In the diagram above, BD = 9, CE = 7, [ABC] = 30, and [ADE] = 20. Find [ACD].
Area( triangle(ABC) ) = 30
Area( triangle(ADE) ) = 20
Area( triangle(ACD) ) = x
Area( triangle(ABD) ) = Area( triangle(ABC) ) + Area( triangle(ACD) )
Area( triangle(ABD) ) = 30 + x
Area( triangle(ACE) ) = Area( triangle(ADE) ) + Area( triangle(ACD) )
Area( triangle(ACE) ) = 20 + x
Using the formaula that Area = ½ · base · height and calling the height h,
---> Area( triangle(ABD) ) = ½ · 9 · h = 30 + x
---> h = (60 + 2x) / 9
---> Area( triangle(ACE) ) = ½ · 7 · h = 20 + x
---> h = (40 + 2x) / 7
Combining these equations: (60 + 2x) / 9 = (40 + 2x) / 7
Solving: 420 + 14x = 360 + 18x
60 = 4x
x = 15
The area of triangle(AD) = 15.