Question 257796: Find the area of the triangle ABC having vertices A(7,-4) B(3,1) C(-10,-8)
Found 3 solutions by Edwin McCravy, drk, Alan3354:
Answer by Edwin McCravy(20060)   (Show Source):
You can put this solution on YOUR website!
Find the area of the triangle ABC having vertices A(7,-4) B(3,1) C(-10,-8)
The area of a triangle with vertices (a,b), (c,d), and (e,f) is
Do you know how to evaluate that determinant? If not post again
asking how.
Note that if you take the vertices in counter-clockwise order around
the triangle, the absolute value bars are not necessary, because the
area will come out positive anyway and so sometimes the formula is
given without the absolute value bars, and they tell you to always
take the vertices in counter-clockwise order; however if you include
the absolute value bars, then it doesn't matter which order
you take the vertices. Notice that I took them counter-clock-wise above,
and the absolute value bars were not really necessary.
Edwin
Answer by drk(1908)  (Show Source):
You can put this solution on YOUR website! If you graph these out, you may notice that there are no right angles. This means we can use one of three formulas:
(1) Hero's formula - approximate
To use this we need the 3 distances. They are: sqrt(255), sqrt(61), sqrt(261)
The semi perimeter, s, is about 12.79
A = sqrt(s*(s-a)(s-b)(s-c))
A = 50.5
(2) matrices
Lets use matricies. If we apply matrices, we get 50.5
--
(3) 1/2absinC
This requires trig calculations, which I don't know you have seen.
--
So the area is 50.5.
Answer by Alan3354(69443)  (Show Source):
|
|
|
