Parallel Lines for $\triangle ABM$

A and B have been marked such that AB = 5 cm. In other words, the base itself becomes 5 cm for all the triangles.

We know $Ar \triangle$ = $\frac{1}{2}$ x Base x Height

Let $\triangle_{h}$ be the sum of areas of all the triangles ABM that have a height of $h$ cm. Given the conditions in the question, 9 types of triangles can be formed, ignoring their sub-cases.

Therefore,

$\triangle_{9}$ = 2 x $\frac{1}{2}$ x 5 x 9

NOTE: The 2 has been multiplied before the expression of a triangle's area to denote that two such triangles are possible. The same logic will apply to the subsequent expressions below in terms of their respective possibilities. As M is fixed, we can simply observe all the possibilities without the need for any combinatorial mathematics.

$\triangle_{8}$ = 4 x $\frac{1}{2}$ x 5 x 8

$\triangle_{7}$ = 6 x $\frac{1}{2}$ x 5 x 7

$\triangle_{6}$ = 8 x $\frac{1}{2}$ x 5 x 6

$\triangle_{5}$ = 10 x $\frac{1}{2}$ x 5 x 5

$\triangle_{4}$ = 12 x $\frac{1}{2}$ x 5 x 4

$\triangle_{3}$ = 14 x $\frac{1}{2}$ x 5 x 3

$\triangle_{2}$ = 16 x $\frac{1}{2}$ x 5 x 2

$\triangle_{1}$ = 18 x $\frac{1}{2}$ x 5 x 1

The sum of all the areas = $\triangle_{9}$ + $\triangle_{8}$ + $\triangle_{7}$ + $\triangle_{6}$ + $\triangle_{5}$ + $\triangle_{4}$ + $\triangle_{3}$ + $\triangle_{2}$ + $\triangle_{1}$

= (2 x $\frac{1}{2}$ x 5)(9 + 16 + 21 + 24 + 25 + 24 + 21 + 16 + 9)

= 5 x 165 = 825 $cm^{2}$