(0,5) to (2,-2) d^2 = 4+49= 53 d=sqr53 = about 7.280 = length of one side

(2,-2) to (5,1) d^2 = 9+9 = 18 d=sqr18 = about 4.243 = length of a 2nd side

(0,5) to (5,1) d^2 = 25 + 16 = 40 d=sqr40 = about 6.325 = length of 3rd side

sum of the 3 sides = 7.28+4.243 + 6.325 = 17.848 = perimeter

half the perimeter = 17.848/2 = 8.924

8.924-6.325 = 2.599

8.924- 7.28 = 1.664

8.924- 4.243 = 4.681

Heron's formula for area = sqr[(p)(p-a)(p-b)(p-c)] where p= half the perimeter, a,b and c are the 3 sides

Area = sqr(8.924)(2.599)(1.664)(4.681) = 13.441 approximately

= area between the 3 straight lines of the triangle.

Calculus method would find the equations of the 3 lines,

(0,5) to (5,1) has slope = m =-4/5 and y intercept 5: y=-4x/5 + 5

(0,5) to (2,-2) has slope = m = -7/2 and y intercept 5, y=-7x/2 + 5

(5,1) to (2,-2) has slope = m = 3/3 = 1 y=x+b

solve for b by plugging in either point -2 =2+b, b= -4, y=x-4

find x intercept of y=x-4: x=4

find x intercept of y=-7x/2 +5, 7x/2 =5, x=10/7

evaluate integral of -4x/5+5 +7x/2 -5 = 27x/10 from x=0 to x=10/7

evaluate integral of -4x/5 + 5 from x=10/7 to x=4

evaluate integral of -4x/5+5 - x+4 = -9x/5 +9 from x=4 to x=5

evaluate another integral to get the area below the x axis which is a triangle of base 4-10/7 = 18/7 and height = 2, with area = 18/7 if you do it with calculus make sure you get the sign positive

add up the integrals and you did it right if you get close to the heron formula estimate

430/49 + 8/7 + 9/5 = (15050+1960+3087)/(49)(7)(5) = 20097/1715 = 11.72

+ 17/7, that triangle area below the x axis = 11.72 + 2.43 = 14.15 which is "close" to 13.44 from Heron's formula estimate. "just" 0.71 difference. I defer to anyone else trying to do the tedious calculations, or finding an easier way to do it. punch in a slightly wrong number on a calculator and you miss the moving target.

odds are it's somewhere between 14.15 and 13.44. I'd guess around 13.8 give or take 0.35