Find [OAB].

We are given O = (0,0), A = (5,2) and B = (6, -8). Find [OAB], the area of triangle OAB.

Use Pythagorean theorem to find sides of a triangle OAB

Use Heron's formula to calculate the area of Δ OAB

We don't have  a right triangle here, so the Pythagorean Theorem is no good to us

OA =   sqrt ( 5^2 + 2^2)  = sqrt  (29)

OB  = sqrt ( 6^2  + 8^2)  = sqrt (100)  = 10

AB = sqrt [ ( 6 - 5)^2  + (-8-2)^2  ]  =  sqrt [ 1^2  + 10^2 ]  = sqrt (101)

Using the Law of Cosines , we have

[ sqrt (101)]^2  - 10^2  - [sqrt (29)]^2

______________________________   = cos (AOB)

  -2 (10)  sqrt (29)

101 - 100 - 29

____________  = cos (AOB)

-20sqrt (29)

-28

_________  =  cos (AOB)

-20sqrt (29)

7 / ( 5 sqrt (29) )  = cos (AOB) 

arccos  [  (7/(5sqrt (29)  ] ≈  74.9°

[ AOB ]  =  (1/2) (10) (sqrt (29)_  sin ( 74.9°)      =    26 units^2

You said: "We don't have  a right triangle here, so the Pythagorean Theorem is no good to us."

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

We do have the right triangles here and they're outside the triangle OAB.

OA = sqrt(2² + 5²) = √29

BO = sqrt(6² + 8²) = √100

AB = sqrt(10² + 1²) = √101