Question 837626
These are the first ideas that came to my mind.
The possibilities are probably endless, but it is hard enough to draw these three.
 
Six congruent triangles:
{{{drawing(400,300,-0.4,8.4,-0.3,6.3,
rectangle(0,0,8,6),line(0,2,8,2),
line(0,4,8,4),line(0,0,8,2),
line(0,4,8,2),line(0,4,8,6),locate(3.8,6,8m),
locate(0.1,1.2,2m),locate(0.1,3.2,2m),locate(0.1,5.2,2m),
locate(7.5,1.2,2m),locate(7.5,3.2,2m),locate(7.5,5.2,2m)
)}}} Six right triangles with legs measuring 2m and 8m.
 
Six congruent trapezoids:
{{{drawing(400,300,-0.4,8.4,-0.3,6.3,
rectangle(0,0,8,6),line(0,2,8,2),
line(0,4,8,4),line(3,4,5,6),
line(3,4,5,2),line(3,0,5,2),
locate(2.3,6,5m),locate(6.3,6,3m),
locate(1.3,4,3m),locate(5.3,4,5m),
locate(2.3,2,5m),locate(6.3,2,3m),
locate(1.3,0.4,3m),locate(5.3,0.4,5m),
locate(0.1,1.2,2m),locate(0.1,3.2,2m),locate(0.1,5.2,2m),
locate(7.5,1.2,2m),locate(7.5,3.2,2m),locate(7.5,5.2,2m)
)}}} Six trapezoids with bases measuring 5m and 3m, and a 2m side perpendicular to the bases, for a 2m height.
 
Six triangles with equal area but none congruent:
(I would draw this on grid paper. I will draw a grid of 1m by 1m squares).
{{{drawing(400,300,-0.4,8.4,-0.3,6.3,
rectangle(0,0,8,6),line(0,2,8,2),
line(0,4,8,4),line(0,1,8,1),
line(0,3,8,3),line(0,5,8,5),
line(1,0,1,6),line(2,0,2,6),
line(3,0,3,6),line(4,0,4,6),
line(5,0,5,6),line(6,0,6,6),
line(7,0,7,6),red(line(0,6,8,2)),
red(line(4,4,8,6)),red(line(8,0,0,6)),
red(line(0,2,8,0)),red(line(0,2,4,3)),
locate(4.1,5,red(A)),locate(3.1,0.7,red(B)),locate(6.1,2.5,red(C)),
locate(6.1,4,red(D)),locate(1.3,3.7,red(E)),locate(4.1,2,red(F))
)}}} The red lines divide the whole 8m by 6m rectangle (area=(8m)(6m)=48 square meters) into 6 triangles, each with an area of 8 square meters.
Triangle A is an isosceles triangle with an 8m base and a 2m height, not congruent to any of the other triangles.
You could say that non-congruent triangles B and C have the same base and height (a 2m base and an 8m height). They are obviously not congruent to any of the other triangles.
Triangle D is an isosceles triangle with base and height measuring 4 m.
Triangle E also has a base and a height measuring 4m,but it is not isosceles, and not congruent to any other triangle.
All those triangles have areas that can be calculated as {{{base*height/2}}} , and each has an area of 8 square meters:
{{{8*2/2=8}}}
{{{2*8/2=8}}}
{{{4*4/2=8}}}
Their areas add up to {{{5*8}}}square meters ={{{40}}}square meters.
The remaining triangle (triangle F) must have an area of
48 square meters - 40 square meters = 8 square meters.
If the class assigned triangle F is not convinced enough, you could tell them that E and F together form a triangle with a 4m base and an 8m height, and that E+F triangle is divided into two equal parts by a median, the line segment between triangles E and F.
If they are still not convinced, they can use Heron's formula to calculate the area, using the lengths of the sides:
5m (for the side that is half a diagonal of the rectangle), {{{sqrt(17)}}} meters, and {{{sqrt(17)}}} meters.
In that case, I would use the version that calculates the area as
{{{(1/4)*sqrt(4a^2b^2-(a^2+b^2-c^2)^2)}}} .
(The letters {{{a}}} , {{{b}}} and {{{c}}} are the side lengths).