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Post Solve for Length "X"
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Posted 2003-09-30, 11:15 PM
Can you solve for length "X" here, because if you can, I think this could help give an answer to the last Probability Riddle I posted.
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Posted 2003-10-01, 07:55 AM in reply to Titusfied's post "Solve for Length "X""
whats the lenth of the bases
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Posted 2003-10-01, 08:26 AM in reply to Titusfied's post "Solve for Length "X""
The answer is actually posted in the new Probability Riddle I posted last night. Anyway, this is all the information you are allowed to get. Basically, its guess and check, not very fun.
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Posted 2003-10-01, 11:05 AM in reply to Titusfied's post "Solve for Length "X""
The base if one triangle is X/(root(2)) Therefore, the length of the rectangle is 60(root(2)) - X(root(2)).

Since the height of the triangle is also X/(root(2)), the area of the rectangle is (root(2))(60-X)(X)/root(2) = 60X-X^2

The area of both triangles together is BH which is (X^2)/2

Therefore, the total area (given) 1250 = (X^2)/2 + 60X - X^2
= 60X - (X^2)/2

X = (apprx) 26.835
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Posted 2003-10-01, 11:15 AM in reply to Titusfied's post "Solve for Length "X""
My brute force program works! Thats around what I got with it...around the square root of 720 is the answer..
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Posted 2003-10-01, 12:02 PM in reply to Titusfied's post "Solve for Length "X""
Seems to work really well. Both of your answers are correct.
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Posted 2003-10-01, 12:45 PM in reply to Titusfied's post "Solve for Length "X""
Um... you need to hoox me up with that brute force prog. It may be useful. PM me or somethin will ya?
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Posted 2003-10-01, 12:52 PM in reply to Titusfied's post "Solve for Length "X""
X=26.833752.....

OR

X=93.166248.....

Check my work.

Nevermind, I checked it myself.

Take that you college boys, sophomore coming for ya!

Last edited by !King_Amazon!; 2003-10-01 at 01:02 PM.
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Posted 2003-10-01, 01:24 PM in reply to Titusfied's post "Solve for Length "X""
Now why did I use this to solve that Probability question? That is the question...
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Posted 2003-10-01, 04:42 PM in reply to Titusfied's post "Solve for Length "X""
This isn't too easy to do on a graphing calculator, but I know those 2 are correct answers. There could be more.
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Posted 2003-10-01, 05:23 PM in reply to Titusfied's post "Solve for Length "X""
eh were you trying to figure the measurements of the side of the triangles?
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Posted 2003-10-01, 05:50 PM in reply to Titusfied's post "Solve for Length "X""
Yes.
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Posted 2003-10-01, 06:23 PM in reply to Titusfied's post "Solve for Length "X""
!King_Amazon! said:
X=26.833752.....

OR

X=93.166248.....

Check my work.

Nevermind, I checked it myself.

Take that you college boys, sophomore coming for ya!
*college boy bitch-slaps sophomore*

93 won't work. Check the bottom length. It is given as 84.85
If X is 93, the length of the bottom of one triangle is 65.76
That means, that the total of the bottom is 131.52 + the length of the rectangle.

That would mean that the length of the rectangle is -46.67 which is impossible.

*bitch-slap with other hand*
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Posted 2003-10-01, 06:27 PM in reply to Titusfied's post "Solve for Length "X""
Well it still works with your equation...
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Posted 2003-10-01, 06:31 PM in reply to Titusfied's post "Solve for Length "X""
It is an extraneous solution.

Example: The area of a square is 64.

The formula for the area of a square is Side^2

From the forumla, the sides could be 8 or -8

However, -8 can be ignored because length cannot be negative.
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Posted 2003-10-01, 06:33 PM in reply to Titusfied's post "Solve for Length "X""
Well Bob, I deleted the brute force program after I was done with it...besides, it would only be useful in this situation.
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Posted 2003-10-01, 06:35 PM in reply to Titusfied's post "Solve for Length "X""
Random off topic note: Bitch slaps come from ancient Hebrew tradition and must be done with your left-hand, your ass-wiping hand.
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Posted 2003-10-01, 06:36 PM in reply to Titusfied's post "Solve for Length "X""
Well, I am ambi-ass-wipe-adextrous then.
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Posted 2003-10-01, 09:10 PM in reply to Titusfied's post "Solve for Length "X""
Clever. I like that comeback.
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