ML Aggarwal: Mensuration - 2

 Page 1


 
1. Each sides of a rhombus is 13 cm and one diagonal is 10 cm. Find  
(i) the length of its other diagonal  
(ii) the area of the rhombus 
Solution:  
 
 
(i) Given, 
Side of rhombus = 13 cm. 
 
Length of diagonal AC = 10 cm. 
 
? OC = 5 cm. 
 
Since, the diagonals of rhombus bisect each other at right angles 
 
So, ?BOC is rt. angled. 
Page 2


 
1. Each sides of a rhombus is 13 cm and one diagonal is 10 cm. Find  
(i) the length of its other diagonal  
(ii) the area of the rhombus 
Solution:  
 
 
(i) Given, 
Side of rhombus = 13 cm. 
 
Length of diagonal AC = 10 cm. 
 
? OC = 5 cm. 
 
Since, the diagonals of rhombus bisect each other at right angles 
 
So, ?BOC is rt. angled. 
 
Then, by Pythagoras Theorem we have 
BC
2
 = OC
2
 + OB
2
 
 
13
2
 = 5
2
 + OB
2
 
 
OB
2
 = 169 – 25 = 144 
 
? OB = v144 = 12 cm 
 
Hence, 
Diagonal BD = 2 × OB = 2 × 12 = 24 cm 
 
(ii) Area of rhombus = 


 × d
1
 × d
2
 
 
= 


 × 10 × 24 = 120cm
2
 
  
2. The cross-section ABCD of a swimming pool is a trapezium. Its 
width AB = 14 m, depth at the shallow end is 1-5 m and at the deep 
end is 8 m. Find the area of the cross-section. 
 
Page 3


 
1. Each sides of a rhombus is 13 cm and one diagonal is 10 cm. Find  
(i) the length of its other diagonal  
(ii) the area of the rhombus 
Solution:  
 
 
(i) Given, 
Side of rhombus = 13 cm. 
 
Length of diagonal AC = 10 cm. 
 
? OC = 5 cm. 
 
Since, the diagonals of rhombus bisect each other at right angles 
 
So, ?BOC is rt. angled. 
 
Then, by Pythagoras Theorem we have 
BC
2
 = OC
2
 + OB
2
 
 
13
2
 = 5
2
 + OB
2
 
 
OB
2
 = 169 – 25 = 144 
 
? OB = v144 = 12 cm 
 
Hence, 
Diagonal BD = 2 × OB = 2 × 12 = 24 cm 
 
(ii) Area of rhombus = 


 × d
1
 × d
2
 
 
= 


 × 10 × 24 = 120cm
2
 
  
2. The cross-section ABCD of a swimming pool is a trapezium. Its 
width AB = 14 m, depth at the shallow end is 1-5 m and at the deep 
end is 8 m. Find the area of the cross-section. 
 
Solution: 
Here, AD and BC are the two parallel sides of trapezium 
 
And, distance between them is 14 m. 
 
? Area of trapezium = 


 (1·5 + 8) × 14 
 
= 


 × 9·5 × 14 
= 66 × 5 m
2
 
 
3. The area of a trapezium is 360 m
2
, the distance between two parallel 
sides is 20 m and one of the parallel side is 25 m. Find the other 
parallel side. 
Solution: 
 
Given, 
Area of a trapezium = 360 m
2
 
 
Distance between two parallel lines = 20 m 
 
One parallel side = 25 m 
Page 4


 
1. Each sides of a rhombus is 13 cm and one diagonal is 10 cm. Find  
(i) the length of its other diagonal  
(ii) the area of the rhombus 
Solution:  
 
 
(i) Given, 
Side of rhombus = 13 cm. 
 
Length of diagonal AC = 10 cm. 
 
? OC = 5 cm. 
 
Since, the diagonals of rhombus bisect each other at right angles 
 
So, ?BOC is rt. angled. 
 
Then, by Pythagoras Theorem we have 
BC
2
 = OC
2
 + OB
2
 
 
13
2
 = 5
2
 + OB
2
 
 
OB
2
 = 169 – 25 = 144 
 
? OB = v144 = 12 cm 
 
Hence, 
Diagonal BD = 2 × OB = 2 × 12 = 24 cm 
 
(ii) Area of rhombus = 


 × d
1
 × d
2
 
 
= 


 × 10 × 24 = 120cm
2
 
  
2. The cross-section ABCD of a swimming pool is a trapezium. Its 
width AB = 14 m, depth at the shallow end is 1-5 m and at the deep 
end is 8 m. Find the area of the cross-section. 
 
