1. Problem statement: We are given an L-shaped polygon with a top horizontal edge of $12$ mm, an interior vertical located $4.3$ mm from the left that drops $2$ mm from the top to an interior joint and continues $9.1$ mm to the bottom, and a slanted segment of length $8$ mm connecting that interior joint to the top-right corner. 2. Coordinate placement: Let the top-left corner be $(0,0)$ and the top-right corner be $(12,0)$. The interior vertical is at $x=4.3$, so the interior joint is at $(4.3,2)$ and the bottom lies at $y=2+9.1=11.1$, giving bottom-left $(0,11.1)$ and interior bottom $(4.3,11.1)$. 3. Consistency check: Compute the slanted length to verify the placement. $$\sqrt{(12-4.3)^2+(0-2)^2}=\sqrt{7.7^2+2^2}=\sqrt{63.29}\approx7.957\approx8$$. 4. Decompose and compute areas: Decompose the shape into a rectangle (left part) and a right triangle (top-right notch). The rectangle has width $4.3$ and height $11.1$, so its area is $4.3\times11.1=47.73$. The triangle has base $12-4.3=7.7$ and height $2$, so its area is $\frac{1}{2}\times7.7\times2=7.7$. Therefore $$A=4.3\times11.1+\frac{1}{2}\times7.7\times2=47.73+7.7=55.43$$. 5. Final answer: The area of the shape is $55.43\,\text{mm}^2$.