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Hi there! Have questions about your homework? At Brainly, there are 60 million students who want to help each other learn. Questions are usually answered in less than 10 minutes. Try it for yourself by posting a question! :D

let the side of square ABCD be x and of square EFGH be y. therefore difference of perimeter =16 so , 4x-4y=16 x-y=4....................(1) x²+y²=400 (x-y)²+2xy=400 (since (x-y)²=x²+y²-2xy) 4²+2xy=400 2xy=384 xy=192 y=192/x.....................(2)

(2) in (1) x-192/x=16 x²-192=16x x²-16x-192=0 after factorisin we get , x=24/ -8 since sides cant be negative x=24 therefore y=8.

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.