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Math Help - Ratios of areas of triangles..

  1. #1
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    Ratios of areas of triangles..

    this is my problem... and i cannot even think where to begin. isnt there not enough information given??

    In the figure below, AD is the perpendicular bisector of BC, AE = 9, and DE = 3.
    What is the ratio of the area of triangle ABC to the area of triangle BCE?
    (Hint: Area of triangle ABC = 1/2 (AD)(BC) and Area of triangle BCE = 1/2 (DE)(BC)
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  2. #2
    Junior Member Spimon's Avatar
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    The ratio of area ABC to BCE is 4:1. That is, ABC is 4 times the area of BCE.
    Yes, it can be read straight off the diagram (12:3) but here's the proof. I hope it makes sense to you. If you're still confused let me know and I'll try to simplify it more.

    Last edited by Spimon; September 15th 2007 at 07:51 PM.
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  3. #3
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    Hello, JoDan!

    There's plenty of information . . .

    They even spelled-out the areas of the two triangles for you.


    \Delta ABC \:=\:\frac{1}{2}(BC)(AD)
    . . And we know that AD = 12
    Hence: . \Delta ABC \:=\:\frac{1}{2}(BC)(12) \:=\:6(BC)

    \Delta EBC \:=\:\frac{1}{2}(BC)(ED)
    . . And we know that ED = 3
    Hence: . \Delta EBC \:=\:\frac{1}{2}(BC)(3) \:=\:\frac{3}{2}(BC)

    The ratio of the areas is: . \frac{\Delta ABC}{\Delta EBC} \:=\:\frac{6}{\frac{3}{2}} \:=\:4

    . . Therefore, the ratio is: . 4:1

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