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#1
04-21-2011, 07:07 AM
 dallen2408 Guest Join Date: Mar 2011 Posts: 8
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
04-21-2011, 07:30 AM
 MikeS Charter Member Join Date: Oct 2001 Location: New London, CT Posts: 3,783
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.
#3
04-21-2011, 05:29 PM
 ZenBeam Guest Join Date: Oct 1999 Location: I'm right here! Posts: 8,858
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.
#4
04-21-2011, 06:14 PM
 Lance Turbo Guest Join Date: Aug 1999 Location: Asheville, NC Posts: 2,629
I'm curious what the "four basic types" of triangles are.
#5
04-21-2011, 06:21 PM
 pulykamell Charter Member Join Date: May 2000 Location: SW Side, Chicago Posts: 43,140
Quote:
 Originally Posted by Lance Turbo 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 06:24 PM.
#6
04-21-2011, 06:47 PM
 friedo Guest Join Date: May 2000 Location: Brooklyn Posts: 23,653
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
04-22-2011, 08:08 AM
 MikeS Charter Member Join Date: Oct 2001 Location: New London, CT Posts: 3,783
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.
#8
04-22-2011, 09:27 AM
 Lance Turbo Guest Join Date: Aug 1999 Location: Asheville, NC Posts: 2,629
Something like that is my guess as well, but if the OP doesn't come back and clarify we may never know.
#9
04-26-2011, 05:32 PM
 dallen2408 Guest Join Date: Mar 2011 Posts: 8
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
04-26-2011, 06:12 PM
 ultrafilter Guest Join Date: May 2001 Location: In another castle Posts: 18,988
Quote:
 Originally Posted by dallen2408 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
04-27-2011, 12:31 AM
 Lance Turbo Guest Join Date: Aug 1999 Location: Asheville, NC Posts: 2,629
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
04-27-2011, 07:45 AM
 dallen2408 Guest Join Date: Mar 2011 Posts: 8
Quote:
 Originally Posted by Lance Turbo 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
04-27-2011, 11:33 AM
 Chessic Sense Guest Join Date: Apr 2007 Posts: 6,609
Quote:
 Originally Posted by Lance Turbo 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.

#14
04-27-2011, 11:57 AM
 Chessic Sense Guest Join Date: Apr 2007 Posts: 6,609
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
04-27-2011, 12:50 PM
 Lance Turbo Guest Join Date: Aug 1999 Location: Asheville, NC Posts: 2,629
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 12:55 PM.
#16
04-27-2011, 03:11 PM
 dallen2408 Guest Join Date: Mar 2011 Posts: 8
Quote:
 Originally Posted by Chessic Sense 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 03:16 PM.
#17
04-27-2011, 04:16 PM
 Lance Turbo Guest Join Date: Aug 1999 Location: Asheville, NC Posts: 2,629
Quote:
 Originally Posted by dallen2408 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
04-28-2011, 01:06 PM
 dallen2408 Guest Join Date: Mar 2011 Posts: 8
Quote:
 Originally Posted by Lance Turbo 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
04-28-2011, 02:13 PM
 Lance Turbo Guest Join Date: Aug 1999 Location: Asheville, NC Posts: 2,629
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
04-28-2011, 06:50 PM
 dallen2408 Guest Join Date: Mar 2011 Posts: 8
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
04-28-2011, 07:53 PM
 Lance Turbo Guest Join Date: Aug 1999 Location: Asheville, NC Posts: 2,629
I explained how I derived the formula in post 15.
#22
04-29-2011, 06:56 AM
 dallen2408 Guest Join Date: Mar 2011 Posts: 8
Thanks for reminding me. Could you fill in the missing steps that you indicated you left out?
#23
04-29-2011, 11:48 AM
 Lance Turbo Guest Join Date: Aug 1999 Location: Asheville, NC Posts: 2,629
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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