# triangle ratio

• Dec 10th 2006, 12:28 PM
Ash
triangle ratio
I have been trying to solve this problem, but the answer keeps coming up wrong.
Am I suppose to add 15 to the 10 to create the base number and then divide that into another number? After that how is the equation setup?

http://i140.photobucket.com/albums/r..._Priss/mah.jpg
• Dec 10th 2006, 01:39 PM
ThePerfectHacker
Quote:

Originally Posted by Ash
I have been trying to solve this problem, but the answer keeps coming up wrong.
Am I suppose to add 15 to the 10 to create the base number and then divide that into another number? After that how is the equation setup?

http://i140.photobucket.com/albums/r..._Priss/mah.jpg

Note CD is composed to of two line segments one is 10 the other is of unknown length. But we know that in total it is 15. Thus the second segment is 5. Because of symmtery the other peice (that you seek) is also 5.
• Dec 10th 2006, 01:50 PM
Soroban
Hello, Ash!

Quote:

If CD = 15, what is the length of AE?
Code:

```    A *       |*       | *       |  *       |  *       |    *       |    *       |  10  *     E *-------*B       |      |*     10|    10| *       |      |  *     D *-------*---* C         10  F  5```

Since CD = 15 and DF = 10, then FC = 5.

Triangle AEB is similar to triangle BFC.

. . . . . . . . . . AE . .BF . . . . . AE . .10
So we have: . --- = --- . --> . --- = ---
. . . . . . . . . . EB . .FC . . . . . 10 . . 5

Therefore: .AE = 20

• Dec 10th 2006, 01:52 PM
Ash