#1  
Old 04-21-2011, 08:07 AM
dallen2408 dallen2408 is offline
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Triangles

I have been searching the internet relentlessly to find a solution for calculating the area of a smaller triangle in the four basic types of triangles via a formula I can place into a spreadsheet. The distance between the small and large triangles will not change from 18" on two sides with 36" on the third or 18" on one side with 36" on the other two, however, the size of the larger triangle will change on a regular basis thus changing the size of the smaller. I have tried to find commonality and constant ratios in the differences to no avail because the ratios change as the size changes. Email me for a diagram of said scenarios. Can anyone help?

Thank you,

Darren Allen
dallen@newcenturyeng.com
  #2  
Old 04-21-2011, 08:30 AM
MikeS MikeS is offline
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I'm having trouble visualizing what you're trying to do, but you might look into using the determinant formulas for the area of a triangle (equation (16) here) or Heron's formula.
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Old 04-21-2011, 06:29 PM
ZenBeam ZenBeam is offline
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If a triangle has two sides of 18 inches, and the third is 36 inches, it's not really a triangle, it's a straight line, and has zero area.

Beyond that, not sure what you're trying to say.
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Old 04-21-2011, 07:14 PM
Lance Turbo Lance Turbo is offline
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I'm curious what the "four basic types" of triangles are.
  #5  
Old 04-21-2011, 07:21 PM
pulykamell pulykamell is online now
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Quote:
Originally Posted by Lance Turbo View Post
I'm curious what the "four basic types" of triangles are.
I'd guess: equilateral, isosceles, scalene, and right? (Although "right" doesn't go with the other three. There are also acute and obtuse triangles.)

Last edited by pulykamell; 04-21-2011 at 07:24 PM.
  #6  
Old 04-21-2011, 07:47 PM
friedo friedo is online now
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All right triangles are either isosceles or scalene. The categories of isosceles, equilateral, and scalene are orthogonal to those of right, acute, and obtuse.

Oh, and anything that's not a right triangle is also an oblique triangle.


Triangle maaan, Triangle maaan.
  #7  
Old 04-22-2011, 09:08 AM
MikeS MikeS is offline
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If I'm not mistaken, the OP is asking the following: given a triangle ABC, construct a triangle DEF inside it such that AB is parallel to DE (and separated by 18 inches), BC is parallel to EF (and separated by 18 or 36 inches, depending), and CA is parallel to FD (and separated by 36 inches). I'd guess he's installing a walkway around a lawn and needs to calculate the area of the lawn, or some such.
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Old 04-22-2011, 10:27 AM
Lance Turbo Lance Turbo is offline
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Something like that is my guess as well, but if the OP doesn't come back and clarify we may never know.
  #9  
Old 04-26-2011, 06:32 PM
dallen2408 dallen2408 is offline
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equilateral, isosceles, scalene, and right Triangles. This distance noted above is the distance between the two triangles on each side. A smaller one inside a larger one. Please email me and I will send you an example of each. I would upload one here but not sure how.
  #10  
Old 04-26-2011, 07:12 PM
ultrafilter ultrafilter is offline
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Quote:
Originally Posted by dallen2408 View Post
equilateral, isosceles, scalene, and right Triangles. This distance noted above is the distance between the two triangles on each side. A smaller one inside a larger one. Please email me and I will send you an example of each. I would upload one here but not sure how.
Use Imgur. We need a picture to figure this one out.
  #11  
Old 04-27-2011, 01:31 AM
Lance Turbo Lance Turbo is offline
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I'll take a crack at this but I'm not sure I'm answering the right question.

Suppose I have a triangle, T, of area A with side lengths a, b, and c. If I draw lines parallel to the sides of this triangle distances x, y, and z from the sides of lengths a, b, and c respectively, a smaller triangle, T', is formed in the interior of T provided x, y, and z are not too large with respect to a, b, and c.

If this is indeed the problem we are trying to solve, the area of T' is:

((A - ax/2 - by/2 - cz/2)^2) / A
  #12  
Old 04-27-2011, 08:45 AM
dallen2408 dallen2408 is offline
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Quote:
Originally Posted by Lance Turbo View Post
I'll take a crack at this but I'm not sure I'm answering the right question.

Suppose I have a triangle, T, of area A with side lengths a, b, and c. If I draw lines parallel to the sides of this triangle distances x, y, and z from the sides of lengths a, b, and c respectively, a smaller triangle, T', is formed in the interior of T provided x, y, and z are not too large with respect to a, b, and c.

If this is indeed the problem we are trying to solve, the area of T' is:

((A - ax/2 - by/2 - cz/2)^2) / A
This is great thank you very much!
  #13  
Old 04-27-2011, 12:33 PM
Chessic Sense Chessic Sense is offline
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Quote:
Originally Posted by Lance Turbo View Post
I'll take a crack at this but I'm not sure I'm answering the right question.

Suppose I have a triangle, T, of area A with side lengths a, b, and c. If I draw lines parallel to the sides of this triangle distances x, y, and z from the sides of lengths a, b, and c respectively, a smaller triangle, T', is formed in the interior of T provided x, y, and z are not too large with respect to a, b, and c.

