# 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?

Question
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?

2021-02-10
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 $$\displaystyle{s}^{{{2}}}$$ where ss is the side length so the larger square has an area of $$\displaystyle{x}^{{{2}}}$$ and the smaller square has an area of $$\displaystyle{y}^{{{2}}}$$. The combined area is then $$\displaystyle{x}{2}+{y}^{{{2}}}$$. If the combined area is 360 square feet then $$\displaystyle{x}^{{{2}}}+{y}^{{{2}}}={360}.$$
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 $$\displaystyle{x}^{{{2}}}+{y}^{{{2}}}={360}$$ and solve for y:
$$\displaystyle{x}^{{{2}}}+{y}^{{{2}}}={360}$$
$$\displaystyle{\left({3}{y}\right)}^{{{2}}}+{y}^{{{2}}}={360}$$
$$\displaystyle{9}{y}^{{{2}}}+{y}^{{{2}}}={360}$$
$$\displaystyle{10}{y}^{{{2}}}={360}$$
$$\displaystyle{y}^{{{2}}}={36}$$
$$\displaystyle\sqrt{{y}}^{{{2}}}=\pm\sqrt{{36}}$$
y=+-6
Since the sides of the squares must be positive, then y=6 so x=3y=3(6)=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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