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?
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.
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.
[sub]Triangle maaan, Triangle maaan.[/sub]
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.
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.
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:
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.
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?
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
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.
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?