Hi Ken,
I produced a diagram much like the one you sent and then I put some labels on it.
The initial triangle has base of length b and height h so the area is given by
A = ^{1}/_{2} b h
The horizontal line cuts off a certain proportion of the area above the line. I want to call this proportion p. (I would rather work with proportions than percentages.) The horizontal cutoff line is of length b' and the triangle above this line has height h'. Thus the area of the triangle above the cutoff line is given by
pA = ^{1}/_{2} b' h'
Substitute A from the equation above to get
p ^{1}/_{2} b h = ^{1}/_{2} b' h'
Thus
p b h = b' h' or
p = ^{b'}/_{b} ^{h'}/_{h}
The original triangle and the triangle above the cutoff line are similar so
^{b'}/_{b} = ^{h'}/_{h}
and hence
p = ^{h'}/_{h} ^{h'}/_{h} = (^{h'}/_{h})^{2}
Finally
h' = √p h
Thus in your example you want 15% of the area above the cutoff line so p = 0.15 and √0.15 = 0.39. Thus draw the cutoff line so that the height of the triangle above the cutoff line is 39% of the height of the original triangle.
The next cutoff line is to leave 40% of the area of the original triangle above it so here p = 0.40 and √0.40 = 0.63. Thus draw the next cutoff line so that the height of the triangle above this cutoff line is 63% of the height of the original triangle.
Penny
