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