Question

What is the area of the inscribed triangle?
Here is a link to the original post: http://1.bp.blogspot.com/-OIhghZmqqVo/UkG7VzQ466I/AAAAAAAAAzM/tER3AtipVFQ/s1600/inner+triangles.png
I need help with the figure 2, I've already used the midpoint theorem to answer figure 1. I will draw a picture...

Answer 1

What I am looking for is a reference (book, blog post, or words to search) rather than the answer. I can use Heron's formula to find the area but I would need the side lengths of the inscribed triangle.

Answer 2

are the arcs important? or is that just some odd way to measure side length?

Answer 3

I don't think the arcs are important. I think they are just showing what lengths are being measured. I already figured out figure 1. I need help with figure 2.

Answer 4

yeah, i cant get an idea developed for 2 :/

Answer 5

i could develop a rather complicated way to approach it .. :)

Answer 6

Any guesses? Words I could search on?

Answer 7

give me your complicated idea. It could help.

Answer 8

transfer it to a coordinant plane and develop the line equations

for the initial figure, 2 circles meeting such that:

\[x^2 + y^2 - 22^2=(x-18)^2+y^2-20^2\]
\[x^2 - 22^2=(x-18)^2-20^2\]
\[x^2 - 22^2=x^2-36x+18^2-20^2\]
\[- 22^2=-36x+18^2-20^2\]
\[\frac{22^2-20^2+18^2}{36}=x\]
plug into the circle to find the top vertex, then the other smaller triangle are a matter of midsegments

Answer 9

my center in my freehand drawing is off, but thats the idea lol

Answer 10

Well, that qualifies for complicated. But a good place to start. I'll spend some time playing around with that idea. Thanks.

Answer 11

ill play with it to see what solution i can develop ... :) im just to stupid to know an easy approach, and to smart to not know anything lol

Answer 12

I'm told that this is a middle school problem. I didn't learn any of this in middle school, but I'm having fun learning it now. I just wish the blog author would give a hint or two...

Answer 13

Higher levels of education does cause us to over complicate things sometimes. :-)

Answer 14

i took a different route, thought about using the law of cosines to find the angles with; then the law of cosines can be used again to find lengths

Answer 15

\[c^2=a^2+b^2-2ab~cos(C)\]
reworking to find C, the angle between the sides:\[\frac{c^2-a^2-b^2}{-2ab}=cos(C)\]arccos finds the angle

knowing the angles we can run the first setup to find a missing side

Answer 16

That makes sense.

Answer 17

another idea that comes to mind is using the sin area formula:



\[A=hb/2\]
\[h=r~\sin\alpha\]
\[A=\frac{rb~\sin\alpha}2\]
might be nice to find the areas of the other 3 triangles