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

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1. anonymous

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2. triciaal

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3. anonymous

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

4. triciaal

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

5. mathstudent55

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6. triciaal

thinking we can use (1/2)(11)(6) for area of ABF agree AF = 3 rt 5 same as FC

7. mathstudent55

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8. mathstudent55

$$\dfrac{11}{3 \sqrt 5} = \dfrac{12}{FC}$$ $$FC = \dfrac{36 \sqrt 5}{11}$$

9. mathstudent55

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10. mathstudent55

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.

11. anonymous

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

12. triciaal

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13. mathstudent55

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14. triciaal

yes!

15. mathstudent55

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16. mathstudent55

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17. mathstudent55

$$\dfrac{11 (6)}{2} + \dfrac{12(6)}{2} =69$$

18. mathstudent55

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.

19. anonymous

: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

20. triciaal

you are welcome pythagorean (3,4,5) factor 2 = (6, 8, 10) FDB

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