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**lakeheadca****Member**- Registered: 2010-01-31
- Posts: 37

BD=4

DC = 6

find AC

EXPLAIN PLEASE.THANKS

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**ZHero****Real Member**- Registered: 2008-06-08
- Posts: 1,889

BC = √(4² + 6²) = 10

Let AB = x

10² + x² = (6 + √(x² - 4²))²

Solve for 'x'

If two or more thoughts intersect, there has to be a point!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,115

Hi lakeheadca;

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.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**lakeheadca****Member**- Registered: 2010-01-31
- Posts: 37

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 (2010-06-29 03:12:32)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,115

Hi lakeheadca;

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.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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