Question

Sum of the areas of two squares is $400c{m}^2.$ If the difference of their perimeters is 16cm, find the sides of two 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:- | or:-

y+16=0 | y-12=0
y= -16 | 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.