# Strange Problem

• Feb 28th 2010, 09:02 PM
norifurippu
Strange Problem
Hello, I'm trying to solve this problem, I really didn't know if it was Algebra or Trigonometry, so if this is the wrong section, I'm sorry.

On the next figure, the segments m1 and m2 are parallel, what's the value of X?

http://img199.imageshack.us/img199/33/problemuj.th.jpg

I tried solving using a system of ecuations, but I can't cause I have 3 unknowns, m1, m2 and x. Tried lots of things but couldn't solve it.

• Feb 28th 2010, 09:33 PM
Soroban
Hello, norifurippu!

Quote:

In the figure, the segments $m1$ and $m2$ are parallel.
What is the value of $x$ ?
Code:

      *       |  *    8   m1 |      *       |          *            9       * - - - - - - - * - - - - - - - - - *               x          *              |                               *          | m2                             2x  *      |                                       *  |                                           *

We don't need the lengths of m1 and m2.

Since $m1 \parallel m2$, the two triangles are similar.

Hence: . $\frac{x}{8} \:=\:\frac{9}{2x} \quad\Rightarrow\quad 2x^2 \:=\:72 \quad\Rightarrow\quad x^2 \:=\:36 \quad\Rightarrow\quad x \:=\:6$

• Mar 1st 2010, 10:49 AM
norifurippu
Thank you Soroban.
So we know the 2 triangles are similar only because m1 and m2 are parallel? Is there anything else we need to know 2 triangles are similar?
• Mar 1st 2010, 01:02 PM
Soroban
Hello again, norifurippu!

Quote:

So we know the 2 triangles are similar only because m1 and m2 are parallel?
Is there anything else we need to know 2 triangles are similar?

Here's a more complete explanation . . .

Code:

          B           o         /    *     m1 /        *       /              *     A o - - - - - - - - - o - - - - - - - - - - - - - - o E                           C  *                        /                                   *                  /                                       *              / m2                                           *        /                                               *    /                                                   o                                                   D

Since $m1 \parallel m2\!:\;\;\angle A \,=\,\angle E$ . (alternate-interior angles)

Also: . $\angle BCA \,=\,\angle DCE$ . (vertical angles)

Hence: . $\angle B \,=\,\angle D$

Therefore: . $\Delta ABC \:\sim\: \Delta CDE$ . (a.a.a.)

• Mar 1st 2010, 04:16 PM
norifurippu
Now I get it! Thanks a lot Soroban!