Originally posted by joe shmoActually, I was wrong, and my mistake was somewhere in the second triangle...
that looks to technical to be false, so you da person geepamoogle.
I did find the easiest method and it says 108 as the answer.
First I find the smallest rectangular window that fits all 4 points, such that each boundary has at least 1 point on it. That rectangle in this case will be 9x16 rectangle with an area of 144.
X from -4 to +5, Y from -9 to +7
Now I can chop off rectangles and right triangles until all the remaining area is within the quadrilateral indicated.
Now (-3,5) is the only point not on the edge, so I take off the 1x2 rectangle from it to the outside corner, leaving an area of 142 left.
That leaves 4 right triangles to carve away. The 1x11 and 7x2 both border the rectangle we took out. There is also a 1x16 and a 9x3.
Total area of these four are (11+14+16+27)/2 = 34, which leaves an Area of 108 in the quadrilateral.
I did relook at my numbers from before, and I still get 37.5 for my first Heron's triangle, but the second should be 70.5, so I must have messed up somewhere in it.
Are we talking only of this quadrilateral? Surely there is a general method, isn't there?
How to find out an area inside a polygon consisting of the points (a(i),b(i)) where i is from 1 to n, the number of vertices is this polygon ?
Then we can fill in his rout points and see what the result is. Drawing the figure on a paper and count triangles, is like cheating to me.
I have reexamined my 1st attempt, and I see where I went wrong..
My second triangle overlapped the first, and left some uncounted..
I should have used sqrt(233), sqrt(257) and sqrt(90) for the sides, and then I probably would have gotten it right..
Just goes to show you you have to check your reasoning thoroughly..
(It's a simple matter to find the length of sides and the "diagonals" connect opposing points, should anyone wish to look into it.)
Originally posted by geepamoogleI must have messed up a good bit yo come up with 125.5...so no biggy
Actually, I was wrong, and my mistake was somewhere in the second triangle...
I did find the easiest method and it says 108 as the answer.
First I find the smallest rectangular window that fits all 4 points, such that each boundary has at least 1 point on it. That rectangle in this case will be 9x16 rectangle with an area of 144.
X from -4 to + ...[text shortened]... irst Heron's triangle, but the second should be 70.5, so I must have messed up somewhere in it.
Originally posted by joe shmoIt doesn't. However, I think that playing at 1700 proves a sound grasp of logic. Determining the area of a quadrangle is pretty basic mathematics. To be honest, I don't know the most elegant solution, I have a fuzzy recollection of a nice complex variable solution. However, even a very pedestrian approach should be obvious in my opinion: two triangles.
How does playing Chess well automaticaly make's you a mathematician?🙄
Originally posted by MrHandSo, you mean that by learnig Chess, we learn Math? Ok, Explain to me why a Knight /King vs King ends in a stalemate.......MATHEMATICALLY..
It doesn't. However, I think that playing at 1700 proves a sound grasp of logic. Determining the area of a quadrangle is pretty basic mathematics. To be honest, I don't know the most elegant solution, I have a fuzzy recollection of a nice complex variable solution. However, even a very pedestrian approach should be obvious in my opinion: two triangles.
And further more, by YOUR logic, your opinion on mathematics has no merit, as you are just slightly above average as a Chess player...(As far as the numbers say..I really don't know)
You are making extrapolations about my statements. I will try to make it as succinct as possible.
Proficiency at chess requires facility with logical thinking and problem solving -- cause and effect.
Mathematics is very much the same, it is necessary to have sound logical thinking skills.
However, it does not necessarily follow that someone who is a good mathematician will be good at chess without work at it, nor does vice-versa follow.
My point is that if someone has achieved a rating of 1700, then they have sound logical thinking and problem solving skills. This is all that is required to find the area of a linear quadrangle.
I am not saying that a mathematician is automatically good at chess. I am not saying that a chess player is automatically good at math.
As far as my chess skills...they're crap! But I'm improving...slowly.
Originally posted by MrHandHey,..I'm sorry if I came off like an a-hole. I tend to run into a large amount of oversized egos on this site...ego's so large, there is no room in their heads for common courtesy....so I guess I was just venting...as i said, sorry🙂
You are making extrapolations about my statements. I will try to make it as succinct as possible.
Proficiency at chess requires facility with logical thinking and problem solving -- cause and effect.
Mathematics is very much the same, it is necessary to have sound logical thinking skills.
However, it does not necessarily follow that someone who is a ...[text shortened]... ically good at math.
As far as my chess skills...they're crap! But I'm improving...slowly.
Originally posted by geepamoogleThis method looks correct at a quick glance. You can find the sides by making each side into a right triangle with two sides parallel to the x and y axes and then use Pythagorean's Theorem to find the quadrilateral side. Then split it into two triangles, find the length of the new side separating the triangles, and use SSS to find the area of each.
I broke it into 2 triangles.
One with sides of sqrt(122), sqrt(53), sqrt(233)...
One with sides of sqrt(257), sqrt(53), sqrt(260).
Using the formula A = 1/4 * sqrt( (A^2 + B^2 + C^2)^2 - 2*(A^4 + B^4 + C^4) ), I get the following..
Triangle 1
Area = 1/4 * Sqrt[(122 + 53 + 233)^2 - 2*(122^2 + 53^2 + 233^2)]
Area = 1/4 * Sqrt[408^2 - 2*(14,884 ...[text shortened]... (324,900 - 2*136,458)
Area = 1/4 * Sqrt(51,984) = 228 / 4 = 57
Total Area = 37.5 + 57 = 94.5
Originally posted by AThousandYoungThe method, by and large was good. It's just that you have to verify the two triangles you select cover the entire area without overlapping, and I forgot to check that. Ends up there was overlap and lack.
This method looks correct at a quick glance. You can find the sides by making each side into a right triangle with two sides parallel to the x and y axes and then use Pythagorean's Theorem to find the quadrilateral side. Then split it into two triangles, find the length of the new side separating the triangles, and use SSS to find the area of each.
Originally posted by AThousandYoungThe answer is 108.. of course first u must roughly jot down the coordinates of the vertices as to divide the quadrilateral in to 2 non over lapping triangles..Lets say,
Just cut the quadrilateral from one corner to the other!
vertice 1 = (-4,-6) ; vertice 2 = (-3,5) ;vertice 3 = (4,7) ;vertice 4 = (5,-9)
As u can see the points (1,2,3) & points (1,3,4) form 2 non-overlapping triangles..Now u can use the basic coordinate geometry formula for the area of a triangle:
Area = 1/2 mod[(y1-y2)*x3+(y2-y3)*x1+(y3-y1)*x2]
Add up the 2 triangles and the answer is 108. There is no need to use Pythagoras theorem or any other theorem which involves sqrts.. it makes it difficult to solve..
Hope this helps..
Cheers! :-)