Question

sum of areas of two squares is 400 cm 2 .If the difference in their perimeters is 16cm. find the sides of the two squares. (In quadratic equation method)

Answer 1

Let the side of one square is x cm
therefore the side of other square is x- 16/4=x-4
Now according to question
x^2+(x-4)^2=400
Or, x^2+x^2-8x+16=400
Or, 2x^2-8x+16-400=0
Or, 2x^2-8x-384=0
Or, 2(x^2-4x-192)=0
Or, x^2-4x-192=0
Or, x^2-16x+12x-192=0
Or, x(x-16)+12(x-16)=0
Or, (x-16)(x+12)=0
Either x-16=0
Or,x=16
/OR x+12=0
Or, x=-12
but x=-12 is is not acceptable because side can not be negative
Hence x=16
there side of one square is 16 cm
and side of other square is 16-4=12 cm