Question

in this diagram, △bac ~ △edf. if the area of △bac = 6 in², what is the area of △edf? area = ? in² enter a decimal rounded to the tenths

Explanation:

Step1: Recall the ratio - of - areas formula for 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 ratio of the side lengths of $\triangle EDF$ to $\triangle BAC$ be $k=\frac{2}{3}$.

Step2: Set up the proportion for the areas

Let $A_1$ be the area of $\triangle EDF$ and $A_2 = 6$ be the area of $\triangle BAC$. Then $\frac{A_1}{A_2}=k^{2}$. Substituting $k = \frac{2}{3}$, we get $\frac{A_1}{6}=(\frac{2}{3})^{2}$.

Step3: Solve for $A_1$

First, calculate $(\frac{2}{3})^{2}=\frac{4}{9}$. Then, from $\frac{A_1}{6}=\frac{4}{9}$, we can cross - multiply to get $9A_1=4\times6$. So $9A_1 = 24$, and $A_1=\frac{24}{9}=\frac{8}{3}\approx2.7$.

Answer:

$2.7$