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.)

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.

Okay so for every triangle has three angles that always add up to 180 degrees. and to tell the difference between whether they are secure, right or obtuse is that acute: has one angle that is less than 90 degrees. right triangle means there is at least one angle that is exactly 90 degrees. and an obtuse angle has an angle that is bigger than 180 degrees. i’ll do #13 as an example: 13. 40 and 50 so if every triangle has 180 degrees 40+50= 90 degrees then180-90= 90 degrees so the third angle is 90 degrees and because this triangle has a 90 degree angle it means it is a right triangle

The first one is 2/3the second is 1/160the third is 4/125