Question

The sum of the areas of two squares is 400 sq.m. If the difference
between their perimeters is 16 m, find the sides of two squares.​

Answer 1

let sides be x and y be sides of square

where x< y

given

x^2+ y^2= 400 ...... 1

4 x- 4 y = 16

4( x- y) = 16

x- y = 4

x=y+4.... 2

substituting value 2 in 1

( y+4) ^2+ y^2 = 400

y^2+16 +8 y + 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 y= 12

since y cannot be negative y=12

x=y+4 , x=16

therefore sides if square are 12 and 16