Solution: 
Here, AD and BC are the two parallel sides of trapezium 
 
And, distance between them is 14 m. 
 
? Area of trapezium = 


 (1·5 + 8) × 14 
 
= 


 × 9·5 × 14 
= 66 × 5 m
2
 
 
3. The area of a trapezium is 360 m
2
, the distance between two parallel 
sides is 20 m and one of the parallel side is 25 m. Find the other 
parallel side. 
Solution: 
 
Given, 
Area of a trapezium = 360 m
2
 
 
Distance between two parallel lines = 20 m 
 
One parallel side = 25 m 
Now, 
 
Let’s assume the second parallel side to be x m 
 
So, Area = (25 + x) × 20 
 
? 360 = (25 + x) × 20 
 
? x = 36 – 25 = 11 m 
 
Therefore, the second parallel side is 11 m. 
 
 
4. Find the area of a rhombus whose side is 6.5 cm and altitude is 5 
cm. If one of its diagonal is 13 cm long, find the length of other 
diagonal. 
Solution: 
Given, 
Side of rhombus = 6.5 cm 
 
and altitude = 5 cm 
 
So,  
Page 5


 
1. Each sides of a rhombus is 13 cm and one diagonal is 10 cm. Find  
(i) the length of its other diagonal  
(ii) the area of the rhombus 
Solution:  
 
 
(i) Given, 
Side of rhombus = 13 cm. 
 
Length of diagonal AC = 10 cm. 
 
? OC = 5 cm. 
 
Since, the diagonals of rhombus bisect each other at right angles 
 
So, ?BOC is rt. angled. 
 
Then, by Pythagoras Theorem we have 
BC
2
 = OC
2
 + OB
2
 
 
13
2
 = 5
2
 + OB
2
 
 
OB
2
 = 169 – 25 = 144 
 
? OB = v144 = 12 cm 
 
Hence, 
Diagonal BD = 2 × OB = 2 × 12 = 24 cm 
 
(ii) Area of rhombus = 


 × d
1
 × d
2
 
 
= 


 × 10 × 24 = 120cm
2
 
  
2. The cross-section ABCD of a swimming pool is a trapezium. Its 
width AB = 14 m, depth at the shallow end is 1-5 m and at the deep 
end is 8 m. Find the area of the cross-section. 
 
Solution: 
Here, AD and BC are the two parallel sides of trapezium 
 
And, distance between them is 14 m. 
 
? Area of trapezium = 


 (1·5 + 8) × 14 
 
= 


 × 9·5 × 14 
= 66 × 5 m
2
 
 
3. The area of a trapezium is 360 m
2
, the distance between two parallel 
sides is 20 m and one of the parallel side is 25 m. Find the other 
parallel side. 
Solution: 
 
Given, 
Area of a trapezium = 360 m
2
 
 
Distance between two parallel lines = 20 m 
 
One parallel side = 25 m 
Now, 
 
Let’s assume the second parallel side to be x m 
 
So, Area = (25 + x) × 20 
 
? 360 = (25 + x) × 20 
 
? x = 36 – 25 = 11 m 
 
Therefore, the second parallel side is 11 m. 
 
 
4. Find the area of a rhombus whose side is 6.5 cm and altitude is 5 
cm. If one of its diagonal is 13 cm long, find the length of other 
diagonal. 
Solution: 
Given, 
Side of rhombus = 6.5 cm 
 
and altitude = 5 cm 
 
So,  
Area of a rhombus = Side × Altitude = 6.5 × 5 = 32.5 cm
2
 
We have, one diagonal = 13 cm 
Hence, 
 
Length of other diagonal = 
 × 


 
 
= 
.×

 
= 5 cm 
  
5. From the given diagram, calculate  
(i) the area of trapezium ACDE  
(ii) the area of parallelogram ABDE  
(iii) the area of triangle BCD.  
 
Solution: 
(i) Area of trapezium ACDE = 


 × (AC + DE) × h 
= 


 × (13 + 7) × 6.5 
= 


 × 20 × 6.5 
= 65 m
2