Question

what is the length of $overline{ac}$? a. 144 b. 108 c. 128 d. 136

Explanation:

Step1: Identify similar triangles

Triangles $ABC$ and $CDE$ are similar because $\angle BAC=\angle CED = 90^{\circ}$ and $\angle ABC+\angle ACB = 90^{\circ}$, $\angle DCE+\angle ACB = 90^{\circ}$, so $\angle ABC=\angle DCE$. Then, $\frac{AB}{CE}=\frac{AC}{DE}$.
We know that $AB = 84$, $DE = 7$, $AE=156$, $AC = 156 - x$ and $CE=x$. So, $\frac{84}{x}=\frac{156 - x}{7}$.

Step2: Cross - multiply

Cross - multiplying the proportion $\frac{84}{x}=\frac{156 - x}{7}$ gives us $84\times7=x(156 - x)$.
$588 = 156x-x^{2}$.
Rearranging to standard quadratic form: $x^{2}-156x + 588=0$.

Step3: Solve the quadratic equation

Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$. Here, $a = 1$, $b=-156$, $c = 588$.
First, calculate the discriminant $\Delta=b^{2}-4ac=(-156)^{2}-4\times1\times588=24336-2352 = 21984$.
$x=\frac{156\pm\sqrt{21984}}{2}=\frac{156\pm148.27}{2}$.
We get two solutions for $x$, but we can also use another approach.
Since the triangles are similar, we know that the ratio of the sides is constant. Let's assume the ratio of similarity is $k$.
We know that $\frac{AB}{DE}=\frac{84}{7}=12$. So, $AC = 12\times CE$. And $AC+CE=156$. Let $CE=x$, then $12x+x=156$, $13x = 156$, $x = 12$.
So $AC=156 - 12=144$.

Answer:

A. 144