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Maple_Tiger
09-07-2004, 06:47 PM
A triangle has a height of 6cm and a base of 8cm. If the height and the base are both 'decreased' by the same amount, the area of the new triangle is 20cm squard.

What is the base and height of the new triangle?

My proublom is that there where no examples for this exercise. Therefore, I'm having trouble figuring out the expression.


I know the Area would be 1/2(base)(height)

I have tride using 1/2(8-2x)(6-2x)=4 or 20 or

1/2(8-2x)(6-X)=4 or 20.

But I just can't get it to work lol.

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Maple_Tiger
09-07-2004, 06:47 PM
A triangle has a height of 6cm and a base of 8cm. If the height and the base are both 'decreased' by the same amount, the area of the new triangle is 20cm squard.

What is the base and height of the new triangle?

My proublom is that there where no examples for this exercise. Therefore, I'm having trouble figuring out the expression.


I know the Area would be 1/2(base)(height)

I have tride using 1/2(8-2x)(6-2x)=4 or 20 or

1/2(8-2x)(6-X)=4 or 20.

But I just can't get it to work lol.

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steve_v
09-07-2004, 06:50 PM
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JG7_Rall
09-07-2004, 06:53 PM
ugh, I just did 5 hours of calculus, chemestry, physics, french, and ap us history homework. Maybe I'll feel up to it later.

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SkyChimp
09-07-2004, 06:54 PM
Make a special request for an appearance by Magister_Ludi. If anything, he's a math freak. In addition to answering your question, he'll show you why the answer proves Germany won the war.

Seriously, he's your best "math" bet. Or maybe Aaron_GT will help as well.

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Maple_Tiger
09-07-2004, 06:57 PM
I realy hope so. This is driving me nuts.

I can work it out if it was a right trianlge or a squar lol.

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pterodactyl77
09-07-2004, 07:00 PM
Area = 1/2 (base) x (height).

Let y = percentage by which both are equally decreased, so y < 100% by definition. The original problem is:

A = 1/2 (8)*(6) = 24

For the new scenario, A=20, reduce both 8 and 6 by the same percentage,y%:

1/2 (8y) * (6y) = 20

Solve for y...

(8)*(6) y^2 = 40

y^2 = 40/48

y = sqrt(40/48) = sqrt(5/6)

y = 0.912870929

Thus, reduce both 8 and 6 by approximately a factor of 0.087129071 or exactly 8.7129071%.

pterodactyl77
09-07-2004, 07:05 PM
oh, sorry...the new base and height are:

height = 6*(0.912870929) = 5.477225574
base = 8*(0.912870929)= 7.302967432

Check it...it seems to work....

p1ngu666
09-07-2004, 07:08 PM
6x8 = 48, - 2 = 24
its 1/6th bigger so 1/3rd from each
5.5 and 7.3333 = 20.166666666666

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p1ngu666
09-07-2004, 07:09 PM
pinnnsssss
moments too late, sigh

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Maple_Tiger
09-07-2004, 07:09 PM
That might work. However, that is not the way I was shown and i just can't follow what your doing. lol

I nead it as an expresion so that i can expand it into a genral formula.

1/2(8-X)(6-X)=20

Example: x^2 -14x +22=0

Then i use,

A=4 B=-28 C=22 28+_squar root of 28^2-4(4)(22)....divided by 2(4)

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[This message was edited by Maple_Tiger on Tue September 07 2004 at 06:17 PM.]

klower
09-07-2004, 07:17 PM
Maple_Tiger,

What you have here is a quadratic equation. you are on the right track with

1/2*(6-x)*(8-x)=20

this can be simplified to

(6-x)*(8-x)=40
or
48 - 14x +x^2 = 40
or
x^2 - 14x + 8 = 0
this is a quadratic equation of the form
ax^2 + bx + c = 0
there are two solutions
x=(-b+(b^2-4ac)^0.5)/2a and
x=(-b-(b^2-4ac)^0.5)/2a

the solutions for x are approximately 0.6 and 13.4. Both will give the correct mathematical result, but you can argue that a negative side length for a triangle makes no sense. Try typing QUADRATIC EQUATION

klower
09-07-2004, 07:18 PM
oops, hit the wrong button. What I was saying was, type QUADRATIC EQUATION into GOOGLE.

cheers

steve_v
09-07-2004, 07:21 PM
http://www.geocities.com/g8tr45/billy.txt

pterodactyl77
09-07-2004, 07:21 PM
In order to write your expression as a general formula as you have done, you cannot have 'X' be identical for the reduction done for 8 and 6. If you do so have 'X' represent the same number, you will skew the triangle, meaning, it will no longer hold its same proportinate shape.

If changing the shape of the triangle is in fact allowed, then you have the correct expression.

If however changing the shape is not allowed, then you should use the equation I posted before, and write it as:

1/2 * (8y) * (6y) = 20

and solve for y:

y = sqrt[(20*2)/(8*6)].

pterodactyl77
09-07-2004, 07:24 PM
In my solution, the general quadratic formula would be written as:

ax^2 + bx + c = 0

Here, a = 48, b = 0, c = -40.

The general solution for quadratic equation holds in the case, as b=0.

