1. Statement of the problem: Find the area of the polygon composed of a right triangle with base 8 km and height 5 km attached to an orthogonal region with labeled sides as described. 2. Formula and important rules: For a rectangle use $$A_{\text{rect}}=w\times h$$ For a triangle use $$A_{\triangle}=\frac{1}{2}bh$$ Important rule: decompose the figure into non-overlapping simple shapes, add their areas, and subtract any cutouts. 3. Decompose the figure into simpler parts and list dimensions: - Main rectangle: width 8 km and height 6 km, so area $$A_1=8\times6$$. - Right-hand rectangular extension: width 2 km and height 7 km, so area $$A_2=2\times7$$. - Small cutout (lower-right) rectangle: width 4 km and height 4 km, so area $$A_3=4\times4$$. - Right triangle attached: base 8 km and height 5 km, so area $$A_{\triangle}=\frac{1}{2}\times8\times5$$. 4. Compute each area (showing intermediate work): Compute main rectangle: $$A_1=8\times6=48$$ Compute extension rectangle: $$A_2=2\times7=14$$ Compute cutout rectangle: $$A_3=4\times4=16$$ Compute triangle area with cancellation when simplifying the fraction: $$A_{\triangle}=\frac{1}{2}\times8\times5$$ $$=\frac{1\times\cancel{8}}{\cancel{2}}\times5=4\times5=20$$ 5. Add areas of pieces and subtract the cutout: $$A= A_1 + A_2 - A_3 + A_{\triangle}$$ $$A=48 + 14 - 16 + 20 = 66$$ 6. Final answer: $$A=66\ \text{km}^2$$