triangle math problem?

Question

the vertices of triangle ABC are A(6, 5) B(9,9) and C(5,12). Find the area of the triangle. anyone know?? thanks!!

Answer 1

I know.

Using your input coordinates ONLY, the area is

A = ½(XaYb + XbYc + XcYa - YaXb - YbXc - YcXa)

These are the results of solving ½ the determinant

│Xa Ya 1│
│Xb Yb 1│
│Xc Yc 1│

Answer 2

a million. B. Any triangle with a ninety degree attitude is an actual triangle inspite of the dimensions of the different 2 angles. 2. D. A triangle with all 3 aspects an identical length ought to really have all 3 angles an identical (60 ranges). 3. B. Triangles A & D are coated by technique of questions 2 & a million, respectively. An isosceles triangles has 2 equivalent aspects.

Answer 3

to find the area of a triangle, multiply the base times the height of the triangle (the height is not equal to the length of a leg of the triangle unless the triangle is a right triangle) and divide by 2

im think this is correct:
first, to find the length of the base of the triangle, find the distace between and two points

distance formula:

d = square root ((x2 - x1)^2 + (y2 - y1)^2)

d = sqr((9 - 6)^2 + (9 - 5)^2)
d = sqr(3^2 + 4^2)
d = sqr(9 + 16)
d = sqr(25)
d = 5

now find the midpoint of those to points to find a point at the middle of the base, and use the point opposite of that and the new point to find the height of the triangle

midpoint:

((x1 + x2) / 2), ((y1 + y2) / 2)

((6 + 9) / 2), ((5 + 9) / 2)
(15 / 2), (14 / 2)
(15/2, 7)

so use this point and point C (the point opposite this point) to find the height:

d = sqr((x2 - x1)^2 + (y2 - y1)^2)
d = sqr((5 - (15/2))^2 + (12 - 7)^2)
d = sqr((-5/2)^2 + 5^2)
d = sqr((25/4) + 25)
d = sqr(125/4)

now multiply the base times the height and divide by 2:

(5sqr(125/4)) / 2
sqr(3125/4) / 2
sqr(3125 / 16)

im pretty sure thats the area, it might be totally wrong...hope it helped though

Answer 4

The easiest way to do this is by graphing the triangle, drawing a square around it, and then subtracting the area of three right triangles from the area of the square.

You can also solve the problem by calculating the lengths of each side, picking one of them to be your base, and then calculating a height for that base - but why make your life harder than it needs to be?