# Thread: Real Hard Geo Question, need help

1. ## 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.

2. ## Wording

The wording of your question is a bit confusing. By the distance from midpoint of $\displaystyle AB$ to $\displaystyle BC$ and $\displaystyle AC$, do you mean the distance $\displaystyle DE$ and $\displaystyle DF$ as I have in my diagram?

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

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

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

If my assumption is wrong, just let me know.

3. wait give me a minute.

4. 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

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

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

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

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

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

NOTE:
This is being done ( not specificly or exactly for your benefit ) so that I can practice using Latex.

5. Why did you result using 2x in the first place

6. 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:

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

Again, sorry for the obfuscation.

7. 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:

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

Again, sorry for the obfuscation.
You are very humble, though how can you use trig?

8. 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.