In the figure shown, ∠ABF=∠CBF, AB=11", AD=3", EC=4", and BF=10". Find the area of ΔABC.

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In the figure shown, ∠ABF=∠CBF, AB=11", AD=3", EC=4", and BF=10". Find the area of ΔABC.

Mathematics
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Wouldn't BE=7" instead of 8"?

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I was thinking that since the angle was bisected so ABC would be isosceles but changed using triangle FDB and FEB then stopped not sure
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thinking we can use (1/2)(11)(6) for area of ABF agree AF = 3 rt 5 same as FC
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\(\dfrac{11}{3 \sqrt 5} = \dfrac{12}{FC} \) \(FC = \dfrac{36 \sqrt 5}{11} \)
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Add AF and FC to find the length of side AC. Then using the lengths of sides AB, AC, and BC, use Heron's formula to find the area.
I would like to let you guys know my textbook said the answer to this problem was: Area = 69 in^2. I believe AC= 13.8
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yes!
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\(\dfrac{11 (6)}{2} + \dfrac{12(6)}{2} =69\)
The problem with this figure is that the lengths of the sides do not work out. If the other given dimensions are correct, EC cannot be 4.
:D Thank you so much! I have never thought of 6 as the height of both triangles. This makes so much sense now. Thank you very much @mathstudent55 @triciaal
you are welcome pythagorean (3,4,5) factor 2 = (6, 8, 10) FDB

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