0) × 1
Area of trapezium = 1/2 × 3.2 = 1.6
So, Area of trapezium = 1.6m
2
(ii) Given that,
Length of bases of trapezium = 28 cm and 3 dm
Length of altitude = 25 cm
We know that, 10 dm = 1 m
? Length of bases in m = 0.28 m and 0.3 m
Similarly, length of altitude in m = 0.25 m
Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude
Area of trapezium = 1/2 (0.28 + 0.3) × 0.25
Area of trapezium = 1/2 × 0.58× 0.25 = 0.0725
So, Area of trapezium = 0.0725m
2
(iii) Given that,
Length of bases of trapezium = 8 m and 60 dm
Length of altitude = 40 dm
We know that, 10 dm = 1 m
? Length of bases in m = 8 m and 6 m
Similarly, length of altitude in m = 4 m
Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude
Area of trapezium = 1/2 (8 + 6) × 4
RD Sharma Solutions for Class 8 Maths Chapter 20 –
Mensuration – I (Area of a Trapezium and a Polygon)
Area of trapezium = 1/2 × 56 = 28
So, Area of trapezium = 28m
2
(iv) Given that,
Length of bases of trapezium = 150 cm and 30 dm
Length of altitude = 9 dm
We know that, 10 dm = 1 m
? Length of bases in m = 1.5 m and 3 m
Similarly, length of altitude in m = 0.9 m
Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude
Area of trapezium = 1/2 (1.5 + 3) × 0.9
Area of trapezium = 1/2 × 4.5 × 0.9 = 2.025
So, Area of trapezium = 2.025m
2
2. Find the area of trapezium with base 15 cm and height 8 cm, if the side parallel to
the given base is 9 cm long.
Solution:
Given that,
Length of bases of trapezium = 15 cm and 9 cm
Length of altitude = 8 cm
We know that,
Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude
Area of trapezium = 1/2 (15 + 9) × 8
Area of trapezium = 1/2 × 192 = 96
So, Area of trapezium = 96m
2
3. Find the area of a trapezium whose parallel sides are of length 16 dm and 22 dm
and whose height is 12 dm.
Solution:
Given that,
Length of bases of trapezium = 16 dm and 22 dm
Length of altitude = 12 dm
We know that, 10 dm = 1 m
? Length of bases in m = 1.6 m and 2.2 m
Similarly, length of altitude in m = 1.2 m
Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude
Area of trapezium = 1/2 (1.6 + 2.2) × 1.2
Area of trapezium = 1/2 × 3.8 × 1.2 = 2.28
So, Area of trapezium = 2.28m
2
RD Sharma Solutions for Class 8 Maths Chapter 20 –
Mensuration – I (Area of a Trapezium and a Polygon)
4. Find the height of a trapezium, the sum of the lengths of whose bases (parallel
sides) is 60 cm and whose area is 600 cm
2
.
Solution:
Given that,
Length of bases of trapezium = 60 cm
Area = 600 cm
2
We know that,
Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude
600 = 1/2 (60) × altitude
600 = 30 × altitude
Which implies, altitude = 600/30 = 20
? Length of altitude is 20 cm
5. Find the altitude of a trapezium whose area is 65 cm
2
and whose base are 13 cm
and 26 cm.
Solution:
Given that,
Length of bases of trapezium = 13 cm and 26 cm
Area = 65 cm
2
We know that,
Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude
65 = 1/2 (13 + 26) × altitude
65 = 39/2 × altitude
Which implies, altitude = (65×2) /39 = 130/39 = 10/3
? Length of altitude = 10/3 cm
6. Find the sum of the lengths of the bases of a trapezium whose area is 4.2 m
2
and
whose height is 280 cm.
Solution:
Given that,
Height of trapezium = 280 cm = 2.8m
Area = 4.2 m
2
We know that,
Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude
To calculate the length of parallel sides we can rewrite the above equation as,
Sum of lengths of parallel sides = (2 × Area) / altitude
Sum of lengths of parallel sides = (2 × 4.2) / 2.8 = 8.4/2.8 = 3
? Sum of lengths of parallel sides = 3 m
RD Sharma Solutions for Class 8 Maths Chapter 20 –
Mensuration – I (Area of a Trapezium and a Polygon)
7. Find the area of a trapezium whose parallel sides of lengths 10 cm and 15 cm are
at a distance of 6 cm from each other. Calculate this area as,
(i) the sum of the areas of two triangles and one rectangle.
(ii) the difference of the area of a rectangle and the sum of the areas of two triangles.
