Question

in this diagram, $\triangle bacsim\triangle edf$. if the area of $\triangle bac = 8$ in², what is the area of $\triangle edf?
area = ? in²
enter a decimal rounded to the tenths

Explanation:

Step1: Recall ratio - of - areas of similar triangles

For two similar triangles $\triangle BAC$ and $\triangle EDF$, if the ratio of their corresponding side - lengths is $k$, the ratio of their areas is $k^{2}$. Let the side - length of $\triangle BAC$ be $a = 3$ in and the side - length of $\triangle EDF$ be $b = 4$ in. The ratio of the side - lengths of $\triangle EDF$ to $\triangle BAC$ is $k=\frac{b}{a}=\frac{4}{3}$.

Step2: Set up the area ratio formula

Let $A_1$ be the area of $\triangle BAC$ and $A_2$ be the area of $\triangle EDF$. We know that $\frac{A_2}{A_1}=k^{2}$. Given $A_1 = 8$ in² and $k=\frac{4}{3}$, then $\frac{A_2}{8}=(\frac{4}{3})^{2}$.

Step3: Solve for the area of $\triangle EDF$

First, calculate $(\frac{4}{3})^{2}=\frac{16}{9}$. Then, from $\frac{A_2}{8}=\frac{16}{9}$, we can cross - multiply to get $9A_2=8\times16$. So, $9A_2 = 128$, and $A_2=\frac{128}{9}\approx14.2$ in².

Answer:

$14.2$