What is the perimeter of Δ ABC ?

[AAPL]

Note: Figure not drawn to scale.

In the figure above, ED = 1, CD = 2, and AE =6√3. What is the perimeter of ΔABC ?

A. 3√3
B. 10+5√3
C. 10√3
D. 15+5√3
E. 25√3

The OA is D.

I need help with this PS question. Please, can any expert explain it for me? Thanks.

[dabral]

Hi AAPL,

First thing to recognize is that triangles ABC and CDE are similar(Why?).

Also, notice that triangle CED is a 30-60-90 triangle because short side is of length 1 and the hypotenuse is 2. That means CE= sqrt{3} and therefore AC= 5*sqrt{3} . From the ratios of sides in a 30-60-90 triangle, we obtain AB=5 and BC=10, and the perimeter of triangle ABC is equal to 15+5*sqrt{3} .

Cheers,
Dabral

[EconomistGMATTutor]

Hello.

Using the Pythagora's Theorem we can find CE. $$CE=\sqrt{2^2-1^2}=\sqrt{3}.$$

Now, AE=AC+CE, this implies that: $$AC=AE-CE=6\sqrt{3}-\sqrt{3}=5\sqrt{3}.$$

To calculate AB, we will use that triangles ABC and EDC are similar. So, $$\frac{AB}{ED}=\frac{AC}{CE},\ this\ implies\ that,\ \frac{AB}{1}=\frac{5\sqrt{3}}{\sqrt{3}},\ that\ is,\ AB=5.$$

Now, we can calculate BC using the Pythagora's Theorem again. We will get, $$BC=\sqrt{5^2+\ \left(5\sqrt{3}\right)^2}=\sqrt{25+75}=\sqrt{100}=10.$$

In conclusion, the perimeter of ABC is $$15+5\sqrt{3}.$$

The correct answer is D.

I hope this can help you.

Feel free to ask me again if you have any doubt.