Solution:
We know that, Area of a trapezium ABCD
= area (?DFA) + area (rectangle DFEC) + area (?CEB)
= (1/2 × AF × DF) + (FE × DF) + (1/2 × EB × CE)
= (1/2 × AF × h) + (FE × h) + (1/2 × EB × h)
= 1/2 × h × (AF + 2FE + EB)
= 1/2 × h × (AF + FE + EB + FE)
= 1/2 × h × (AB + FE)
= 1/2 × h × (AB + CD) [Opposite sides of rectangle are equal]
= 1/2 × 6 × (15 + 10)
= 1/2 × 6 × 25 = 75
? Area of trapezium = 75 cm
2
8. The area of a trapezium is 960 cm
2
. If the parallel sides are 34 cm and 46 cm, find
the distance between them.
Solution:
We know that,
Area of trapezium = 1/2 (Sum of lengths of parallel sides) × distance between parallel
sides
i.e., Area of trapezium = 1/2 (Sum of sides) × distance between parallel sides
To calculate the distance between parallel sides we can rewrite the above equation as,
Distance between parallel sides = (2 × Area) / Sum of sides
= (2 × 960) / (34 + 46)
= (2 × 960) / 80 = 1920/80 = 24
? Distance between parallel sides = 24 cm
Page 5
RD Sharma Solutions for Class 8 Maths Chapter 20 –
Mensuration – I (Area of a Trapezium and a Polygon)
EXERCISE 20.2 PAGE NO: 20.22
1. Find the area, in square metres, of the trapezium whose bases and altitudes are as
under:
(i) bases = 12 dm and 20 dm, altitude = 10 dm
(ii) bases = 28 cm and 3 dm, altitude = 25 cm
(iii) bases = 8 m and 60 dm, altitude = 40 dm
(iv) bases = 150 cm and 30 dm, altitude = 9 dm
Solution:
(i) Given that,
Length of bases of trapezium = 12 dm and 20 dm
Length of altitude = 10 dm
We know that, 10 dm = 1 m
? Length of bases in m = 1.2 m and 2 m
Similarly, length of altitude in m = 1 m
Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude
Area of trapezium = 1/2 (1.2 + 2.0) × 1
Area of trapezium = 1/2 × 3.2 = 1.6
So, Area of trapezium = 1.6m
2
(ii) Given that,
Length of bases of trapezium = 28 cm and 3 dm
Length of altitude = 25 cm
We know that, 10 dm = 1 m
? Length of bases in m = 0.28 m and 0.3 m
Similarly, length of altitude in m = 0.25 m
Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude
Area of trapezium = 1/2 (0.28 + 0.3) × 0.25
Area of trapezium = 1/2 × 0.58× 0.25 = 0.0725
So, Area of trapezium = 0.0725m
2
(iii) Given that,
Length of bases of trapezium = 8 m and 60 dm
Length of altitude = 40 dm
We know that, 10 dm = 1 m
? Length of bases in m = 8 m and 6 m
Similarly, length of altitude in m = 4 m
Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude
Area of trapezium = 1/2 (8 + 6) × 4
RD Sharma Solutions for Class 8 Maths Chapter 20 –
Mensuration – I (Area of a Trapezium and a Polygon)
Area of trapezium = 1/2 × 56 = 28
So, Area of trapezium = 28m
2
(iv) Given that,
Length of bases of trapezium = 150 cm and 30 dm
Length of altitude = 9 dm
We know that, 10 dm = 1 m
? Length of bases in m = 1.5 m and 3 m
Similarly, length of altitude in m = 0.9 m
Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude
Area of trapezium = 1/2 (1.5 + 3) × 0.9
Area of trapezium = 1/2 × 4.5 × 0.9 = 2.025
So, Area of trapezium = 2.025m
2
2. Find the area of trapezium with base 15 cm and height 8 cm, if the side parallel to
the given base is 9 cm long.
Solution:
Given that,
Length of bases of trapezium = 15 cm and 9 cm
Length of altitude = 8 cm
We know that,
Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude
Area of trapezium = 1/2 (15 + 9) × 8
Area of trapezium = 1/2 × 192 = 96
So, Area of trapezium = 96m
2
3. Find the area of a trapezium whose parallel sides are of length 16 dm and 22 dm
and whose height is 12 dm.
Solution:
Given that,
Length of bases of trapezium = 16 dm and 22 dm
Length of altitude = 12 dm
We know that, 10 dm = 1 m
? Length of bases in m = 1.6 m and 2.2 m
Similarly, length of altitude in m = 1.2 m
Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude
Area of trapezium = 1/2 (1.6 + 2.2) × 1.2
Area of trapezium = 1/2 × 3.8 × 1.2 = 2.28
So, Area of trapezium = 2.28m
2
RD Sharma Solutions for Class 8 Maths Chapter 20 –
Mensuration – I (Area of a Trapezium and a Polygon)
4. Find the height of a trapezium, the sum of the lengths of whose bases (parallel
sides) is 60 cm and whose area is 600 cm
2
.
