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**Prove It** The large triangle is of dimensions 6, 8, 10.

The top triangle is of dimensions $\displaystyle \begin{align*} x, 6 - y, \sqrt{ x^2 + (6- y)^2} \end{align*}$.

The bottom triangle is of dimensions $\displaystyle \begin{align*} 8-x,y,\sqrt{(8-x)^2+y^2} \end{align*}$.

Thus $\displaystyle \begin{align*} \sqrt{x^2 + (6-y)^2} + \sqrt{(8-x)^2+y^2} = 10 \end{align*}$.

We are also told $\displaystyle \begin{align*} xy = 12 \end{align*}$, so $\displaystyle \begin{align*} y = \frac{12}{x} \end{align*}$. Substituting in gives

$\displaystyle \begin{align*} \sqrt{x^2 + \left( 6 - \frac{12}{x} \right) ^2 }+ \sqrt{ (8-x)^2 + \left( \frac{12}{x} \right) ^2 } &= 10 \\ \sqrt{ x^2 + \frac{ (6x - 12)^2}{x^2} } + \sqrt{(8-x)^2 + \frac{144}{x^2}} &= 10 \\ \sqrt{ \frac{x^4 + (6x-12)^2}{x^2} } + \sqrt{ \frac{x^2 \left( 8 - x \right) ^2 + 144}{x^2} } &= 10 \\ \frac{\sqrt{ x^4 + (6x-12)^2}}{x} + \frac{\sqrt{x^2 \left( 8-x \right) ^2 + 144}}{x} &= 10 \\ \sqrt{ x^4 + (6x-12)^2 } + \sqrt{ x^2 \left( 8-x \right) ^2 + 144 } &= 10x \\ \sqrt{ x^4 + 36x^2 - 144x + 144 } + \sqrt{ x^4 - 16x^3 + 64x^2 + 144} &= 10x \end{align*}$

Now try to solve this equation.