Question 846879
{{{drawing(300,300,-2,10,-2,10,grid(1),
circle(0,0,0.2),
circle(5,3,0.2),
circle(2,6,0.2),
line(0,0,5,3),
line(5,3,2,6),
line(2,6,0,0))}}}
We need to find the altitude of the triangle.
Calculate the angle that the (0,0) to (5,3) line makes with the x-axis.
{{{tan(theta)=3/5}}}
{{{theta=0.54042}}} radians
Calculate the angle that the (0,0) to (2,6) line makes with the x-axis.
{{{tan(theta)=6/2}}}
{{{theta=1.24906}}} radians
So then the angle between the (2,6) line and the (5,3) line is
{{{1.24906-0.54042=0.708626}}} radians 
The sine of that angle and the hypotenuse would give us the altitude of the triangle.
The hypotenuse is the distance from (0,0) to (2,6)
{{{H^2=(2-0)^2+(6-0)^2=40}}}
{{{H=sqrt(40)}}}
So then,
{{{A=sqrt(40)*sin(0.708626)}}}
The base of the triangle is the distance from (0,0) to (5,3),
{{{B^2=5^2+3^2=34}}}
{{{B=sqrt(34)}}}
So then the area of the triangle is,
{{{A[t]=(1/2)A*B=(1/2)sqrt(40)sqrt(34)sin(0.708626)}}}
{{{A[t]=(1/2)A*B=(1/2)sqrt(1360)*0.650791}}}
{{{A[t]=12}}}