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Math Help - need help with similar trinagle problem

  1. #1
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    need help with similar trinagle problem


    Given thatYM = 3a, MZ = a+2, LM = 18 and XZ = 30, find a in the following diagram
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  2. #2
    Member eXist's Avatar
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    Cause they are similar triangles you can setup a ratio:

    <br />
\frac{3a}{3a + (a + 2)} = \frac{18}{30} \Rightarrow \frac{3a}{4a + 2} = \frac{18}{30}<br />


    Cross multiply:
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  3. #3
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    Quote Originally Posted by eXist View Post
    Cause they are similar triangles you can setup a ratio:

    <br />
\frac{3a}{3a + (a + 2)} = \frac{18}{30}<br />
    yeah that is what i had but the other way round, in not sure if that is correct? Or can a figure be worked out for a?
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  4. #4
    Member eXist's Avatar
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    Cross multiply from there:

    <br />
(3a)(30) = (4a + 2)(18)<br />

    <br />
90a = 72a + 36<br />

    <br />
18a = 36<br />

    <br />
a = 2<br />
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  5. #5
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    Quote Originally Posted by eXist View Post
    Cross multiply from there:

    <br />
(3a)(30) = (4a + 2)(18)<br />

    <br />
90a = 72a + 36<br />

    <br />
18a = 36<br />

    <br />
a = 2<br />
    ah i c thanks for detailed explanation
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