Question

find the value of x that makes △abc ~ △def .
triangle abc: ab=6, bc=9, ac=x+11.
triangle def: de=18, ef=9(x+2), df=36.
x = \square

Explanation:

Step1: Identify Corresponding Sides

Since $\triangle ABC \sim \triangle DEF$, their corresponding sides are proportional. So, $\frac{AB}{DE} = \frac{AC}{DF} = \frac{BC}{EF}$. Let's match the sides: $AB = 6$, $DE = 18$; $AC = x + 11$, $DF = 36$; $BC = 9$, $EF = 9(x + 2)$. We can use $\frac{AB}{DE} = \frac{AC}{DF}$.

Step2: Set Up Proportion

Substitute the values: $\frac{6}{18} = \frac{x + 11}{36}$. Simplify $\frac{6}{18}$ to $\frac{1}{3}$. So, $\frac{1}{3} = \frac{x + 11}{36}$.

Step3: Solve for x

Cross - multiply: $3(x + 11)=36$. Divide both sides by 3: $x + 11 = 12$. Subtract 11 from both sides: $x=12 - 11=1$. We can also check with the other ratio $\frac{BC}{EF}=\frac{9}{9(x + 2)}=\frac{1}{x + 2}$. Since $\frac{AB}{DE}=\frac{1}{3}$, then $\frac{1}{x + 2}=\frac{1}{3}$, so $x + 2 = 3$, $x = 1$.

Answer:

$x = 1$