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.

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- Jun 10th 2009, 02:42 AMVinsanityReal 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 AMI-ThinkWording
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.

Answer

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. - Jun 10th 2009, 03:29 AMVinsanity
wait give me a minute.

can you give your final answer as well. - Jun 10th 2009, 03:56 AMaidan

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. - Jun 10th 2009, 11:19 PMVinsanity
Why did you result using 2x in the first place

- Jun 11th 2009, 06:36 AMaidan
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. - Jun 11th 2009, 11:58 PMVinsanity
- Jun 12th 2009, 04:54 AMaidan
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.