If this is indeed the problem we are trying to solve, the area of T' is:

((A - ax/2 - by/2 - cz/2)^2) / A
I'm not following you here, Lance. I get the ax/2 quantities (It's one sidewalk plus the two corners), and that would make the smallest parenthesis almost the area of the smaller triange (A'), but you've double-subtracted each "corner diamond". And I don't know where your ^2/A operation is coming from. I assume this fixes the corner problem, but I can't see how.

Proof, please?
  #14  
Old 04-27-2011, 12:57 PM
Chessic Sense Chessic Sense is offline
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I realize now that the corner diamonds are just parallelograms, and thus they're just xy, yz, and zx. So how about T' = T - (ax/2) - (by/2) - (cz/2) + xy + xz + yz?
  #15  
Old 04-27-2011, 01:50 PM
Lance Turbo Lance Turbo is offline
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I'm posting from my phone so I'm going to leave some steps out of this proof.

Triangles T and T' are similar so:

A' = ((a'/a)^2) * A

Some fairly straightforward trig (and a diagram) gives:

a' = a - x cot(beta) - x cot(gamma) - y csc(gamma) - z csc(beta)

Here alpha is the angle opposite a etc.

Now get rid of trig functions using the following two equations:

cos(alpha) = (b^2 + c^2 - a^2) / (2bc)

and

sin(alpha) = 2A/(bc)

Permuting variables where necessary.

Put that all together to get my formula.

Last edited by Lance Turbo; 04-27-2011 at 01:55 PM.
  #16  
Old 04-27-2011, 04:11 PM
dallen2408 dallen2408 is offline
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Quote:
Originally Posted by Chessic Sense View Post
I realize now that the corner diamonds are just parallelograms, and thus they're just xy, yz, and zx. So how about T' = T - (ax/2) - (by/2) - (cz/2) + xy + xz + yz?
What are you saying; that the formulas given earlier will not work and that this formula replaces the other one? Which completed formula do I use?

Last edited by dallen2408; 04-27-2011 at 04:16 PM.
  #17  
Old 04-27-2011, 05:16 PM
Lance Turbo Lance Turbo is offline
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Quote:
Originally Posted by dallen2408 View Post
Which completed formula do I use?
You should use the correct one.

Chessic, if you need to break this up geometrically you can expand and group my formula as:

A - ax(1 - ax/4A) - by(1 - by/4A) - cz(1 - cz/4a) + abxy/2A + acxz/2A + bcyz/2A

This can be read as the area of the original triangle, minus the areas of three trapezoids, and then add back in the area of the three corner diamonds.
  #18  
Old 04-28-2011, 02:06 PM
dallen2408 dallen2408 is offline
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Quote:
Originally Posted by Lance Turbo View Post
You should use the correct one.

Chessic, if you need to break this up geometrically you can expand and group my formula as:

A - ax(1 - ax/4A) - by(1 - by/4A) - cz(1 - cz/4a) + abxy/2A + acxz/2A + bcyz/2A

This can be read as the area of the original triangle, minus the areas of three trapezoids, and then add back in the area of the three corner diamonds.
I have use the original formula given and it works for all of the triangle samples I could do except for right triangles. Will the above quoted formula work for right triangles and could you email me a for clear picture (description)? dallen2408@centurytel.net or dallen@newcenturyeng.com
  #19  
Old 04-28-2011, 03:13 PM
Lance Turbo Lance Turbo is offline
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Are you claiming that the formula works for scalene and isosceles triangles but not for right triangles? If that's the case, you are incorrect. Please post the details of the triangle the formula doesn't work for.
  #20  
Old 04-28-2011, 07:50 PM
dallen2408 dallen2408 is offline
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ta[little triangle area]=((TA[Big Triangle Area] - (side a*distance x)/2 - (side b*distance y)/2 - (side c*distance z)/2)^2) / TA

This formula works for Equilateral, Isosceles and Scalene Triangles. I have drawn out each of these in CAD with properties of each triangle area. The numbers crunch correctly. I even changed the sizes of the outside triangles and the numbers match what the CAD had given for the smaller triangle's area. The proof is in the CAD. I just repeated the formula for a right triangle where a=20, b=10, c=(a^2+b^2+c^2) or 22.36, x=1.5, y=3, z=1.5. It does work for all four triangles. You are the best! Thanks

Now; how did you know this? Could you please explain?
  #21  
Old 04-28-2011, 08:53 PM
Lance Turbo Lance Turbo is offline
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I explained how I derived the formula in post 15.
  #22  
Old 04-29-2011, 07:56 AM
dallen2408 dallen2408 is offline
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Thanks for reminding me. Could you fill in the missing steps that you indicated you left out?
  #23  
Old 04-29-2011, 12:48 PM
Lance Turbo Lance Turbo is offline
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a' = a - x cot(beta) - x cot(gamma) - y csc(gamma) - z csc(beta)
= a - x (a^2+ c^2 - b^2)/(4A) - x (a^2+ b^2 - c^2)/(4A) - aby/(2A) - acz/(2A)
= a - a^2 x/(2A) - aby/(2A) - acz/(2A)
= a(A - ax/2 - by/2 - cz/2)/A

So...

A' = A (a'/a)^2
= A ((A - ax/2 - by/2 - cz/)/A)^2)
= ((A - ax/2 - by/2 - cz/2)^2)/A

Keep in mind that I am posting from my phone so you may need to correct some typos here and there.
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