Question

Suppose the triangles shown are similar with angle A = angle D, angle B = angle E, and angle C = angle F. Answer the question.

Answer 1

Similar triangles have the same ratio between corresponding sides:

$\frac{AB}{DE}=\frac{BC}{EF}=\frac{CA}{FD}$

We have the sizes of the sides:

• AB = 17

,

• BC = 22

,

• CA = 2x-7

,

• DE = 34

,

• EF = 44

,

• FD = 2x+4

We use the ratio property to find x:

$\begin{gathered} \frac{BC}{EF}=\frac{CA}{FD} \\ \frac{22}{44}=\frac{2x-7}{2x+4} \\ \frac{1}{2}=\frac{2x-7}{2x+4} \end{gathered}$

And now we clear x:

$\begin{gathered} \frac{1}{2}=\frac{2x-7}{2x+4} \\ 2x+4=2(2x-7) \\ 2x+4=4x-14 \\ x(2-4)=-14-4 \\ x=\frac{-14-4}{2-4}=\frac{-18}{-2}=9 \end{gathered}$

Now that we have x = 9, we can find the lenght of side DF (DF and FD are the same side):

$FD=2x+4=2\cdot9+4=18+4=22$

The answer is option C, FD = 22