1. **Problem Statement:** Determine the surface area of the 3-D object made up of two right pyramids with rectangular bases. 2. **Understanding the Object:** The object consists of two right pyramids joined at their rectangular bases. Each pyramid has a rectangular base and triangular faces. 3. **Formula for Surface Area of a Pyramid:** The surface area $SA$ of a right pyramid is given by: $$SA = B + \frac{1}{2} P l$$ where $B$ is the area of the base, $P$ is the perimeter of the base, and $l$ is the slant height of the pyramid. 4. **Calculate Base Area $B$:** The base is a rectangle with sides 6.9 cm and 6.6 cm. $$B = 6.9 \times 6.6 = 45.54 \text{ cm}^2$$ 5. **Calculate Perimeter $P$ of the Base:** $$P = 2(6.9 + 6.6) = 2(13.5) = 27 \text{ cm}$$ 6. **Calculate Slant Height $l$:** The slant height is given as 10 cm. 7. **Calculate Surface Area of One Pyramid:** $$SA_{one} = B + \frac{1}{2} P l = 45.54 + \frac{1}{2} \times 27 \times 10 = 45.54 + 135 = 180.54 \text{ cm}^2$$ 8. **Calculate Total Surface Area of Two Pyramids Joined at Base:** Since the two pyramids share the base, the total surface area is twice the lateral area plus one base area: $$SA_{total} = 2 \times \left( SA_{one} - B \right) + B = 2 \times (180.54 - 45.54) + 45.54 = 2 \times 135 + 45.54 = 270 + 45.54 = 315.54 \text{ cm}^2$$ **Final Answer:** The surface area of the object is $$\boxed{315.54 \text{ cm}^2}$$.