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## Factoring Quadratics 1.11 - Solution

We have been given a right triangle with an area of $35 \text{ m}^2.$

Recall that area of a right triangle is half the product of its legs. With this, we can write an equation to model the area of the given triangle. $\begin{gathered} \dfrac{(x-11)(x-8)}{2}=35 \end{gathered}$ To find the dimensions of the triangle, we should solve the equation above. Let's start by writing it in standard form.
$\dfrac{(x-11)(x-8)}{2}=35$
Write in slope-intercept form
$(x-11)(x-8)=70$
$x(x-8)-11(x-8)=70$
$x^2-8x-11(x-8)=70$
$x^2-8x-11x+88=70$
$x^2-19x+88=70$
$x^2-19x+18=0$
Next, we will factor the resulting equation.
$x^2-19x+18=0$
Factor
$x^2-18x-x+18=0$
$x(x-18)-x+18=0$
$x(x-18)-(x-18)=0$
$(x-18)(x-1)=0$
Now, we can apply the Zero Product Property to solve it.
$(x-18)(x-1)=0$
$\begin{array}{lc}x-18=0 & \text{(I)}\\ x-1=0 & \text{(II)}\end{array}$
$\begin{array}{l}x=18 \\ x-1=0 \end{array}$
$\begin{array}{l}x_1=18 \\ x_2=1 \end{array}$
Because the legs of a right triangle cannot have negative values, the value of $x$ is $18.$ From here, we can determine the dimensions as shown below. \begin{aligned} \textbf{Leg I: }& {\color{#0000FF}{18}}-11=\phantom{1}7 \text{ m}\\ \textbf{Leg II: }& {\color{#0000FF}{18}}-\phantom{1}8=10 \text{ m} \end{aligned}