Each triangle is a 30-60-90 triangle, and the hypotenuse of one triangle is the longer leg of an adjacent triangle. The hypotenuse of the largest triangle is 8 centimeters. What is the number of centimeters in the length of the longer leg of the smallest triangle? Express your answer as a common fraction.

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  1. Draw the triangles. Recall that the sides are in the ratio 1:√3:2

    Label the longest hypotenuse, and then work down to get the desired leg's size.

    I get 6.

    I am assuming there are only two triangles.

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  2. You should know the ratio of sides of your triangle to be:
    1 : √3 : 2 or
    x : √3x : 2x, where x is a constant

    so in the next one:

    a : 2x : c

    then c/2x = 2x/√3x
    √3 cx = 4x^2
    c = 4x/√3

    a/x = 2x/√3x
    a = 2x/√3

    general triangle: (2x/√3) : 2x : (4x/√3)

    check: is a^2 + b^2 = c^2
    LS = 4x^2/3 + 4x^2 = 16x^2/3
    RS = (14x/√3)^2 = 16x^2/3

    so if 4x/√3 = 8
    x = 2√3
    our largest triangle is 4 : 4√3 : 8

    our hypotenuse increased from 2 to 8, that is by a factor of 4, so to get to our smallest triangle largest leg we have to divide 4√3 by 4 to get √3

    (which of course we knew at the beginning)

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  3. There are 4 triangles

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  4. All four right triangles are 30-60-90 triangles. Therefore, the length of the shorter leg in each triangle is half the hypotenuse, and the length of the longer leg is sqrt3 times the length of the shorter leg. We apply these facts to each triangle, starting with triangle AOB and working clockwise.
    From $\triangle AOB$, we find $AB = AO/2 = 4$ and $BO = AB\sqrt{3}=4\sqrt{3}$.

    From $\triangle BOC$, we find $BC = BO/2 =2\sqrt{3}$ and $CO = BC\sqrt{3} =2\sqrt{3}\cdot\sqrt{3} = 6$.

    From $\triangle COD$, we find $CD = CO/2 = 3$ and $DO = CD\sqrt{3} = 3\sqrt{3}$.

    From $\triangle DOE$, we find $DE = DO/2 = 3\sqrt{3}/2$ and $EO =DE\sqrt{3} = (3\sqrt{3}/2)\cdot \sqrt{3} = (3\sqrt{3}\cdot \sqrt{3})/2 = \boxed{9/2}$

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  5. the answer is 9/2

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