Question

prove that △abc ~ △def using the sss similarity criterion, where ab = 6, bc = 8, ac = 10, de = 9, ef = 12, and df = 15. fill in the missing ratio.
statement
△abc ~ △def
\frac{ab}{de}=\frac{bc}{ef}=\frac{ac}{df}
\frac{ab}{de}=\frac{6}{9}=\frac{2}{3}
\frac{bc}{ef}=\frac{8}{12}=\frac{2}{3}
\frac{ac}{df}=
reason
given side lengths
given side lengths
given side lengths
ratios of corresponding sides are equal
sss similarity criterion
a. \frac{10}{15}=\frac{2}{3}
b. \frac{10}{20}=\frac{1}{2}
c. \frac{10}{30}=\frac{1}{3}
d. \frac{10}{12}=\frac{5}{6}

Explanation:

Step1: Recall the SSS - Similarity Criterion

For two triangles $\triangle ABC$ and $\triangle DEF$ to be similar by SSS (Side - Side - Side) similarity criterion, the ratios of their corresponding sides must be equal, i.e., $\frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}$.
We are given $AB = 6$, $BC = 8$, $AC = 10$, $DE = 9$, $EF = 12$, and $DF = 15$.
We already know $\frac{AB}{DE}=\frac{6}{9}=\frac{2}{3}$, $\frac{BC}{EF}=\frac{8}{12}=\frac{2}{3}$.
We need to find $\frac{AC}{DF}$.

Step2: Calculate the ratio $\frac{AC}{DF}$

Substitute $AC = 10$ and $DF = 15$ into the ratio $\frac{AC}{DF}$.
$\frac{AC}{DF}=\frac{10}{15}=\frac{2}{3}$

Answer:

$\frac{AC}{DF}=\frac{10}{15}=\frac{2}{3}$