RS Aggarwal Solutions: Mensuration - 2 | Mathematics (Maths) Class 7 PDF Download

 Page 1


RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration  (Ex 20B) Exercise 20B 
  
 
  
   
Question 1.
 
 
Solution: 
Outer length of plot (L) = 75 m 
and breadth (B) = 60 m 
Width of path inside = 2 m 
 
Inner length (l) = 75 – 2 x 2 = 75 – 4 = 71 m 
and width (b) = 60 – 2 x 2 = 60 – 4 = 56 m 
Area of the path = L x B – l x b = (75 x 60 – 71 x 56) m² = 4500 – 3976 = 
524 m² 
Rate of constructing it = Rs. 125 per m² 
Total cost = Rs. 524 x 125 = Rs. 65500 
Question 2.
 
 
Solution: 
Outer length of the plot (L) = 95 m 
and breadth (B) = 72 m 
Page 2


RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration  (Ex 20B) Exercise 20B 
  
 
  
   
Question 1.
 
 
Solution: 
Outer length of plot (L) = 75 m 
and breadth (B) = 60 m 
Width of path inside = 2 m 
 
Inner length (l) = 75 – 2 x 2 = 75 – 4 = 71 m 
and width (b) = 60 – 2 x 2 = 60 – 4 = 56 m 
Area of the path = L x B – l x b = (75 x 60 – 71 x 56) m² = 4500 – 3976 = 
524 m² 
Rate of constructing it = Rs. 125 per m² 
Total cost = Rs. 524 x 125 = Rs. 65500 
Question 2.
 
 
Solution: 
Outer length of the plot (L) = 95 m 
and breadth (B) = 72 m 
RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration  (Ex 20B) Exercise 20B 
  
 
  
   
Width of path = 3.5 m 
 
Inner length (l) = 95 – 2 x 3.5 = 95 – 7 = 88 m 
and breadth = 72 – 2 x 3.5 = 72 – 7 = 65 m 
Outer area = L x B = 95 x 72 m² = 6840 m² 
and inner area = l x b = 88 x 65 m² = 5720 m² 
Area of path = outer area – inner area = 6840 – 5720 = 1120 m² 
Rate of constructing it = Rs. 80 per m² 
Total cost = Rs. 1120 x 80 = Rs. 89600 
and rate of laying grass = Rs. 40 per m² 
Total cost = Rs. 40 x 5720 = Rs. 228800 
Total cost = Rs. 89600 + Rs. 228800 = Rs. 318400 
Question 3. 
 
Solution: 
Length of saree (L) = 5 m 
and breadth (B) = 1.3 m 
Page 3


RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration  (Ex 20B) Exercise 20B 
  
 
  
   
Question 1.
 
 
Solution: 
Outer length of plot (L) = 75 m 
and breadth (B) = 60 m 
Width of path inside = 2 m 
 
Inner length (l) = 75 – 2 x 2 = 75 – 4 = 71 m 
and width (b) = 60 – 2 x 2 = 60 – 4 = 56 m 
Area of the path = L x B – l x b = (75 x 60 – 71 x 56) m² = 4500 – 3976 = 
524 m² 
Rate of constructing it = Rs. 125 per m² 
Total cost = Rs. 524 x 125 = Rs. 65500 
Question 2.
 
 
Solution: 
Outer length of the plot (L) = 95 m 
and breadth (B) = 72 m 
RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration  (Ex 20B) Exercise 20B 
  
 
  
   
Width of path = 3.5 m 
 
Inner length (l) = 95 – 2 x 3.5 = 95 – 7 = 88 m 
and breadth = 72 – 2 x 3.5 = 72 – 7 = 65 m 
Outer area = L x B = 95 x 72 m² = 6840 m² 
and inner area = l x b = 88 x 65 m² = 5720 m² 
Area of path = outer area – inner area = 6840 – 5720 = 1120 m² 
Rate of constructing it = Rs. 80 per m² 
Total cost = Rs. 1120 x 80 = Rs. 89600 
and rate of laying grass = Rs. 40 per m² 
Total cost = Rs. 40 x 5720 = Rs. 228800 
Total cost = Rs. 89600 + Rs. 228800 = Rs. 318400 
Question 3. 
 
