# need help with similar trinagle problem

• Sep 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
• Sep 1st 2009, 11:46 PM
eXist
Cause they are similar triangles you can setup a ratio:

$\displaystyle \frac{3a}{3a + (a + 2)} = \frac{18}{30} \Rightarrow \frac{3a}{4a + 2} = \frac{18}{30}$

Cross multiply:
• Sep 1st 2009, 11:49 PM
realistic
Quote:

Originally Posted by eXist
Cause they are similar triangles you can setup a ratio:

$\displaystyle \frac{3a}{3a + (a + 2)} = \frac{18}{30}$

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?
• Sep 1st 2009, 11:51 PM
eXist
Cross multiply from there:

$\displaystyle (3a)(30) = (4a + 2)(18)$

$\displaystyle 90a = 72a + 36$

$\displaystyle 18a = 36$

$\displaystyle a = 2$
• Sep 2nd 2009, 12:13 AM
realistic
Quote:

Originally Posted by eXist
Cross multiply from there:

$\displaystyle (3a)(30) = (4a + 2)(18)$

$\displaystyle 90a = 72a + 36$

$\displaystyle 18a = 36$

$\displaystyle a = 2$

ah i c thanks for detailed explanation (Bow)