Calculate the area of a triangle who's vertices are the points (0,16) , (0,5) and (8,2)

To calculate the area of the triangle whose vertices are
(0,16),(0,5) and(8,2).

We know that if (x,y1), (x2,y2) and
(x3,y3) are the vertices of the triangle, then the area of the triangle is given
by:

Area of the triangle = (1/2) |{(x2-x1)(y2+y1)
+(x3-x2)(y3+y1)+(x1-x3)(y1+y2)}|

Therefore the area of the
triangle whose vertices are (0,16), (0,5), (8 ,2) is given
by:

Area of the triangle = (1/2) {(0-0)(16+5) +(8-0)((2+5)
+(0-8)(2+16)}

Area of triangle = (1/2) |{0 + 56 -
144}|

Area of the given triangle = (1/2) * 88 =
44 sq units.

W can calculate
the area by Heron's formula also:

Area of the triangle =
sqrt{s(s-a)(s-b)(s-c)}, a, b ,c are sides of triangle, s =
(a+b+c)/2.

a = sqrt(0+(16-5)^2 )=
11

b = sqrt[8^2+ (2-14)^2] = sqrt260 =
16.125..

c =sqrt[(8-0)*2+(2-5)^2] = sqrt73 =
8.544..

Therefore s = (11+sqrt260+sqrt73)/2 =
17.834..

Area of the triangle =
sqrt{(17.834..)(6.834...)(1.7097..)(9.2902...)}

Area of the
triangle = sqrt1935.999998

Area of the triangle =
43.999...sq units