Question

Sum of the areas of 2 squares is 400 cm^2.if the difference of their perimeters is 16 cm,find the sides of the 2 squares.

Answer 1

Let side of first square be x and the side of another square be y.
difference in perimeter of two squares:-
4x-4y=16
4(x-y)=16
x-y=16/4
therefore, x-y=4
and x=4+y---take this(1)

sum of areas of two squares:-
x²+y²=400
Put (1) here
(4+y)²+y²=400

(4)²+(y)²+2×4×y+y²=400

16+y²+8y+y²=400

2y²+8y+16-400=0

2y²+8y-384=0

2(y²+4y-192)=0
(here we have taken 2 common and take to another side. then it becomes:-)
y²+4y-192=0
we have done with factorisation method:-
y²-12y+16y-192=0

y(y-12)+16(y-12)=0

(y+16) (y-12)=0
either
y+16=0
or
y-12=0
y= -16
or
y=12

we will take y=12 because y being side cannot be negative.
So, y=12
put y=12 in (1)
x=4+y

x=4+12

therefore, x=16

and hence, Side of first square= x = 16

and Side of another square= y = 12.