Step 1: Break the figure into three slanted "panels."
The two nearly vertical segments split the long slanted shape into three panels. Each panel is crossed by two parallel slanted lines, creating small triangular pieces at the top and bottom of each panel, and additional triangles when adjacent panels are combined.
Step 2: Count the smallest triangles.
In each of the three panels there are four smallest triangles (two up–pointing near the top edge and two down–pointing near the bottom edge).
\(\Rightarrow\) Smallest triangles \(= 3 \times 4 = 12\).
Step 3: Count medium triangles (formed by merging two adjacent small ones within a panel).
Within each panel, the two small top triangles merge to give one medium triangle, and the two small bottom triangles merge to give another. So each panel contributes \(2\) medium triangles.
\(\Rightarrow\) Medium triangles \(= 3 \times 2 = 6\).
Step 4: Count cross-panel medium triangles (formed by using one side from each of two neighbouring panels).
Across each pair of adjacent panels (left–middle and middle–right), the slanted parallels line up so that we can form one triangle along the top strip and one along the bottom strip. There are two neighbouring pairs.
\(\Rightarrow\) Cross-panel medium triangles \(= 2 \text{ pairs} \times 2 = 4\).
Step 5: Count the largest triangles (spanning the full width).
Using the outer boundary with each of the two slanted internal lines gives two large triangles on the top side and two on the bottom side overall.
\(\Rightarrow\) Largest triangles \(= 2+2 = 4\).
Step 6: Total.
\[
\text{Total triangles} = 12\ (\text{small}) + 6\ (\text{medium in-panel}) + 4\ (\text{cross-panel}) + 4\ (\text{largest}) = \boxed{24}.
\]

Given an open-loop transfer function \(GH = \frac{100}{s}(s+100)\) for a unity feedback system with a unit step input \(r(t)=u(t)\), determine the rise time \(t_r\).
Consider a linear time-invariant system represented by the state-space equation: \[ \dot{x} = \begin{bmatrix} a & b -a & 0 \end{bmatrix} x + \begin{bmatrix} 1 0 \end{bmatrix} u \] The closed-loop poles of the system are located at \(-2 \pm j3\). The value of the parameter \(b\) is: