Question 980471
Your wording is not right.  You are suggesting to draw a perpendicular segment from the 20 cm side to meet the 18 cm side, and this perpendicular segment must cut the TRIANGLE into two equal areas.  One of these pieces will be a triangle and the other piece will be a trapezoid.


An objective is to find the actual altitude of the original triangle, using 20 cm as the base, and you may pick alpha as the angle between the 16 cm and the 20 cm sides.  I skip some of the steps here, but {{{cos(alpha)=83/144}}}, and Law Of Cosines was used.


The Pythagorean trigonometry identify for unit circle lets us do this:
{{{cos^2(alpha)+sin^2(alpha)=1}}}
{{{sin(alpha)=sqrt(1-cos^2(alpha))}}}
{{{sin(alpha)=sqrt(1-(83/144)^2)}}}
{{{highlight_green(sin(alpha)=sqrt(13847/20736))}}} not simplifiable; you might want to use scientific or graphing calculator or table of trig values.


If let a=altitude from "vertex" to the base,
then 
{{{a=16sin(alpha)}}}
{{{highlight_green(a=16*sqrt(13847/20736))}}}, and you use this for finding the area of the original triangle.


If let A be area for the original uncut triangle, then  {{{A=(1/2)20*(16*sqrt(13847/20736))}}}


---not finished---