# Real Hard Geo Question, need help

• Jun 10th 2009, 02:42 AM
Vinsanity
Real Hard Geo Question, need help
Let ABC be a triangle with right angle C. The mid-point of the hypotenuse AB is twice as far from the side BC as it is from side AC. Find the length of the side AC, if AB = 10cm.
• Jun 10th 2009, 03:23 AM
I-Think
Wording
The wording of your question is a bit confusing. By the distance from midpoint of $AB$ to $BC$ and $AC$, do you mean the distance $DE$ and $DF$ as I have in my diagram?

This is how I did the calculation, if this is wrong, just let me know.

Let $DE=x, DF=2x$
As point, by letting $\angle{B}=\theta$ and $\angle{A}=90-\theta$, we see that the triangles $AED$and $DFB$are the same. Thus
$AE=EC=2x$, $CF=FB=x$

Calculate for x with Pythagoras' theorem, and you'll find AC.

If my assumption is wrong, just let me know.
• Jun 10th 2009, 03:29 AM
Vinsanity
wait give me a minute.
• Jun 10th 2009, 03:56 AM
aidan
Quote:

Originally Posted by Vinsanity
Let ABC be a triangle with right angle C. The mid-point of the hypotenuse AB is twice as far from the side BC as it is from side AC. Find the length of the side AC, if AB = 10cm.

The 'Pythagoras' reference:
The distance BC is 2x and the disance CA is 4x

$(2x)^2 + (4x)^2 = 10^2$

$4x^2 + 16x^2 = 10^2$

$x^2 (4+16) = 10^2$

$x^2 = \frac {10^2}{(4+16)}$

$x = \sqrt { \frac {10^2}{(4+16)} }$

NOTE:
This is being done ( not specificly or exactly for your benefit ) so that I can practice using Latex.
• Jun 10th 2009, 11:19 PM
Vinsanity
Why did you result using 2x in the first place
• Jun 11th 2009, 06:36 AM
aidan
Quote:

Originally Posted by Vinsanity
Why did you result using 2x in the first place

I'm sorry that wasn't clear.
My apologies.

Obviously my attempt at clarity wasn't.
You probably used this to answer the question anyway:

$\bar{AC} = 10 \times \sin \left ( \tan^{-1} \left ( \frac{2x}{x} \right ) \right )$

Again, sorry for the obfuscation.
• Jun 11th 2009, 11:58 PM
Vinsanity
Quote:

Originally Posted by aidan
I'm sorry that wasn't clear.
My apologies.

Obviously my attempt at clarity wasn't.
You probably used this to answer the question anyway:

$\bar{AC} = 10 \times \sin \left ( \tan^{-1} \left ( \frac{2x}{x} \right ) \right )$

Again, sorry for the obfuscation.

You are very humble, though how can you use trig?
• Jun 12th 2009, 04:54 AM
aidan
Quote:

Originally Posted by Vinsanity
..., though how can you use trig?

I can use trig because I had a very good 6th grade teacher. The teacher was adamant that we know what a trig function did -- not merely able to write that tangent = sine/cosine or that sine = opposite / hypotenuse.

I can use trig because I practiced solving and creating trig problems.

There are many reasons that I can use trig, but those are the primaries.