SOLUTION: Two squares are such that the perimeter of one square is 24 greater than the other. If the sum of the areas of the squares is 596 square units, find the perimeter of the larger sq

Question 647192: Two squares are such that the perimeter of one square is 24 greater than the other. If the sum of the areas of the squares is 596 square units, find the perimeter of the larger square.
This is what I got so far: square A perimeter = x+x+x+x +24 or 4x +24 - simplified x+6. Area of square B (x+6) squared = 596
(x+6)????
Answer by aaronwiz(69) (Show Source): You can put this solution on YOUR website!
Hi, my name is Aaron I am in 10th grade and im in honors trig/pre-calc.
Lets call a side of square A (the smaller square) X.
Lets call a side of square b (the bigger square) Y.
Lets set up a system of equations.
Since we know one side of square B is 6 bigger, we can say
x+6=y
also we know the area of them added together is 596
x^2+y^2=596

substitute y=x+6 into the equation above
x^2+ (x+6)^2=596
foil...
x^2+x^2 +12x+36=596
simplify it down
2x^2+12x=560
x^2+6x-280=0
now plug into quadratic formula

result: x= 14 or-20
since distance cant be positive x has to be 14. we now know the sides of square A is 14
so plug back into y=x+6 to find the side length of square b
y=20
multiply by 4, because a square has 4 sides, so we can find permitter.
Square B's permitter is 80units.
Hope this helped. Please leave me positive feedback, this took over 20 mins to type
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