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Math Help - ratio of area

  1. #1
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    ratio of area

    Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the triangle is divided into a square with two smaller right triangles. The area of one of the two small right triangle is m times the area of the square. The ratio of the area of the other small right triangle to the area of the square is

    a) 1/2m+1 b) m c) 1-m d) 1/4m e) 1/8msquared

    I got the answer d) but I am not sure how I got the same answer for the different triangles.

    Could someone please take a look at my work?

    Vicky.
    Attached Files Attached Files
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  2. #2
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    Hello, Vicky!

    I couldn't follow your work, but it looks reasonable.


    Through a point on the hypotenuse of a right triangle,
    lines are drawn parallel to the legs of the triangle so that the triangle
    is divided into a square with two smaller right triangles.
    The area of one of the two small right triangle is m times the area of the square.
    The ratio of the area of the other small right triangle to the area of the square is

    . . a)\;\frac{1}{2m+1} \qquad b)\;m\qquad c)\;1-m\qquad d)\;\frac{1}{4m}\qquad e)\;\frac{1}{8m^2}
    Code:
        - *
          |::*
        b |:::::*
          |:::A2:::*
        - *-----------*
          |     x     |::*
          |           |:::::*
        x |     S   x |::::::::*
          |           |:::::A1::::*
          |           | :::::::::::::*
        - *-----------*-----------------*
          :     x     :        a        :

    The side of the square is x.
    . . Its area is: . S \:=\:x^2

    The area of A_1 is: . A_1 \:=\:\tfrac{1}{2}ax

    We are told: . A_1 \:=\:m\cdot S \quad\Rightarrow\quad \tfrac{1}{2}ax \:=\:mx^2 \quad\Rightarrow\quad a \:=\:2mx .[1]


    The area of A_2 is: . A_2 \:=\:\tfrac{1}{2}xb .[2]


    From the similar right triangles: . \frac{b}{x} \:=\:\frac{x}{a} \quad\Rightarrow\quad b \:=\:\frac{x^2}{a}

    Substitute into [2]: . A_2 \:=\:\tfrac{1}{2}x\left(\frac{x^2}{a}\right) \:=\:\frac{x^3}{2a}


    The ratio of A_2 to S is: . \frac{A_2}{S} \:=\:\frac{\frac{x^3}{2a}}{x^2} \:=\:\frac{x}{2a}

    Substitute [1]: . \frac{A_2}{S} \;=\;\frac{x}{2(2mx)} \;=\;\frac{1}{4m}

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  3. #3
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    Quote Originally Posted by Vicky1997 View Post
    Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the triangle is divided into a square with two smaller right triangles. The area of one of the two small right triangle is m times the area of the square. The ratio of the area of the other small right triangle to the area of the square is

    a) 1/2m+1 b) m c) 1-m d) 1/4m e) 1/8msquared

    I got the answer d) but I am not sure how I got the same answer for the different triangles.

    Could someone please take a look at my work?

    Vicky.
    Yes, (d) is correct.
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  4. #4
    Senior Member nikhil's Avatar
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    Lightbulb aliter

    ok so let ABC be a right angle triangle where B is 90 degrees.
    take any point (T) on AC and drop perpendicular on AB and AC and name them D and E
    (this is just the construction that you have done)
    let AB=a and BC=b
    now let area of triangle TEC =m times area of square or
    (1/2) (b-x)x=m(x^2) or
    b-x=2mx or
    x=b/(2m+1)..........................................1
    now
    area of triangle ABC=ar triangle ADT +ar of square +ar of triangle TEC
    (1/2)ab=(1/2)(b-x)x +x^2+(1/2)(a-x)x or
    ab=x(a+b) or
    x=ab/(a+b)........................................2
    from 1 and 2 we have
    ab/(a+b)=b/(2m+1)
    or
    a/b=1/2m
    now we are suppose to calculate ratio of area of other triangle to the area of square
    =(a-x)x/[2(x^2)]
    =(a-x)/2x
    now put x=b/(2m+1)
    =(a-[b/(2m+1)])/2(b/(2m+1))
    =[2ma+a-b]/2b
    =m(a/b)+(a/2b)-(1/2)
    now put a/b=1/2m
    =m(1/2m)+(1/4m)-(1/2)
    =1/4m
    Attached Thumbnails Attached Thumbnails ratio of area-.bmp  
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