# The area of a triangle within a triangle **URL:** \ **Category:** Factual Questions\ **Created:** [April 10, 2008, 5:33pm UTC](https://boards.straightdope.com/t/the-area-of-a-triangle-within-a-triangle/444779 "2008-04-10T17:33:29Z")\ **Posts on this page:** 8\ **Page:** 2
**Author:** ![CJJ](https://avatars.discourse-cdn.com/v4/letter/c/ecc23a/32.png) [@CJJ](https://boards.straightdope.com/u/CJJ)\ **Post date:** [April 11, 2008, 5:46pm UTC](https://boards.straightdope.com/t/the-area-of-a-triangle-within-a-triangle/444779/21 "2008-04-11T17:46:43Z")
[QUOTE=Omphaloskeptic] Almost; you’re missing a square root in your application of the law of cosines. [/QUOTE] Yes; for anyone reading this thread in the future, I’ll re-write the proof: [QUOTE=A much more careful CJJ\*] Let’s assume ABC is equilateral. Triangle ABG is then similar to ADI (although flipped, as Omphaloskeptic pointed out). Now in triangle ABG, is the length of side AB is X, the length of side BG is (1/3)X, and between them is the known angle of 60 degrees. We can therefore compute the length of side AG; it is [sqrt(7)/3]X. We earlier determined if ABC is equilateral that ABG and ADI are similar (though one is flipped in relation to the other). Comparing the two, AG corresponds to AI, which by definition is (1/3)X. Therefore, the two similar triangles have sides in the ratio of sqrt(7):1. Using this information, one can either compute the length of the sides in the embedded equilateral triangle DEF (the answer is [1/sqrt(7)]X) or use the fact discovered earlier that the total area of the little corner triangles (for any ABC, not just an equilateral) sum to the area of the central triangle. Either way, the area of DEF will be (1/7) the area of ABC. [/quote] The answer is also confirmed by [Routh’s Theorem](http://en.wikipedia.org/wiki/Routh%27s_theorem) with r=s=t=2. There was a time long ago when I wouldn’t make such elementary mathematical mistakes in a proof; I hope it’s just an off day…and appreciate the indulgence of the board. I think this was more valuable for me than the OP, as the exercise demonstrates that “fighting ignorance” isn’t always an external battle. ---
**Author:** ![spinky](https://avatars.discourse-cdn.com/v4/letter/s/59ef9b/32.png) [@spinky](https://boards.straightdope.com/u/spinky)\ **Post date:** [April 11, 2008, 6:00pm UTC](https://boards.straightdope.com/t/the-area-of-a-triangle-within-a-triangle/444779/22 "2008-04-11T18:00:04Z")
Damn you! I was just about to post the shocking revelation that I had figured out a better way to space my [triangular grid](http://sdmbexample.nfshost.com/triangles2.png) which made it easy to precisely see the area of both triangles for the equilateral case, and that it proved you were all wrong and that the answer was 1/7 (one has an area of 63 and the other has an area of 441) 🙂 ---
**Author:** ![Omphaloskeptic](https://avatars.discourse-cdn.com/v4/letter/o/bcef8e/32.png) [@Omphaloskeptic](https://boards.straightdope.com/u/Omphaloskeptic)\ **Post date:** [April 11, 2008, 7:04pm UTC](https://boards.straightdope.com/t/the-area-of-a-triangle-within-a-triangle/444779/23 "2008-04-11T19:04:59Z")
[QUOTE=ntucker] Damn you! I was just about to post the shocking revelation that I had figured out a better way to space my [triangular grid](http://sdmbexample.nfshost.com/triangles2.png) which made it easy to precisely see the area of both triangles for the equilateral case, and that it proved you were all wrong and that the answer was 1/7 (one has an area of 63 and the other has an area of 441) 🙂 [/QUOTE] That’s the idea I had in mind, though [my solution](http://img187.imageshack.us/img187/5353/thirdsjj0.png) is a little smaller. I like how it reduces the problem to one of counting triangles. ---
