Answers

2016-02-28T20:04:04+05:30
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.



hope this helps.........................
0
2016-02-28T20:22:46+05:30
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.

AND DONE.
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