Question 1115402

The area of a triangle with vertices ({{{x[1]}}},{{{y[1]}}}), ({{{x[2]}}},{{{y[2]}}}), ({{{x[3]}}},{{{y[3]}}}) is:


{{{A = (x[1]*y[2] - x[2]*y[1] + x[2]*y[3] - x[3]*y[2] + x[3]*y[1] - x[1]*y[3]) / 2}}}


In this case, A ({{{2}}},{{{5}}})=({{{x[1]}}},{{{y[1]}}}), B({{{-3}}},{{{10}}})=({{{x[2]}}},{{{y[2]}}}) and C ({{{-5}}},{{{2}}})=({{{x[3]}}},{{{y[3]}}}), we have


{{{A = (2*10- (-3)*5 + (-3)*2 - (-5)*10 + (-5)*5 - 2*2) / 2}}}

{{{A = (20+15 -6 +50 -25 - 4) / 2}}}

{{{A = 50 / 2}}}

{{{A = 25}}}


A similar formula can be used to find the area of any polygon. 

 If you prefer not to use this formula, then another method is to use Distance Formula to find the length of each side:

side {{{AB}}}, {{{AC}}}, and {{{BC}}} 
if
A ({{{2}}},{{{5}}})
B({{{-3}}},{{{10}}})
C ({{{-5}}},{{{2}}})

{{{AB=(x[1]-x[2])^2+(y[1]-y[2])^2}}}   

{{{AB= sqrt(2-(-3))^2+(5-10)^2)}}}   
{{{AB= sqrt(2+3)^2+(-5)^2)}}} 
  {{{AB=sqrt(5^2+(-5)^2)}}}   
  {{{AB=sqrt(25+25)}}}  
{{{AB=sqrt(2*25)}}} 
{{{AB=5sqrt(2)}}} 
{{{AB=7.1 }}} 


{{{AC=(x[1]-x[2])^2+(y[1]-y[2])^2}}}   

{{{AC= sqrt(2-(-5))^2+(5-2)^2)}}}   
{{{AC= sqrt(2+5)^2+3^2)}}} 
  {{{AC=sqrt(7^2+3^2)}}}   
  {{{AC=sqrt(49+9)}}} 


{{{AC=sqrt(58)}}} 
{{{AC=7.6}}} 



 {{{BC=(x[1]-x[2])^2+(y[1]-y[2])^2}}}
 {{{BC= sqrt(-3-(-5))^2+(10-2)^2)}}}
{{{BC= sqrt(2^2+8^2)}}}
{{{BC= sqrt(68)}}}
{{{BC= 8.25}}}




since you have scalene triangle (no equal sides), you can also calculate the area from the lengths of all three sides using Heron's Formula

Calculate "s" (half of the triangles perimeter):

{{{s = (a+b+c)/2}}}
{{{s = (7.1+7.6+8.25)/2}}}
{{{s = 11.475}}}

Then calculate the Area:
{{{A =sqrt(s(s-AB)(s-AC)(s-BC))}}}
{{{A =sqrt(11.475(11.475-7.1)(11.475-7.6)(11.475-8.25))}}}
{{{A =25.0476}}} round it to whole number
{{{A =25}}}