# Help with solving.

• Dec 7th 2006, 06:00 PM
Ash
Help with solving.
What are the steps to solving this problem???
The second problem was written wrong. It should have been, If CD= 15, what is the length of AE?
I now understand how to do the first problem. But how do I apply it to solving the second one?
I changed the image to shows the correct question.
http://i140.photobucket.com/albums/r...riss/qinfo.jpg
• Dec 7th 2006, 06:14 PM
topsquark
Quote:

Originally Posted by Ash
What are the steps to solving these problems???
http://i140.photobucket.com/albums/r...riss/qinfo.jpg

Take the problem on the left.

Triangles CBA and CED are similar since angle C is shared, angles CED and CBA are right, and angles EDC and BAC are equal.

So $\frac{BA}{ED} = \frac{BC}{BE}$

$\frac{30}{10} = \frac{CE + 30}{CE}$

$3 CE = CE + 30$

$2 CE = 30$

$CE = 15$

The second problem is supposed to be done using the same principle, but the diagram seems to conflict with the problem? (CD is the whole base, but 10 seems to be only a part of it?)

-Dan
• Dec 7th 2006, 06:21 PM
AfterShock
Quote:

Originally Posted by topsquark
Take the problem on the left.

Triangles CBA and CED are similar since angle C is shared, angles CED and CBA are right, and angles EDC and BAC are equal.

So $\frac{BA}{ED} = \frac{BC}{BE}$

$\frac{30}{10} = \frac{CE + 30}{CE}$

$3 CE = CE + 30$

$2 CE = 30$

$CE = 15$

The second problem is supposed to be done using the same principle, but the diagram seems to conflict with the problem? (CD is the whole base, but 10 seems to be only a part of it?)

-Dan

Indeed, as with the first I think he meant BA = 30, not BE.

In the second, if DE = 10, then CD can't possibly be 10, so something is wrong.
• Dec 7th 2006, 07:17 PM
Ash
Quote:

Originally Posted by AfterShock
Indeed, as with the first I think he meant BA = 30, not BE.

In the second, if DE = 10, then CD can't possibly be 10, so something is wrong.

On the second figure It was written wrong. It's suppose to be If CD=15 what is the length of AE?