Question 148398
If "ACB" forms a right triangle, you should have the following sides:
"AC"= opposite side
"CB"= adjacent side
"AB"= hypotenuse side
Now, there's a point in line "AC" called "D" and forms an <font><font size=4><b>obtuse triangle</font> "ADB"-----> unknown Area?
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<font><font size=4><b>Area=(1/2)*base*height</font> -----> working equation
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We have to remember that the "base" is always perpendicular to the "height" for the equation to work. The "base" is given which is "AD"= 6, but we can't use "DB" as height bec. is not perpendicular to "AD" right? 
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If you extend line AD to C meets line "BC" and forms 90 degree. Very good! As it is given, "BC"= 8 and we'll be the height of triangle "ADB".
Going back to our working eq'n,
{{{A=(1/2)base*height=(1/2)(AD)(BC)}}}
{{{A=(1/2)(6)(8)}}}
{{{A=24 square units}}}
<font><font size=4>Thank you,
Jojo</font>