Question 203869
<font face="Tahoma" size=1.5>
</pre><font size=4><b>
To begin with, let us see the triangle,
{{{drawing(400,400,-10,10,-5,6,grid(1),graph(400,400,-10,10,-5,6),triangle(4,-2,-6,4,8,-2), red(locate(3.5,-2.15,A)),red(locate(-6.5,4.55,B)),red(locate(8.2,-2.15,C)))}}}---> Obtuse Triangle


We know finding Area for Triangle, {{{A=(1/2)bh}}}, Working Eqn.


In the given Obtuse Triangle, the perpendicular Height lies outside.
{{{drawing(400,400,-10,10,-5,6,graph(400,400,-10,10,-5,6),triangle(4,-2,-6,4,8,-2), red(locate(3.5,-2.15,A)),red(locate(-6.5,4.55,B)),red(locate(8.2,-2.15,C)),green(line(-6,-2,-6,4)),green(locate(-6.5,-2.15,D)),red(locate(-8.5,2,height)),green(line(-6,-2,4,-2)),red(line(4,-2,8,-2)),red(locate(5.5,-2.15,base)),red(locate(-5.7,-1.2,90^o)))}}}


To find <font color=red>height</font>, we take distance between points "B" & "D"
by taking their coordinates.


Applying Distance Formula = {{{h=sqrt((x[2]-x[1])^2+(y[2]-y[1])^2))}}}
{{{h=sqrt((-6-(-6))^2+(-2-4)^2)}}}
{{{h=sqrt((-6+6)^2+(-6)^2)=sqrt(0+36)=sqrt(36)}}}
{{{red(h=6units)}}}


Finding for the <font color=red>base</font>, we use the same Distance Formula between points "A" & "C",
{{{b=sqrt((8-4)^2+(-2-(-2))^2)}}}
{{{b=sqrt(4^2+0^2)=sqrt(16)}}}
{{{red(b=4units)}}}


Therefore, going back to our Working Eqn,
{{{A=(1/2)bh=(1/2)(4)(6)=24/2}}}
{{{red(A=12square_units)}}}, Answer


Thank you,
Jojo</font>