Solution:
Given that,
Length of bases of trapezium = 60 cm
Area = 600 cm
2
We know that,
Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude
600 = 1/2 (60) × altitude
600 = 30 × altitude
Which implies, altitude = 600/30 = 20
? Length of altitude is 20 cm
5. Find the altitude of a trapezium whose area is 65 cm
2
and whose base are 13 cm
and 26 cm.
Solution:
Given that,
Length of bases of trapezium = 13 cm and 26 cm
Area = 65 cm
2
We know that,
Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude
65 = 1/2 (13 + 26) × altitude
65 = 39/2 × altitude
Which implies, altitude = (65×2) /39 = 130/39 = 10/3
? Length of altitude = 10/3 cm
6. Find the sum of the lengths of the bases of a trapezium whose area is 4.2 m
2
and
whose height is 280 cm.
Solution:
Given that,
Height of trapezium = 280 cm = 2.8m
Area = 4.2 m
2
We know that,
Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude
To calculate the length of parallel sides we can rewrite the above equation as,
Sum of lengths of parallel sides = (2 × Area) / altitude
Sum of lengths of parallel sides = (2 × 4.2) / 2.8 = 8.4/2.8 = 3
? Sum of lengths of parallel sides = 3 m
RD Sharma Solutions for Class 8 Maths Chapter 20 –
Mensuration – I (Area of a Trapezium and a Polygon)
7. Find the area of a trapezium whose parallel sides of lengths 10 cm and 15 cm are
at a distance of 6 cm from each other. Calculate this area as,
(i) the sum of the areas of two triangles and one rectangle.
(ii) the difference of the area of a rectangle and the sum of the areas of two triangles.
Solution:
We know that, Area of a trapezium ABCD
= area (?DFA) + area (rectangle DFEC) + area (?CEB)
= (1/2 × AF × DF) + (FE × DF) + (1/2 × EB × CE)
= (1/2 × AF × h) + (FE × h) + (1/2 × EB × h)
= 1/2 × h × (AF + 2FE + EB)
= 1/2 × h × (AF + FE + EB + FE)
= 1/2 × h × (AB + FE)
= 1/2 × h × (AB + CD) [Opposite sides of rectangle are equal]
= 1/2 × 6 × (15 + 10)
= 1/2 × 6 × 25 = 75
? Area of trapezium = 75 cm
2
8. The area of a trapezium is 960 cm
2
. If the parallel sides are 34 cm and 46 cm, find
the distance between them.
Solution:
We know that,
Area of trapezium = 1/2 (Sum of lengths of parallel sides) × distance between parallel
sides
i.e., Area of trapezium = 1/2 (Sum of sides) × distance between parallel sides
To calculate the distance between parallel sides we can rewrite the above equation as,
Distance between parallel sides = (2 × Area) / Sum of sides
= (2 × 960) / (34 + 46)
= (2 × 960) / 80 = 1920/80 = 24
? Distance between parallel sides = 24 cm
RD Sharma Solutions for Class 8 Maths Chapter 20 –
Mensuration – I (Area of a Trapezium and a Polygon)
9. Find the area of Fig. 20.35 as the sum of the areas of two trapezium and a
rectangle.
Solution:
From the figure we can write,
Area of figure = Area of two trapeziums + Area of rectangle
Given that,
Length of rectangle = 50 cm
Breadth of rectangle = 10 cm
Length of parallel sides of trapezium = 30 cm and 10 cm
Distance between parallel sides of trapezium = (70–50)/2 = 20/2 = 10
So, Distance between parallel sides of trapezium = 10 cm
Area of figure = 2 × 1/2 (Sum of lengths of parallel sides) × altitude + Length × Breadth
Area of figure = 2 × 1/2 (30+10) × 10 + 50 × 10
Area of figure = 40 × 10 + 50 × 10
Area of figure = 400 + 500 = 900
? Area of figure = 900 cm
2
10. Top surface of a table is trapezium in shape. Find its area if its parallel sides are
1 m and 1.2 m and perpendicular distance between them is 0.8 m.
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Document Description: Mensuration - I (Area of a Trapezium and a Polygon) - EXERCISE 20.2 for Class 8 2024 is part of Mathematics (Maths) Class 8 preparation.
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