Question

The sum of the areas of two squares is 400 m^2. If the difference between their perimeter is 16m find the sides of two squares​

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