SAT Congruence and Similarity

Correct Answer: D

Solution:

Tip: Know how to prove triangles are similar.
We're given that $\overline{AB} \parallel \overline{DE}.$ Note that $\overline{BD}$ and $\overline{AE}$ are transversals and therefore $\angle CAB \cong \angle CED$ and $\angle CBA \cong \angle CDE.$ Since $\angle ACB$ and $\angle DCE$ are vertical angles, it follows they are congruent also.

By AAA, $\triangle ABC \sim \triangle EDC.$ Note that since $\triangle ABC$ is isosceles, so is $\triangle DEC.$

Corresponding parts in similar triangles are proportional. So,

$\begin{array}{r c l} \frac{AB}{ED} &=& \frac{BC}{DC}\\ AB \cdot DC &=& BC \cdot ED\\ DC &=& \frac{BC \cdot ED}{AB}\\ DC &=& \frac{4.5 \cdot 6}{4}\\ DC &=& 6.75 \end{array}$

Therefore, $CE = DC = 6.75.$

The perimeter of $\triangle EDC = ED + DC + CE = 6 + 6.75 + 6.75 = 19.5.$

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Incorrect Choices:

(A)
If you find the length of $\overline{DC}$ or $\overline{CE},$ you will get this wrong answer.

(B)
If you set up the wrong proportion, you may get this wrong answer. There are several ways to achieve this. Here is one example:

If you start with

$\frac{AB}{DC} =\frac{DE}{AC}$

you will get $DC=3,$ and the perimeter of $\triangle EDC = ED + DC + CE = 6 + 3+ 3= 12.$

(C)
If you find the perimeter of $\triangle ABC,$ you will get this wrong answer.

(E)
If you find the perimeter of the entire figure, you will get this wrong answer.