Counting the complex triangle structure yields 21 triangles.

There are 28 triangles in the given figure.

Number of triangles in $\triangle ABC$ = (1+2+3+4) = 10. Number of triangles in $\triangle ADE$ = (1+2+3+4) = 10. Here, no new triangle is formed by connecting the two figures. Total triangles = 10+10 = 20

Dividing the figure into segments and counting the triangles yields a total of 11 triangles.

The total number of triangles in the figure is 17.

Number of triangles in $\triangle ABC$ = (1+2+3+4) = 10. Number of triangles in $\triangle DBC$ = (1+2+3+4) = 10. Here, no new triangle is formed by connecting the two figures. Hence, total triangles = 10+10 = 20

A rectangle has $n$ = 4 sides. The number of diagonals = $\frac{n(n-3)}{2} = \frac{4(4-3)}{2} = \frac{4}{2} = 2$.

Total rectangles = (Total rectangles in the original figure) + (other rectangles) = 9+12=21.

This is a rectangle like a $3 \times 2$ grid. To find the total number of rectangles, we use combinations of rows and columns. In total, there are 9 rectangles.

Here the figure given has eight different triangles. Hence, number of total triangles = 8×2=16

The figure is in the shape of a cross or 'plus' with internal lines. If we count each part systematically, it forms 16 triangles.

Counting all possible combinations in the given complex figure results in a total of 30 triangles.

The given figure has 3 horizontal lines along the base. Using the standard formula, the total number of triangles is 20.

Counting all the small and large combined triangles in the given complex figure results in a total of 70 triangles.

The figure contains two main triangles $\triangle ABD$ and $\triangle BCD$. In $\triangle ABD$, there are $1+2=3$ triangles. In $\triangle BCD$, there are $1+2=3$ triangles. Connecting these two forms one larger $\triangle ABC$. Hence, total triangles = $3+3+1=7$.

The total number of squares in the given figure is 12.

The total number of triangles formed by the diagonals and segments within the square is 16.

Here, we will get (5×5)= 25 squares from such five figures '田'. But while counting these squares we have included some squares two times. In this figure, there are two shaded squares which are common and have been counted twice. Therefore, we have to subtract such squares one time. Hence, total squares = 25-4 = 21.

Counting the small triangles within the main triangle results in 15 triangles.

Total squares = (64)+(53)+(42)+(31) = 24+15+8+3 = 50