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sdas [7]
3 years ago
6

Find the area of the triangle whose vertices are given below. ​a(0​,0​) ​b(negative 5​,3​) ​c(3​,2​) the area of triangle abc is

nothing square units. ​(simplify your​ answer.)
Mathematics
1 answer:
beks73 [17]3 years ago
6 0
There are a number of ways the area can be figured. One of them is to create points b'(-5, 0) and c'(3, 0). Then subtract the areas of triangles bb'a and cc'a from trapezoid bb'c'c. Here is that computation.
  (area of bb'c'c) = (1/2)(3 + 2)(3-(-5)) = 20
  (area of bb'a) = (1/2)3*5 = 15/2
  (area of cc'a) = (1/2)(2*3) = 3

Area of abc = 20 - 15/2 - 3 = 19/2 = 9 1/2 square units.

_____
The equivalent computation to that above is
  area = |bx*cy -by*cx|/2 = |-5*2-3*3|/2 = 19/2 . . . square units
The formula for this derives from considering points pairwise as defining trapezoids with respect to the x-axis. Summing the areas of those proceeding clockwise around the figure will give a simple formula for area from vertices.
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iVinArrow [24]
Yay, derivitives
I'mma ignore that x is the shorter side because I don't know which one has to be shorter yet


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and the relatiionship between them is
hmm,
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erm
I think it is f(x)=1200x-2x²
takind the derivitive yeilds none of the others


so ya, you are right
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