Solution: 
Length of saree (L) = 5 m 
and breadth (B) = 1.3 m 
RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration  (Ex 20B) Exercise 20B 
  
 
  
   
Width of border = 25 cm 
 
Inner length (l) = 5 – 50/100 = 5 – 0.5 = 4.5 m 
and inner breadth (b) = 1.3 – 50/100 = 1.3 – 0.5 = 0.8 m 
Now area of the boarder = L x B – l x b 
= (5 x 1.3 – 4.5 x 0.8) m² 
= 6.50 – 3.60 
= 2.90 m² 
= 2.90 x 100 x 100 = 29000 cm² 
Cost of 10 cm² = Re. 1 
Total cost = Rs. 2900 x 1= Rs. 2900 
Question 4. 
 
Solution: 
Inner length of lawn (l) = 38 m 
and breadth (b) = 25 m. 
Width of path = 2.5 m. 
Outer length (L) = 38 + 2 x 2.5 = 38 + 5 = 43 m 
and outer breadth (B) = 25 + 2 x 2.5 = 25 + 5 = 30 m 
Page 4


RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration  (Ex 20B) Exercise 20B 
  
 
  
   
Question 1.
 
 
Solution: 
Outer length of plot (L) = 75 m 
and breadth (B) = 60 m 
Width of path inside = 2 m 
 
Inner length (l) = 75 – 2 x 2 = 75 – 4 = 71 m 
and width (b) = 60 – 2 x 2 = 60 – 4 = 56 m 
Area of the path = L x B – l x b = (75 x 60 – 71 x 56) m² = 4500 – 3976 = 
524 m² 
Rate of constructing it = Rs. 125 per m² 
Total cost = Rs. 524 x 125 = Rs. 65500 
Question 2.
 
 
Solution: 
Outer length of the plot (L) = 95 m 
and breadth (B) = 72 m 
RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration  (Ex 20B) Exercise 20B 
  
 
  
   
Width of path = 3.5 m 
 
Inner length (l) = 95 – 2 x 3.5 = 95 – 7 = 88 m 
and breadth = 72 – 2 x 3.5 = 72 – 7 = 65 m 
Outer area = L x B = 95 x 72 m² = 6840 m² 
and inner area = l x b = 88 x 65 m² = 5720 m² 
Area of path = outer area – inner area = 6840 – 5720 = 1120 m² 
Rate of constructing it = Rs. 80 per m² 
Total cost = Rs. 1120 x 80 = Rs. 89600 
and rate of laying grass = Rs. 40 per m² 
Total cost = Rs. 40 x 5720 = Rs. 228800 
Total cost = Rs. 89600 + Rs. 228800 = Rs. 318400 
Question 3. 
 
Solution: 
Length of saree (L) = 5 m 
and breadth (B) = 1.3 m 
RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration  (Ex 20B) Exercise 20B 
  
 
  
   
Width of border = 25 cm 
 
Inner length (l) = 5 – 50/100 = 5 – 0.5 = 4.5 m 
and inner breadth (b) = 1.3 – 50/100 = 1.3 – 0.5 = 0.8 m 
Now area of the boarder = L x B – l x b 
= (5 x 1.3 – 4.5 x 0.8) m² 
= 6.50 – 3.60 
= 2.90 m² 
= 2.90 x 100 x 100 = 29000 cm² 
Cost of 10 cm² = Re. 1 
Total cost = Rs. 2900 x 1= Rs. 2900 
Question 4. 
 
Solution: 
Inner length of lawn (l) = 38 m 
and breadth (b) = 25 m. 
Width of path = 2.5 m. 
Outer length (L) = 38 + 2 x 2.5 = 38 + 5 = 43 m 
and outer breadth (B) = 25 + 2 x 2.5 = 25 + 5 = 30 m 
RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration  (Ex 20B) Exercise 20B 
  
 
  
   
 
Area of path = Outer area – Inner area 
= (43 x 30 – 38 x 25) m² 
=(1290 – 950) m² = 340 m² 
Rate of gravelling the path = Rs. 120 per m² 
Total cost = Rs. 120 x 340 = Rs. 40800 
Question 5. 
 
Solution: 
Length’ of room (l) = 9.5m 
Breadth (b) = 6m 
Width of outer verandah = 1.25m 
 
Page 5


RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration  (Ex 20B) Exercise 20B 
  
 
  
   
Question 1.
 
