Original area = xy.
New length =  x  . 
2 
New breadth = 3y.
New area =  ❨  x  x 3y  ❩  =  3  xy. 
2  2 
∴ Increase % =  ❨  1  xy x  1  x 100  ❩%  = 50%. 
2  xy 
Decrease in area 





∴ Decrease % =  ❨  7  xy x  1  x 100  ❩%  = 28%. 
25  xy 
Then, length = (x + 20) metres.
Perimeter =  ❨  5300  ❩  m = 200 m. 
26.50 
∴ 2[(x + 20) + x] = 200
⟹ 2x + 20 = 100
⟹ 2x = 80
⟹ x = 40.
Hence, length = x + 20 = 60 m.
5  1 
4 
13  1 
2 
Other side  = 

ft  
= 

ft  
= 

ft  
=  6 ft. 
∴ Area of closet = (6 x 4.5) sq. ft = 27 sq. ft.
So, b = 34 ft.
∴ Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft.
∴ A_{1} = (100 x 100) cm^{2} and A_{2} (102 x 102) cm^{2}.
(A_{2}  A_{1}) = [(102)^{2}  (100)^{2}]
= (102 + 100) x (102  100)
= 404 cm^{2}.
∴ Percentage error =  ❨  404  x 100  ❩%  = 4.04% 
100 x 100 
Then, AB + BC = 2x metres.
AC = √2x = (1.41x) m.
Saving on 2x metres = (0.59x) m.
Saving % =  ❨  0.59x  x 100  ❩%  = 30% (approx.) 
2x 
Perimeter = Distance covered in 8 min. =  ❨  12000  x 8  ❩m = 1600 m. 
60 
Let length = 3x metres and breadth = 2x metres.
Then, 2(3x + 2x) = 1600 or x = 160.
∴ Length = 480 m and Breadth = 320 m.
∴ Area = (480 x 320) m^{2} = 153600 m^{2}.
Also, lb = 20.
(l + b)^{2} = (l^{2} + b^{2}) + 2lb = 41 + 40 = 81
⟹ (l + b) = 9.
∴ Perimeter = 2(l + b) = 18 cm.
Area of the lawn = 2109 m^{2}.
∴ Area of the crossroads = (2400  2109) m^{2} = 291 m^{2}.
Let the width of the road be x metres. Then,
60x + 40x  x^{2} = 291
⟹ x^{2}  100x + 291 = 0
⟹ (x  97)(x  3) = 0
⟹ x = 3.
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