A third alternative to the formulas I suggested in a comment is to use the vector cross product. Recast the three given points as points in 3D space with 0 z-coordinate:
A(20, 21, 0), B(8, 29, 0), and C(13, 1, 0)
Now consider two vectors,
u = B - A = 〈8, 29, 0〉 - 〈20, 21, 0〉 = 〈-12, 8, 0〉
v = C - A = 〈13, 1, 0〉 - 〈20, 21, 0〉 = 〈-7, -20, 0〉
Then recall the cross product identity,
||u × v|| = ||u|| ||v|| |sin(θ)|
where θ is the angle between the two vectors. The magnitude of u × v is equal to the area of the parallelogram spanned by u and v, and hence twice the area of the triangle of interest.
So we have
area = ||1/2 〈-12, 8, 0〉 × 〈-7, -20, 0〉||
area = 1/2 ||〈-12, 8, 0〉 × 〈-7, -20, 0〉||
area = 1/2 ||〈0, 0, 296〉||
area = 296/2
area = 148