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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
    • one year ago
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    |dw:1436683918388:dw|

  2. triciaal
    • one year ago
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    |dw:1436684090522:dw|

  3. anonymous
    • one year ago
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    Wouldn't BE=7" instead of 8"?

  4. triciaal
    • one year ago
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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

  5. mathstudent55
    • one year ago
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    |dw:1436685144499:dw|

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

  7. mathstudent55
    • one year ago
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    |dw:1436685510443:dw|

  8. mathstudent55
    • one year ago
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    \(\dfrac{11}{3 \sqrt 5} = \dfrac{12}{FC} \) \(FC = \dfrac{36 \sqrt 5}{11} \)

  9. mathstudent55
    • one year ago
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    |dw:1436686267192:dw|

  10. mathstudent55
    • one year ago
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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.

  11. anonymous
    • one year ago
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    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
    • one year ago
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  13. mathstudent55
    • one year ago
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    |dw:1436686587186:dw|

  14. triciaal
    • one year ago
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    yes!

  15. mathstudent55
    • one year ago
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    |dw:1436686608952:dw|

  16. mathstudent55
    • one year ago
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    |dw:1436686654476:dw|

  17. mathstudent55
    • one year ago
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    \(\dfrac{11 (6)}{2} + \dfrac{12(6)}{2} =69\)

  18. mathstudent55
    • one year ago
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    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
    • one year ago
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    :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
    • one year ago
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    you are welcome pythagorean (3,4,5) factor 2 = (6, 8, 10) FDB

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