SOLUTION: A wire 80 cm in length is cut into two parts and each part is bent to form a square. If the sum of the areas of the squares is 300 cm?, find the lengths of the sides of the two squ

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Question 1202975: A wire 80 cm in length is cut into two parts and each part is bent to form a square. If the sum of the areas of the squares is 300 cm?, find the lengths of the sides of the two squares
Answer by math_helper(2461) About Me  (Show Source):
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x = length of each side of first square
y = length of each side of second square
We get:
(1) 4x + 4y = 80 <--- from length information
(2) x^2 + y^2 = 300 <--- from area information
Eq (1) can be solved for y; 4y = 80-4x ==> y=20-x
Substitute "20-x" for "y" in (2), to get:
+x%5E2+%2B+%2820-x%29%5E2+=+300+
+x%5E2+%2B+%28x%5E2-40x%2B400%29+=+300+
++2x%5E2+-+40x+%2B+400+=+300+
++2x%5E2+-+40x+%2B+100+=+0+
+++x%5E2+-+20x+%2B+50+=+0+
From WolframAlpha:
++%28x-5sqrt%282%29-10%29%28x%2B5sqrt%282%29-10%29+=+0+
x = +10%2B5sqrt%282%29+ or x = +10-5sqrt%282%29+
If x = +10%2B5sqrt%282%29+, we get y = 20-10-5sqrt%282%29+=+10-5sqrt%282%29
(which makes sense, it is the other choice for x above, as the assignments of x and y are arbitrary).
Answer:
The sides of the two squares measure +10%2B5sqrt%282%29 and 10-5sqrt%282%29