Trapezium Artists

We can make a valid configuration with any odd number of equilateral triangles by extending this pattern:

We can get any multiple of four with this pattern:

In general, note that the number of triangles in one of these shapes is the difference of the squares of the lengths of the parallel sides; for instance, $21=5^2-2^2$ and $16=5^2-3^2$ .

All square numbers are either of the form $4k$ or $4k+1$ ; so the difference of two squares can never be of the form $4k+2$ .

So the areas we can't make are $102,106,110,\ldots,198$ , whose total (found by summing the arithmetic progression) is $\boxed{3750}$ .

Just to verify the claim that the number of triangles in any trapezium is the difference of the squares of the lengths of the trapezium’s bases, we add

$\begin{aligned} & n+2\left( n-1 \right)+2\left( n-2 \right)+\ldots +2\left( n-k \right)+\left( n-\left( k+1 \right) \right) \\ & =\underbrace{2n+2n+\ldots +2n}_{k+1\text{ terms}}-2\left( 1+2+\ldots +k \right)-\left( k+1 \right) \\ & =2n\left( k+1 \right)-2\dfrac{k\left( k+1 \right)}{2}-\left( k+1 \right) \\ & =\left( k+1 \right)\left( 2n-k-1 \right) \\ & =\left[ n-\left( n-\left( k+1 \right) \right) \right]\left[ n+\left( n-\left( k+1 \right) \right) \right] \\ & ={{n}^{2}}-{{\left( n-\left( k+1 \right) \right)}^{2}} \\ \end{aligned}$

Obvious restrictions for integers $n$ and $k$ : $n\ge 2$ , $0\le k\le n-2$ .

The rest is best explained by @Chris Lewis .

The possible trapeziums have areas of the difference between two perfect squares (because the nth layer counting from the top of this triangle corresponds to the nth triangular number, and the first n of them sum up to n²), so the impossible areas should be even numbers between 100 and 200 that are not divisible by 4.

Sum of impossible areas
= (100 / 4) × (102 + 198) / 2
= 3750