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  1. #1
    Member helloying's Avatar
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    geometry problem

    how to solve part (c)?
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  2. #2
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    Quote Originally Posted by helloying View Post
    how to solve part (c)?
    Use proportions: (I assume that x = 10 is correct)

    \dfrac{z+k}{k} = \dfrac{10}6~\implies~z=\dfrac{10}7 k - k~\implies~z=\dfrac23k
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  3. #3
    Member helloying's Avatar
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    Quote Originally Posted by earboth View Post
    Use proportions: (I assume that x = 10 is correct)

    \dfrac{z+k}{k} = \dfrac{10}6~\implies~z=\dfrac{10}7 k - k~\implies~z=\dfrac23k

    Thank you. Your ans is correct but i don understand why.Can u explain to me please?
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  4. #4
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    Quote Originally Posted by helloying View Post
    Thank you. Your ans is correct but i don understand why.Can u explain to me please?
    You have 2 similar triangles: DEA and BCA.

    You know that DE \parallel BC

    Then the segments of sides must correspond to the lengths of the parallels:

    The side AB correspond to BC
    The side AD correspond to DE

    \dfrac{AB}{BC} = \dfrac{AD}{DE}

    Since AB = z + k and the measures of the parallels are known to you this proportion becomes:

    \dfrac{z+k}{10} = \dfrac{k}{6}

    Multiply both sides by 60 to get rid of the denominators:

    6(z+k)=10k~\implies~6z = 4k~\implies~z=\dfrac23k
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