EPTI'm just looking at this work and going wtf!?!
Maths help
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Spenguin
Australia3316 Posts
I'm just looking at this work and going wtf!?! | ||
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Muirhead
United States556 Posts
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Spenguin
Australia3316 Posts
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Seraphim
United States4467 Posts
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Spenguin
Australia3316 Posts
You really have nothing better to do but laugh at me while I struggle? I can get 2 perpendicular bisectors. Here's what I've done, with the given triangle i have put in a dot, from there I have drawn two lines from that dot which meet at two of the sides of the triangle, the angle of the line from the dot and the triangle side is 90 degrees, i can't seem to get a third line or make sense of this fucking shit. the triangle is isosceles btw. EDIT: got three lines that have 90 degrees, now i have to draw a circle around the triangle, with its vertices on the circumference, massive wtf still, no one can help me. | ||
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Spenguin
Australia3316 Posts
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Dromar
United States2145 Posts
On March 08 2008 14:42 Spenguin wrote: [insert witty retort] You really have nothing better to do but laugh at me while I struggle? I can get 2 perpendicular bisectors. Here's what I've done, with the given triangle i have put in a dot, from there I have drawn two lines from that dot which meet at two of the sides of the triangle, the angle of the line from the dot and the triangle side is 90 degrees, i can't seem to get a third line or make sense of this fucking shit. the triangle is isosceles btw. EDIT: got three lines that have 90 degrees, now i have to draw a circle around the triangle, with its vertices on the circumference, massive wtf still, no one can help me. uh.... weren't you supposed to learn this in class? I've never heard of a "circumcentre," but I'm guessing it's pretty simple. I'm gonna make some visual aids. I'll edit in a bit. | ||
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fight_or_flight
United States3988 Posts
On March 08 2008 14:48 Spenguin wrote: still can't get the compass working Do you have to do this mathematically or can you use a compass? On March 08 2008 14:49 Dromar wrote: uh.... weren't you supposed to learn this in class? I've never heard of a "circumcentre," but I'm guessing it's pretty simple. I'm gonna make some visual aids. I'll edit in a bit. It looks pretty hard to me, lol. | ||
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Spenguin
Australia3316 Posts
EDIT: Wait, here she has us drawing this stuff (waste of time) when you can do it with numbers! | ||
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fight_or_flight
United States3988 Posts
http://www.mathopenref.com/trianglecircumcenter.html | ||
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Dromar
United States2145 Posts
It is as I thought (I think). Wikipedia is your friend ![[image loading]](http://aycu17.webshots.com/image/45776/2004269088987552410_rs.jpg) Ok, looking at it a bit more, the picture shows an equilateral triangle, NOT an isoceles. Anyway, you just take the perpendicular bisection of the three sides of the triangle, and the circumcenter is where they cross. In english, that means, for each side of the triangle, you draw a perpendicular line through the triangle in the middle of that side of the triangle. I'll upload a better paint pic soon. ![[image loading]](http://aycu19.webshots.com/image/45578/2004242605368221167_rs.jpg) The Wikipedia link gives examples of irregular triangles. | ||
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Spenguin
Australia3316 Posts
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micronesia
United States24701 Posts
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Spenguin
Australia3316 Posts
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Polemarch
Canada1564 Posts
the circumcentre is equidistant from all 3 points of the triangle, right? so how about this. call the triangle ABC: 1. set your compass to some random length 2. draw circles of a random (but equal) length centered at A and B such that these circles intersect at two points, say P and Q. draw a line through PQ. this is the line of all points that are equidistant from A and B. 3. repeat the above to get the line of points equidistant from B and C. 4. the intersection point of the two lines above is equidistant from A, B, and C; so it's the circumcentre. | ||
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Spenguin
Australia3316 Posts
On March 08 2008 15:05 micronesia wrote: Does anyone know where to find a derivation of Heron's area of a triangle? All wikpedia has is this, http://en.wikipedia.org/wiki/Heron's_formula#History | ||
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Polemarch
Canada1564 Posts
http://mathforum.org/library/drmath/sets/select/dm_heron.html | ||
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Spenguin
Australia3316 Posts
On March 08 2008 15:09 Polemarch wrote: hmm... the circumcentre is equidistant from all 3 points of the triangle, right? so how about this. call the triangle ABC: 1. set your compass to some random length 2. draw a circle of a random length centered at A and B such that these circles intersect at two points, say P and Q. draw a line through PQ. this is the line of all points that are equidistant from A and B. 3. repeat the above to get the line of points equidistant from B and C. 4. the intersection point of the two lines above is equidistant from A, B, and C; so it's the circumcentre. I have been able to do that without a compass, if it is passible I'll have to check with a working compass. | ||
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fight_or_flight
United States3988 Posts
On March 08 2008 15:12 Polemarch wrote: googling heron's formula derivation: http://mathforum.org/library/drmath/sets/select/dm_heron.html I feel like such a failure. On March 08 2008 15:05 micronesia wrote: Does anyone know where to find a derivation of Heron's area of a triangle? This it? http://mathforum.org/library/drmath/sets/select/dm_heron.html | ||
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micronesia
United States24701 Posts
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