Question

b) Find the side of a square of area 400cm​

Answer 1

Answer:

⇒  Let the sides of the two squares be xcm and ycm, where x>y

⇒  Then, their areas are x  

2

 and y  

2

 and their perimeters are 4x and 4y.

By given condition,

⇒  x  

2

+y  

2

=400           ----- ( 1 )

⇒  And 4x−4y=16

⇒  4(x−y)=16

⇒  x−y=4

⇒  x=y+4

Substituting the value of x from ( 2 ) in ( 1 ), we get,

⇒  (y+4)  

2

+y  

2

=400

⇒  y  

2

+16+8y+y  

2

=400

⇒  2y  

2

+16+8y=400

⇒  y  

2

+4y−192=0

⇒  y  

2

+16y−12y−192=0

⇒  y(y+16)−12(y+16)=0

⇒  (y+16)(y−12)=0

∴  y=−16 or  12

Since, y cannot be negative, y=12

⇒  x=y+4=12+4=16

∴  Sides of the two squares are 16cm and 12cm.

⇒  Required difference =16−12=4cm