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#1 2010-06-24 17:17:56

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

Geometry: grade 11 problem#4

BD=4
DC = 6
find AC

EXPLAIN PLEASE.THANKS

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#2 2010-06-24 17:23:08

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

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.

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#3 2010-06-24 17:43:27

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 86,161

Re: Geometry: grade 11 problem#4

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.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

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#4 2010-06-29 03:11:41

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

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

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#5 2010-06-29 09:25:06

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 86,161

Re: Geometry: grade 11 problem#4

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.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

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