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BD=4
DC = 6
find AC

## #2 2010-06-25 15:23:08

ZHero
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### Re: Geometry: grade 11 problem#4

BC = √(4² + 6²) = 10

Let AB = x

10² + x² = (6 + √(x² - 4²))²

Solve for 'x'

If two or more thoughts intersect with each other, then there has to be a point.

## #3 2010-06-25 15:43:27

bobbym

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### Re: Geometry: grade 11 problem#4

Find triangle BDC by pythagorean theorem.

4^2 + 6^2 = c^2 = 52

Side BC = √52

Now find the other triangle:

Side AB = s, Side AC = 6 + x, x = AD

s^2 + (√52)^2  = (6 +x)^2

4^2 + x^2 = s^2

Add the two equations up.

68 + x^2 = ( 6 + x )^2

x = 8 / 3

So Side AC is 6 + 8 / 3 = 26 / 3

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

## #4 2010-06-30 01:11:41

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### Re: Geometry: grade 11 problem#4

Thank you guys,
but I really have no Idea what ZHERO did? HOW you come with
10² + x² = (6 + √(x² - 4²))² ?????????????

Bobby
4^2 + 6^2 = c^2 = 52 (c should be 10)
4^2 + x^2 = s^2  (4 + x^2 = s^2 <--- should be this)??
I know, I ask so many questions, sorry my teachers told me too.
I hate this question. Help me, Please,

Last edited by lakeheadca (2010-06-30 01:12:32)

## #5 2010-06-30 07:25:06

bobbym

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### Re: Geometry: grade 11 problem#4

side BC =√52

side BC is the bottom of triangle ABC

Now if you call segment AD = x

Now you have 2 equations:

(AB)^2 + (√52)^2  = (6 + AD)^2

4^2 + (AD)^2 = (AB)^2  Both by pythagorean theorem.

Remember AB is side AB. And AD is segement AD. Now you have 2 equations and 2 unknowns. Rearrange:

(AB)^2 + (√52)^2  = (6 + AD)^2

4^2 + (AD)^2 - (AB)^2 =0 If you add both equations notice side (AB)^2 cancels out.

Can you now solve for AD?

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.