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Math Help - geometry problem

  1. #1
    Junior Member
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    geometry problem

    1. pic 7

    A. 7
    B. 21
    C. 26
    D. 52

    2. Pic 8

    A. -1
    B. 2
    C. -2
    D. 1/2
    Attached Thumbnails Attached Thumbnails geometry problem-pic-7.bmp   geometry problem-pic-8.bmp  
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by dgenerationx2 View Post
    1. pic 7

    A. 7
    B. 21
    C. 26
    D. 52
    A median of a triangle is a segment drawn a vertex to the midpoint of the opposite side.

    AD = BD

    5x - 2 = 2x + 7
    3x = 9
    x=3

    AD + BD = AB
    5(3) - 2 + 2(3) + 7=26

    Quote Originally Posted by dgenerationx2 View Post
    2. Pic 8

    A. -1
    B. 2
    C. -2
    D. 1/2
    Did you mean to say the slope of the altitude =4?

    The product of the two slopes must be -1 since the lines are perpendicular.

    4\cdot\frac{-x}{2}=-1

    \frac{-4x}{2}=-1

    -2x=-1

    x={\color{red}\frac{1}{2}}
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  3. #3
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    1) \overline{AD}=\overline{BD}\rightarrow 5x-2=2x+7\leftrightarrow 3x=9\leftrightarrow x=3\rightarrow\overline{AB}=2\cdot(5\cdot 3-2)=\boxed{26}.
    2) From Pythagoras' theorem: 4^2+\overline{DC}^2=\overline{AC}^2 and 4^2+\overline{BD}^2=\overline{AB}^2. From the slope: \tan\alpha=\frac {\overline{AD}}{\overline{BC}}=\frac {4}{\overline{BC}}=\frac {-x}2\rightarrow \overline{BC}=\frac 8{-x}. Can you continue?
    EDIT: I think masters is right as doing this way is hard and stupid.:P
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