1. **State the problem:** Find the surface area of a rectangular pyramid with length $l=8$, width $w=4$, and height $h=2$. The surface area includes the base area plus the area of the four triangular faces. 2. **Formula:** The surface area $SA$ of a rectangular pyramid is given by: $$SA = lw + l s_l + w s_w$$ where $lw$ is the base area, $s_l$ is the slant height along the length side, and $s_w$ is the slant height along the width side. 3. **Find slant heights:** The slant heights are found using the Pythagorean theorem: $$s_l = \sqrt{\left(\frac{w}{2}\right)^2 + h^2} = \sqrt{\left(\frac{4}{2}\right)^2 + 2^2} = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2} \approx 2.83$$ $$s_w = \sqrt{\left(\frac{l}{2}\right)^2 + h^2} = \sqrt{\left(\frac{8}{2}\right)^2 + 2^2} = \sqrt{4^2 + 2^2} = \sqrt{20} = 2\sqrt{5} \approx 4.47$$ 4. **Calculate surface area:** $$SA = lw + l s_l + w s_w = 8 \times 4 + 8 \times 2.83 + 4 \times 4.47$$ $$SA = 32 + 22.64 + 17.88 = 72.52$$ 5. **Final answer:** The surface area of the rectangular pyramid is approximately **72.52** square units, rounded to the nearest hundredth.