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Math Help - Triangle Ratios

  1. #1
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    Triangle Ratios

    Triangle ABC is a 30-60-90 triangle, where A is 30, B is 90, and C is 60. A segment BD is the drawn altitude to AC, forming ΔADB, then DE is the altitude drawn to AB,forming new ∆ADE. Then, EF is the altitude drawn to AC, forming new ∆AFE. 78Then, FG is the altitude drawn to AB, forming new ∆AFG.Then, GH is the altitude drawn to AC, forming new ∆AHG. Then, HI is the altitude drawn to AB, forming new ∆AHI. Find the ratio of the area of ∆AIH to the area of ∆ABC?

    I found that the resulting ratio 729√(3)/8192 : √3/2 or .154 : .866
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  2. #2
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    Hello, warriors837!

    Triangle ABC is a 30-60-90 triangle, where A = 30^o,\:B = 90^o,\:C = 60^o.
    Altitude BD is drawn to AC, forming \Delta ADB
    Altitude DE is drawn to AB, forming \Delta DE.
    Altitude EF is drawn to AC, forming \Delta AFE.
    Altitude FG is drawn to AB, forming \Delta AFG.
    Altitude GH is drawn to AC, forming \Delta HG.
    Altitude HI is drawn to AB, forming \Delta AHI.

    Find the ratio of the area of \Delta AIH to the area of \Delta ABC.

    I found that the resulting ratio \frac{729\sqrt{3}}{8192}
    . .
    Close, but incorrect.

    Note that all triangles are 30-60-90 right triangles.

    In a 30-60-90 triangle, the side opposite 30 is half the hypotenuse.
    The side opposite 60 is \sqrt{3} times the shortest side.

    Let BC = 8, then AB = 8\sqrt{3}
    The area of \Delta ABC is: . A \:=\:\tfrac{1}{2}(8\sqrt{3})(8) \:=\:32\sqrt{3} .[1]


    In \Delta BDC\!:hyp = 8,\; CD =4,\;BD = 4\sqrt{3}

    In \Delta DEB\!:\;hyp = 4\sqrt{3},\;BE = 2\sqrt{3},\;DE = 6


    In \Delta DEA\!:\;DE = 6,\;EA \:=\: 8\sqrt{3} - 2\sqrt{3} \:=\:6\sqrt{3}
    . . The area of \Delta DEA \:=\:\tfrac{1}{2}(6\sqrt{3})(6) \:=\:18\sqrt{3}

    We have: . \frac{\text{area of }\Delta DEA}{\text{area of }\Delta ABC} \;=\;\frac{18\sqrt{3}}{32\sqrt{3}} \;=\;\frac{9}{16}


    Drawing two altitudes, we have two similar right triangles.
    . . The area of the smaller triangle is \tfrac{9}{16} of the larger.

    This procedure is performed three times.
    Hence, the area of the final triangle (\Delta AIH) is: . \left(\tfrac{9}{16}\right)^3 the area of \Delta ABC.

    Therefore: . \frac{\text{area of }\Delta AIH}{\text{area of }\Delta ABC} \;=\;\frac{729}{4096}

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