SOLUTION: The sum of the perimeters of two squares is 80 in., and the area of one is 3 times the area of the other. Find the side length of each square.

Algebra ->  Equations -> SOLUTION: The sum of the perimeters of two squares is 80 in., and the area of one is 3 times the area of the other. Find the side length of each square.      Log On


   

Question 1148799: The sum of the perimeters of two squares is 80 in., and the area of one is 3 times the area of the other. Find the side length of each square.
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Let x be side length small square
and let y be side length of big square.

This means system%284x%2B4y=80%2Cand%2Cy%5E2=3x%5E2%29.

Simpler system x%2By=20%2Cy%5E2%2B3x%5E2

y=20-x
-
y%5E2=%2820-x%29%5E2=400-40x%2Bx%5E2
-
400-40x%2Bx%5E2=3x%5E2
200-20x=x%5E2
x%5E2%2B20x-200=0

x=%28-20%2Bsqrt%28400%2B4%2A200%29%29%2F2

highlight_green%28x=-10%2B10sqrt%283%29%29, edge length of smaller square; use it to find y.














-----------------------------------------------------mistake-------------------
Simplified, system is system%28x%2By=20%2Cand%2Cy%5E2=3x%5E2%29.
Subst for x and simplify,
y%5E2-60y%2B600=0
cross%28y=60-10sqrt%283%29%29 or cross%28y=60%2B10sqrt%283%29%29
and you can find any corresponding x values.

Answer by ikleyn(52778) About Me  (Show Source):
You can put this solution on YOUR website!
.

            The solution by @josgarithmetic is not precisely correct,  producing partly absurdist result.

            So I came to bring the  CORRECT  solution. 


Let x be the side length of the larger square and y be that of the smaller square.


    4x + 4y = 80    (1)

    x^2 = 3y^2      (2)

===========================>   

    x + y = 20      (3)

    x^2 = 3y^2      (4)



From (1),  y = 20-x.  Substitute it into (4).  You will get


    x^2 = 3*(20-x)^2

    x^2 = 3*(400 - 40x + x^2)

    x^2 = 1200 - 120x + 3x^2

    2x^2 - 120x + 1200 = 0

    x^2  -  60x + 600 = 0

    x = %2860+%2B-+sqrt%2860%5E2+-+4%2A600%29%29%2F2 = %2860+%2B-+sqrt%281200%29%29%2F2 = %2860+%2B-+20%2Asqrt%283%29%29%2F2 = 30+%2B-+10%2Asqrt%283%29.



Although the quadratic equation has 2 roots, only smaller value  x= 30+%2B-+10%2Asqrt%283%29  is the solution to the problem,
since the larger value  30+%2B+10%2Asqrt%283%29  produces NEGATIVE value of "y", due to (3).



ANSWER.  The larger square has the side length  30+-10%2Asqrt%283%29 = 12.68 inches, approximately.

         The smaller square has the side length  20+-+%2830-10%2Asqrt%283%29%29 = -10+%2B+10%2Asqrt%283%29 = 7.32 in (approximately).

Solved.