1. **Problem statement:** Given a pyramid with a rectangular base of sides $a=6$ cm and $b=4$ cm, and a height $h=5$ cm. The apex is directly above the intersection of the diagonals of the rectangle. We need to calculate the surface area $O$ of the pyramid. 2. **Formula and explanation:** The surface area $O$ of a pyramid with a rectangular base is the sum of the base area and the areas of the four triangular faces: $$O = A_{base} + A_{side1} + A_{side2} + A_{side3} + A_{side4}$$ Since the pyramid has two pairs of identical triangular faces (due to the rectangular base), we can write: $$O = ab + 2A_{side_a} + 2A_{side_b}$$ where $A_{side_a}$ and $A_{side_b}$ are the areas of the triangular faces with base $a$ and $b$ respectively. 3. **Calculate the base area:** $$A_{base} = a \times b = 6 \times 4 = 24 \text{ cm}^2$$ 4. **Calculate the side heights:** The apex is above the intersection of the diagonals, so the slant heights $h_a$ and $h_b$ correspond to the heights of the triangular faces on sides $a$ and $b$. The half-lengths of the base sides are: $$\frac{a}{2} = 3 \text{ cm}, \quad \frac{b}{2} = 2 \text{ cm}$$ Using the Pythagorean theorem for the triangular faces: $$h_a = \sqrt{h^2 + \left(\frac{b}{2}\right)^2} = \sqrt{5^2 + 2^2} = \sqrt{25 + 4} = \sqrt{29}$$ $$h_b = \sqrt{h^2 + \left(\frac{a}{2}\right)^2} = \sqrt{5^2 + 3^2} = \sqrt{25 + 9} = \sqrt{34}$$ 5. **Calculate the areas of the triangular faces:** $$A_{side_a} = \frac{1}{2} a h_a = \frac{1}{2} \times 6 \times \sqrt{29} = 3\sqrt{29}$$ $$A_{side_b} = \frac{1}{2} b h_b = \frac{1}{2} \times 4 \times \sqrt{34} = 2\sqrt{34}$$ 6. **Calculate the total surface area:** $$O = 24 + 2 \times 3\sqrt{29} + 2 \times 2\sqrt{34} = 24 + 6\sqrt{29} + 4\sqrt{34}$$ 7. **Approximate numerical value:** $$\sqrt{29} \approx 5.385, \quad \sqrt{34} \approx 5.831$$ $$O \approx 24 + 6 \times 5.385 + 4 \times 5.831 = 24 + 32.31 + 23.32 = 79.63 \text{ cm}^2$$ **Final answer:** $$\boxed{O \approx 79.63 \text{ cm}^2}$$