SOLUTION: Okay here goes nothing.. The sum of the perimeters of two squares is 40 inches. The sum of their areas is 58 square inches. What are the dimensions of the squares?

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Question 57491: Okay here goes nothing..
The sum of the perimeters of two squares is 40 inches. The sum of their areas is 58 square inches. What are the dimensions of the squares?
Found 2 solutions by faceoff57, funmath:
Answer by faceoff57(108) About Me  (Show Source):
You can put this solution on YOUR website!
58
9 + 49 = 58 sum of areas
12 + 28 = 40 sum of perimeters
one square has a side of 3 inches.
the other square has a side of 7 inches.

Answer by funmath(2933) About Me  (Show Source):
You can put this solution on YOUR website!
The sum of the perimeters of two squares is 40 inches. The sum of their areas is 58 square inches. What are the dimensions of the squares?
The formula for the perimeter of a square is P=4s.
The formula for the area of a square is A=s%5E2
Let one square's side =x
The other square's side is: y
:
The sum of their perimaters is:
4x%2B4y=40
The sum of their areas is:
x%5E2%2By%5E2=58
:
Solve the sum of the perimeter for x:
4x%2B4y-4y=40-4y
4x=40-4y
4x%2F4=40%2F4-4y%2F4
x=10-y Substitute that value in for x in the sum of the area and solve for y:
%2810-y%29%5E2%2By%5E2=58
100-20y%2By%5E2%2By%5E2=58
100-20y%2B2y%5E2=58
2y%5E2-20y%2B100=58
2y%5E2-20y%2B100-58=58-58
2y%5E2-20y%2B42=0
2%28y%5E2-10y%2B21%29=0
2%28y-3%29%28y-7%29=0
2%2F2%28y-3%29%28y-7%29=0%2F2
%28y-3%29%28y-7%29=0
y-3=0 and y-7=0
y-3+3=0+3 and y-7+7=0+7
y=3 and y=7
If you substitute y=3 into the periemeter equation you get 7, if you substitute the 7 in you get 3.
Therefore one square is 3x3 inches, the other is 7x7 inches.
Happy Calculating!!!