 
Solution: 
Outer length of plot (L) = 75 m 
and breadth (B) = 60 m 
Width of path inside = 2 m 
 
Inner length (l) = 75 – 2 x 2 = 75 – 4 = 71 m 
and width (b) = 60 – 2 x 2 = 60 – 4 = 56 m 
Area of the path = L x B – l x b = (75 x 60 – 71 x 56) m² = 4500 – 3976 = 
524 m² 
Rate of constructing it = Rs. 125 per m² 
Total cost = Rs. 524 x 125 = Rs. 65500 
Question 2.
 
 
Solution: 
Outer length of the plot (L) = 95 m 
and breadth (B) = 72 m 
RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration  (Ex 20B) Exercise 20B 
  
 
  
   
Width of path = 3.5 m 
 
Inner length (l) = 95 – 2 x 3.5 = 95 – 7 = 88 m 
and breadth = 72 – 2 x 3.5 = 72 – 7 = 65 m 
Outer area = L x B = 95 x 72 m² = 6840 m² 
and inner area = l x b = 88 x 65 m² = 5720 m² 
Area of path = outer area – inner area = 6840 – 5720 = 1120 m² 
Rate of constructing it = Rs. 80 per m² 
Total cost = Rs. 1120 x 80 = Rs. 89600 
and rate of laying grass = Rs. 40 per m² 
Total cost = Rs. 40 x 5720 = Rs. 228800 
Total cost = Rs. 89600 + Rs. 228800 = Rs. 318400 
Question 3. 
 
Solution: 
Length of saree (L) = 5 m 
and breadth (B) = 1.3 m 
RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration  (Ex 20B) Exercise 20B 
  
 
  
   
Width of border = 25 cm 
 
Inner length (l) = 5 – 50/100 = 5 – 0.5 = 4.5 m 
and inner breadth (b) = 1.3 – 50/100 = 1.3 – 0.5 = 0.8 m 
Now area of the boarder = L x B – l x b 
= (5 x 1.3 – 4.5 x 0.8) m² 
= 6.50 – 3.60 
= 2.90 m² 
= 2.90 x 100 x 100 = 29000 cm² 
Cost of 10 cm² = Re. 1 
Total cost = Rs. 2900 x 1= Rs. 2900 
Question 4. 
 
Solution: 
Inner length of lawn (l) = 38 m 
and breadth (b) = 25 m. 
Width of path = 2.5 m. 
Outer length (L) = 38 + 2 x 2.5 = 38 + 5 = 43 m 
and outer breadth (B) = 25 + 2 x 2.5 = 25 + 5 = 30 m 
RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration  (Ex 20B) Exercise 20B 
  
 
  
   
 
Area of path = Outer area – Inner area 
= (43 x 30 – 38 x 25) m² 
=(1290 – 950) m² = 340 m² 
Rate of gravelling the path = Rs. 120 per m² 
Total cost = Rs. 120 x 340 = Rs. 40800 
Question 5. 
 
Solution: 
Length’ of room (l) = 9.5m 
Breadth (b) = 6m 
Width of outer verandah = 1.25m 
 
RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration  (Ex 20B) Exercise 20B 
  
 
  
   
Outer length (L) = 9.5 + 2 x 1.25 = 9.5 + 2.5 = 12.0 m 
and breadth (B) = 6 + 2 x 1.25 = 6 + 2.5 = 8.5 m 
Area of verandah = Outer area – Inner area = L x B – l x b 
= (12.0 x 8.5 – 9.5 x 6) m² – (102.0 – 57.0) m² = 45 m² 
Rate of cementing = Rs. 15 per m² 
Total cost = Rs. 80 x 45 = Rs. 3600 
Question 6. 
 
Solution: 
Side of the flower bed = 2 m 80 cm = 2.80 m      [since 100 cm = 1 m] 
 
? Area of the square flower bed = (Side)
2 
= (2.80 m )
2
 = 7.84 m
2
 
Side of the flower bed with the digging strip = 2.80 m + 30 cm + 30 cm 
 = (2.80 + 0.3 + 0.3) m = 3.4 m 
Area of the enlarged flower bed with the digging strip = (Side )
2
 = (3.4 )
2
  
= 11.56 m
2
 
 
? Increase in the area of the flower bed = 11.56 m
2
 - 7.84 m
2