[mathwiz3502]

Here is one of my most favorite questions for Mathcounts, a national math compitition:

Square ABCD has sides of 13 units. Point E is located inside square ABCD where line AE is 5 units and line BE is 12 units. Now, how far, in units, is point E from line AD. Express you answer as a mixed number.
Hint:your work should have nothing to do with any math above 9th grade level (calculus, triginometry).;)

This is not a homework question so I would rather have you simply state your answer rather than explaining it to give others a go at it:D
:confused::confused::confused::confused:

[elscarta]

60/13 = 4 8/13

[Unknown008]

Ok, I got 4 8/13 (60/13) too! :D using only two rules in maths.

EDIT: Oops, found my mistake... I got the length of EX, X being the foot of the triangle with base AB :o. Rectifying that, I get 25/13 which is... 1 12/13. Wait, that's what Perito got!

[Perito]

I'm missing something here. If my diagram is correct (and I tried to do it to scale), there's no way the distance (on my drawing, FE) is over 2 units long (I calculate ). What did I miss?

[mathwiz3502]

Totally correct Perito and Unknown008, elscarta got E to AB
Here is how I do it:
Create a triangle with the points (perito's graph)
see that 5, 12, and 13 makes a right triangle
understand that E to AD is horizontal making a right angle where it meets
multiply 12 and 5 then divide by two to get triangle ABE is 30
see that 13 is another "base" while E to AB is the height (E to AB is F to A)
get 13b divided by two equals 30
b=60 over 13
now you know b and c(5) for a^2+b^2=c^2
unsimplify 5 to 65 over 13
realize that 60/5=12 and 65/5=13
see that it is another 5,12, and 13 triangle but times 5 and divided by 13
get 25/13 or 1 12/13
OK

[elscarta]

Sorry, you are right I misread the question. Actually there is a simpler way to find the answer.

1. recognise that triangle ABE is a right angle triangle (5^2 + 12^2 = 13^2)
2. angle FAE = angle ABE since both a complementary to angle EAB
3. therefore triangle AEF is similar to triangle ABE

4. using ratio of sides in similar triangles you get

EF/5 = 5/13
therefore EF = 25/13 = 1 12/13

[mathwiz3502]

Good way

[Unknown008]

Yup, I too found using pythagoras to lengthy. That's why I said before, only two rules, the pythagorean triples and similar triangles :)

Anyway, if you have more mathwiz, I'll say... Keep 'em coming! :)