**Author:** ![spinky](https://avatars.discourse-cdn.com/v4/letter/s/59ef9b/32.png) [@spinky](https://boards.straightdope.com/u/spinky)\ **Post date:** [April 11, 2008, 7:23pm UTC](https://boards.straightdope.com/t/the-area-of-a-triangle-within-a-triangle/444779/24 "2008-04-11T19:23:47Z")
[QUOTE=Omphaloskeptic] That’s the idea I had in mind, though [my solution](http://img187.imageshack.us/img187/5353/thirdsjj0.png) is a little smaller. [/QUOTE] And about a million times better. Very nice. ---
**Author:** ![lobotomyboy63](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/lobotomyboy63/32/2900_2.png) [@lobotomyboy63](https://boards.straightdope.com/u/lobotomyboy63)\ **Post date:** [April 11, 2008, 11:29pm UTC](https://boards.straightdope.com/t/the-area-of-a-triangle-within-a-triangle/444779/25 "2008-04-11T23:29:01Z")
[QUOTE=Rysto] You don’t need a right angle to use the cosine law. The cosine law reduces to Pythagoras’ Theorem when you have a right angle. [/QUOTE] Thx, I’m still digesting all this. FWIW I asked a math teacher at school who worked it out and got 1/7 of the area of the larger triangle. ---
**Author:** ![dallen2408](https://avatars.discourse-cdn.com/v4/letter/d/7bcc69/32.png) [@dallen2408](https://boards.straightdope.com/u/dallen2408)\ **Post date:** [March 25, 2011, 2:33pm UTC](https://boards.straightdope.com/t/the-area-of-a-triangle-within-a-triangle/444779/26 "2011-03-25T14:33:52Z")
I have a similar issue in that I have been give a triangle of any given size or type. I need to find the area a smaller triangle that is equally smaller than the larger on all sides (IE: 18" smaller). How is this calculated and what would it look like for an Excel spreadsheet? Ruthe’s theorem appears to be a bit difficult to translate for this situation. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [March 25, 2011, 5:37pm UTC](https://boards.straightdope.com/t/the-area-of-a-triangle-within-a-triangle/444779/27 "2011-03-25T17:37:52Z")
That’ll depend on the exact dimensions (all three of them) of the larger triangle. For instance, if any of the larger triangle’s sides is 18", then the area of the smaller triangle vanishes. Once you have all three sides of the larger triangle, then you can easily get the sides of the smaller triangle, and then apply Hero’s Formula. I’ve a hunch, though, that the problem you’re facing actually asks for the sides of the new triangle to be a certain _proportion_ smaller (such as 18%), rather than a certain absolute _amount_ smaller (like 18"). ---
**Author:** ![Hari\_Seldon](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/hari_seldon/32/5173_2.png) [@Hari\_Seldon](https://boards.straightdope.com/u/Hari_Seldon)\ **Post date:** [March 26, 2011, 10:04pm UTC](https://boards.straightdope.com/t/the-area-of-a-triangle-within-a-triangle/444779/28 "2011-03-26T22:04:17Z")
At this point, I don’t know if anyone cares, but here is my solution. Starting from Ultrafilter’s brilliant suggestion, I said to myself, why not take an isoscelese right triangle, specifically the one whose sides are the x-axis, the y-axis and the line x + y + 1, whose area is 1/2. Take the 1/3 points to have coordinates (1/3,0), (2/3,1/3), and (0,2/3). A little bit of work gives the triangle in question as having vertices (2/7,1/7), (4/7,2/7), and (1/7,4/7). The side lengths are then 1/7 times sqrt(5), sqrt(10), and sqrt(13), respectively. Now Heron’s formula for the area is not too useful in its usual form, but it is not hard to show that it is equivalent to (1/4)sqrt(2a^2b^2 + 2b^2c^2 + 2b^2c^2 - a^4 - b^4 -c^4) which gets rid of all but the outside radial. When evaluated you get (1/196)sqrt(196) which is 1/14, exactly 1/7 times the original area. [Previous page](https://boards.straightdope.com/t/the-area-of-a-triangle-within-a-triangle/444779.md?page=1)