 # Suppose that the combined area of two squares is 360 square feet. Each side of the larger square is three times as long as a side of the smaller square. How big is each square? ruigE 2021-02-09 Answered
Suppose that the combined area of two squares is 360 square feet. Each side of the larger square is three times as long as a side of the smaller square. How big is each square?
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Let x be the side length of the larger square and yy be the side length of the smaller square.
The area of a square is ${s}^{2}$ where ss is the side length so the larger square has an area of ${x}^{2}$ and the smaller square has an area of ${y}^{2}$. The combined area is then $x2+{y}^{2}$. If the combined area is 360 square feet then ${x}^{2}+{y}^{2}=360.$
If the side of the larger square is three times as long as the side of the smaller square, then x=3y.
Substitute x=3y into ${x}^{2}+{y}^{2}=360$ and solve for y:
${x}^{2}+{y}^{2}=360$
${\left(3y\right)}^{2}+{y}^{2}=360$
$9{y}^{2}+{y}^{2}=360$
$10{y}^{2}=360$
${y}^{2}=36$
${\sqrt{y}}^{2}=±\sqrt{36}$
$y=±6$
Since the sides of the squares must be positive, then y=6 so $x=3y=3\left(6\right)=18$. The larger square then has side lengths of 18 ft and the smaller square has side lengths of 6 ft.

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