need help with similar trinagle problem

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September 1st 2009, 10:33 PM
realistic

need help with similar trinagle problem

http://img16.imageshack.us/img16/4434/similar.jpg
Given thatYM = 3a, MZ = a+2, LM = 18 and XZ = 30, find a in the following diagram
September 1st 2009, 11:46 PM
eXist

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:
September 1st 2009, 11:49 PM
realistic

Quote:

Originally Posted by eXist

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?
September 1st 2009, 11:51 PM
eXist

Cross multiply from there:

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

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

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

$<br /> a = 2<br />$
September 2nd 2009, 12:13 AM
realistic

Quote:

Originally Posted by eXist

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 (Bow)

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