MathRevolution wrote:
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.
I could solve this question after assuming that DE and GF are perpendicular to side BC, making Triangles DBE and GFC right triangles.
Since we know BE = 3 and BD = 6, we can calculate DE. DE = \(3\sqrt{3}\). Similarly, GF = \(3\sqrt{3}\).
Now, the sides of Triangles DBE and GFC are in a ratio of \(1:\sqrt{3}:2\). Hence, these are 30-60-90 triangles with angle DBE = angle GCF = 60.
This implies that angle BAC =60 and Triangle ABC is an equilateral triangle. AB = BC = AC = 9
AD = AB - DB = 9 - 6 = 3
Similarly, AG = AC - GC = 9 - 6 = 3
So, perimeter of pentagon ADEFG = AD + AG + DE + GF + EF = \(3 + 3 + 3\sqrt{3} + 3\sqrt{3} + 3 = 9 + 6\sqrt{3}\)
Hence, the correct answer is A.
However, the above solution is dependent on the assumption that DE and GF are perpendicular to BC. Probably, the question should provide this information either in the question stem or in the figure.
Is there any way we can conclude that DE and GF are perpendicular to BC based on the given information ? Or are there any methods for calculating the perimeter without relying on the fact that DE and GF are perpendicular to BC?
chetan2u - Can you help?