Maple_Tiger
09-07-2004, 07:28 PM
<BLOCKQUOTE class="ip-ubbcode-quote"><font size="-1">quote:</font><HR>Originally posted by klower:
Maple_Tiger,

What you have here is a quadratic equation. you are on the right track with

1/2*(6-x)*(8-x)=20

this can be simplified to

(6-x)*(8-x)=40
or
48 - 14x +x^2 = 40
or
x^2 - 14x + 8 = 0
this is a quadratic equation of the form
ax^2 + bx + c = 0
there are two solutions
x=(-b+(b^2-4ac)^0.5)/2a and
x=(-b-(b^2-4ac)^0.5)/2a

the solutions for x are approximately 0.6 and 13.4. Both will give the correct mathematical result, but you can argue that a negative side length for a triangle makes no sense. Try typing QUADRATIC EQUATION<HR></BLOCKQUOTE>



I follow this part:

1/2 (8-x)(6-x)=20

But don't i times 1/2 of

x^2 -14x +48=20

x^2 -14x +28=0

1/2x^2 -7x +14=0

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WTE_Galway
09-07-2004, 07:30 PM
a negative side length for the triangle would make perfect sense if the problem was an engineering one to do with forces, a navigational one using vectors or perhaps something involving 3 phase power factors

its only if the problem represents a euclidean trangle in two dimensional space that the negative presents a quandry

klower
09-07-2004, 07:32 PM
pterodactyl77,
The original poster said the sides were both decreased by the same ammount, not the same proportion. If the problem statement said "same proportion", your solution would be correct. Clearly understanding the problem statment is a critical point in getting the correct answer to this type of math problem. Perhaps Maple can clarify the original problem statement.

WTE_Galway
09-07-2004, 07:33 PM
(6-x)*(8-x)=40

is correct .. multiplying both sides by two gets rid of the half

then you simple expand out the LHS

Maple_Tiger
09-07-2004, 07:36 PM
<BLOCKQUOTE class="ip-ubbcode-quote"><font size="-1">quote:</font><HR>Originally posted by klower:
pterodactyl77,
The original poster said the sides were both decreased by the same ammount, not the same proportion. If the problem statement said "same proportion", your solution would be correct. Clearly understanding the problem statment is a critical point in getting the correct answer to this type of math problem. Perhaps Maple can clarify the original problem statement.<HR></BLOCKQUOTE>



Both decreased by the same amount.

In my book here there is three way for solving lol.


Factoring...wich i don't no of.

some other one.

This one -b+_ squar root of B^2-4ac and then divided by 2a.

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pterodactyl77
09-07-2004, 07:37 PM
klower is absolutely correct.

WTE_Galway
09-07-2004, 07:38 PM
klower has the correct answer, if you plug it back into the original problem it gives the correct results

klower
09-07-2004, 07:39 PM
Now if only I could fly and shoot as well as I can master algebra problems.

Cheers

Maple_Tiger
09-07-2004, 07:40 PM
<BLOCKQUOTE class="ip-ubbcode-quote"><font size="-1">quote:</font><HR>Originally posted by WTE_Galway:
(6-x)*(8-x)=40

is correct .. multiplying both sides by two gets rid of the half

then you simple expand out the LHS<HR></BLOCKQUOTE>


Mayby that what i'm doing wronge.

1/2 (8-x)(6-x)=20

right now i end up with

1/2x^2 -7x +14=0 wen i expand

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WTE_Galway
09-07-2004, 07:49 PM
1/2*(6-x)*(8-x)=20


(6-x)*(8-x)=40 (multiplied both sides by two)

48 - 6x - 8x +x^2 = 40 (expanding)

48 - 14x +x^2 = 40 (tidy up)

Maple_Tiger
09-07-2004, 08:08 PM
Why is it when i use

1/2(x^2 -14 +48)=20

1/2x^2 -7 +24-20

1/2x^2 -7 +4=0


Lmao, the reason why i wasn't getting it, was because i'm an idoit.

i messed up on this:

(x^2 -14 +48)=20

(x^2 -14 +28)=0

1/2x^2 -7 +14=0

I didn't simplify the terms first before i brought the 20 over.


You guys helped me see the light.


I did simlpify by multipying both sides by two and it worked also.


Thanks Klower, WTE_Galway, and pterodactly and everyone that chipped in.


I wish i was not such a simple tin.

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I am hear,
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pterodactyl77
09-07-2004, 08:22 PM
Well done! you'll get the hang of it--the only way you learn is to make mistakes and then don't make them again...!

Best of luck! http://ubbxforums.ubi.com/infopop/emoticons/icon_smile.gif

ps - sorry for the (confusing) answer to a similar but different problem--my mistake...as klower very courteously pointed out.

Maple_Tiger
09-07-2004, 09:06 PM
One of the ways i could have learned to solve these proubloms, was factoring. But, i chose the other method.

I'm taking upgrading right now, and it has been awhile since i was last in school. lol

There only telling us to learn one method. Workse for me lol. Mayby i'll be ready for the test tomorow lol.

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The more less you'r travelling, the further back in time you go.


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Dammerung
09-08-2004, 01:52 AM
ick, quadratics. I hate completing the square

Oh, there are no fighter pilots down in hell...
Oh, there are no fighter pilots down in hell...
The whole damn place is full of queers, navigators, and bombadiers...
Oh, there are no fighter pilots down in hell...

Maple_Tiger
09-08-2004, 04:59 AM
Completing the square is one of the easy parts. It's the word probloms that realy make you think.

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I am hear,
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p-39driver
09-08-2004, 10:21 AM
Hmmmmm....
Ever play the old board game "Battle Fleet Mars"
(tactical interface) By SPI?You might an answer there.
It's been a loooooong time since I've had ponder such problems.

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crazyivan1970
09-08-2004, 10:59 AM
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Kasdeya
09-08-2004, 11:25 AM
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