Question 1012945
The triangle before making the cut:
Altitude 40, unknown base b, area 1200.
{{{(1/2)b*40=1200}}}
{{{highlight(b=60)}}}


Same or congruent triangle, now draw a segment through the altitude, this segment parallel to the b=60 base.  The upper region is another isosceles triangle with altitude h and base x.  The lower region is trapezoid with altitude a and TWO bases are x and 60.
-
{{{system(h+a=40,(1/2)x*h=600,((x+60)/2)*a=600)}}}
That is according to the description of the two regions being of equal areas, and the two heights being the original given triangle height  (altitudes).
-
Note this is a system of three equations in three unknown variables.


{{{system(h+a=40,hx=1200,xa+60a=1200)}}}


Use first equation as h=40-a, and substitute into the other two equations, 
{{{system(40-ax=1200,ax+60a=1200)}}}
and observing that both of these have a term, {{{ax}}};
{{{ax=1200-60a}}}
SUBSTITUTE into ...
{{{40-(1200-60a)=1200}}}
doing the steps,
..
..
{{{highlight(a=39&1/3)}}}---height for the trapezoid.


Still want to find h and for this, return to  {{{h+a=40}}}.
{{{highlight(h=2/3)}}}.--------height for the triangle.