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Math Help - Special triangles and ratio of areas help?

  1. #1
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    Special triangles and ratio of areas help?

    Hi Mathforums!
    I've been lurking around here for a round a week and found you guys are really helpful. Just today, I've been hit with two hard problems (to me, at least. The class geniuses had no problems whatsoever with these) I was sent home and was told to bring in the answer on Monday.
    So I've attached the problems, and there are some scribbles on them. Please take a look at them and see if you can help.

    So for the first one, since the diagonals add up to 28, and I discovered it consisted of 4 30:60:90 triangles, I took one of the 4 small triangles. The side facing 30 became x, and the side facing 60 is x(sqrt 3).
    So then x + x (sqrt 3) = 14

    Now, I use pythagorean;s theorem
    x^2 + ( 3 )x^2 = c^2, where C is an outer side.

    Now I get 2 equations - 4x^2= c^2
    and x (1 + (sqrt 3)) = 14
    so I get x = 14/ (1 + (sqrt 3))

    I plug that in, I get c^2= 4x^2
    c^2 = (4 x 196) / (4 + (sqrt 3))
    c^2 = 784 / (4 + (sqrt 3))

    Now I'm stuck. Can anyone help me?
    Attached Thumbnails Attached Thumbnails Special triangles and ratio of areas help?-math-1.jpg   Special triangles and ratio of areas help?-math-2.jpg  
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  2. #2
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    And also, for the second question, my mind is blown.
    Here's all I have.
    The two upper triangle's areas add up to equal the triangle on the bottoms' area.

    The ratio of the areas between the triangles are : 16/25, 9/16, and 9/25.
    Therefore, the ratios of the corresponding sides are their square roots:
    4/5, 3/4, and 3/5 are the ratios of proportionality ( I think thats what its called?)

    Thanks for answering!
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  3. #3
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    Hello, picklepie159!

    Your preliminary work is correct.


    1, The sum of the diagonals of rhombus ABCD is 28
    . . and \angle CBA = 2(\angle DAB).

    Find the perimeter of the rhombus.

    Code:
    
                A
                *
               /|\
              / | \
             / y|  \
            /   |   \
           /    |    \
        B * - - + - - * D
           \ x  |    /
            \   |   /
             \  |  /
              \ | /
               \|/
                *
                 C

    You are correct; we have 30-60 right triangles.


    And we have: . 2x + 2y \:=\:28 \quad\Rightarrow\quad x + y \:=\:14 .[1]

    We also have: . y \,=\,\sqrt{3}\,x .[2]


    Substitute [2] into [1]: . x + \sqrt{3}\,x \:=\:14 \quad\Rightarrow\quad x(1 + \sqrt{3}) \:=\:14

    . . x \:=\:\dfrac{14}{1+\sqrt{3}} \quad\Rightarrow\quad x\:=\: 7(\sqrt{3}-1)

    Substitute into [2]: . y \:=\:\sqrt{3}\cdot7(\sqrt{3}-1) \quad\Rightarrow\quad y \:=\:7(3-\sqrt{3})


    Pythagorus says: . AB^2 \:=\:x^2+y^2

    . . AB^2 \;=\;\left[7(\sqrt{3}-1)\right]^2 + \left[7(3-\sqrt{3})\right]^2

    . . . . . . =\;49(3-2\sqrt{3}+1) + 49(9-6\sqrt{3}+3)

    . . . . . . =\;49(4 - 2\sqrt{3} + 12 - 6\sqrt{3})

    . . . . . . =\; 49(16-8\sqrt{3}) \;=\;392(2-\sqrt{3})

    . . . AB \;=\;\sqrt{392(2-\sqrt{3})} \;=\;14\sqrt{4-2\sqrt{3}}


    \text{Therefore: \,Perimeter}\;=\;4(AB) \;=\;56\sqrt{4-2\sqrt{3}} \;\approx\;41

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  4. #4
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    Thanks a lot! I see what I did wrong. I was hoping someone could nudge me along in the right direction for the 2nd question? Thanks!
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  5. #5
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    1st one is quite simple; seems Soroban took the long way...

    we have x = 14 / (1 + sqrt(3))

    outer side = 2x (property of 30-60-90 triangle)

    so perimeter = 4(2x) = 8x = 112 / (1 + sqrt(3)) = ~41
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  6. #6
    Junior Member Kaloda's Avatar
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    the answer in number 2 is 144!
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  7. #7
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    Agree with 144; easy enough IF XYZ is equilateral triangle:
    diagram sure looks so; but not stated.

    So assuming 4 equilateral triangles (3 inside lines parallel to sides):
    side of XYZ = sum of sides of the other 3;
    Formula: SIDE = 2(AREA) / 3^(1/4)

    k=3^(1/4), a= areaXYZ side, b=area25 side, c=area16 side, d=area9 side

    a = b + c + d : works out to :
    a = 2sqrt(25)/k + 2sqrt(16)/k + 2sqrt(9)/k = ~18.236;
    Makes area = 144.
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