Straight Lines Question - None

Question

Given a triangle whose vertices are at $0, 0$, $4, 4$ and $10, 0$. A square is drawn in it such that its base is on the x-axis and its two corners are on the 2 sides of the triangle. The area of the square is equal to:

Accepted Answer

Solution: Let the square have side length $s$. The square has its base on the x-axis with corners at $(x, 0)$ and $(x+s, 0)$, and upper corners at $(x, s)$ and $(x+s, s)$. The equation of the line from $(0,0)$ to $(4,4)$ is $y = x$. The equation of the line from $(4,4)$ to $(10,0)$ is $y = -\frac{2}{3}(x - 10) = -\frac{2}{3}x + \frac{20}{3}$. For the square to fit with corners on these lines: $s = x$ (from left line) and $s = -\frac{2}{3}(x+s) + \frac{20}{3}$ (from right line). From $s = x$ and the right line condition: $s = -\frac{2}{3}(s + s) + \frac{20}{3} = -\frac{4}{3}s + \frac{20}{3}$. Thus $\frac{7}{3}s = \frac{20}{3}$, so $s = \frac{20}{7}$. Area $= s^2 = \frac{400}{49}$.

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