If a triangle ABC is there such that BE=EF=FC=3 and DB=GC=2B

[Max@Math Revolution]

If a triangle ABC is there such that BE=EF=FC=3 and DB=GC=2BE, what is the perimeter of a pentagon ADEFG?
A. 9+6√3
B. 6+4√3
C. 9+6√2
D. 6+4√2
E. 4+2√3

*An answer will be posted in 2 days.

[Max@Math Revolution]

Since BE=2 and DB=4, degree B = degree C=60 degree. So, DE=GF=3√3 and AD=AG=3. Hence, the perimeter of a pentagon ADEFG is 3+3+3+3√3+3√3=9+6√3.Hence,the correct answer is A.

[danielle07]

I am quite confused. Can you explain it by step? Thanks

[Admin1]

First of all, there is a mistake on the statement. It says that BE=EF=FC=3, this is the mistake. It should say BE=EF=FC=2.

I will explain the solution using the right statement.

As BE=2 we can get that DB=GC=4.

If we look at the right triangle BED, we can use the Pythagoras Theorem and get the DE side.

We have BE=2 and DB=4. So,

DB^2=BE^2+DE^2.
4^2 = 2^2+DE^2
16 = 4 +DE^2
16-4 = DE^2
12=DE^2
DE= rad(12)
DE= rad(4*3)
DE=rad(4)*rad(3)
DE=2rad(3)

In the same way we can get that GF=2rad(3).

Now, we have to calculate AD and AG.

If we draw a line from D to G, we can see that the triangles ABC and ADG are similar, So, we can set the following equations:

BC/ DG = AB / AD = AC / AG

On the other hand,
BC= BE+EF+FC = 2+2+2=6.
AB=AD+DB=AD+4.
DG=EF=2.

Using the left equation: BC/DG = AB/AD we will get 6/2=(AD+4)/AD which implies

3*AD=AD+4
2*AD=4
AD=2

Using the right equation (AB/AD=AC/AG), we will get that AG=2 (the calculus are similar).

Finally, the perimeter of the Pentagon ADEFG is

AD+DE+EF+FG+GA = 2 + 2rad(3) + 2 + 2rad(3) + 2 = 6 + 4rad(3)

The correct answer is B.