1. **State the problem:** Find the surface area of a triangular prism given its net. 2. **Identify the shapes in the net:** The net consists of three rectangles and two triangles. - The central rectangle measures 8 in by 5 in. - Two triangles are attached to the longer side (8 in) of the central rectangle. - Each triangle has a base of 3 in and height of 4 in. 3. **Formula for surface area of a triangular prism:** $$\text{Surface Area} = \text{Perimeter of triangular base} \times \text{length} + 2 \times \text{Area of triangular base}$$ 4. **Calculate the area of one triangle:** $$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3 \times 4 = 6 \text{ in}^2$$ 5. **Find the length of the triangular base sides:** - Base side is 3 in. - Height is 4 in. - The other two sides are equal (isosceles triangle), calculate using Pythagoras: $$\text{side} = \sqrt{4^2 + \left(\frac{3}{2}\right)^2} = \sqrt{16 + 2.25} = \sqrt{18.25}$$ 6. **Calculate the perimeter of the triangular base:** $$P = 3 + 2 \times \sqrt{18.25}$$ 7. **Calculate the surface area:** - Length of prism is 5 in (height of central rectangle). - Surface area of rectangular sides: $$P \times 5 = \left(3 + 2 \times \sqrt{18.25}\right) \times 5$$ - Surface area of two triangular bases: $$2 \times 6 = 12$$ 8. **Combine all parts:** $$\text{Surface Area} = 5 \times \left(3 + 2 \times \sqrt{18.25}\right) + 12$$ 9. **Simplify:** $$= 5 \times 3 + 5 \times 2 \times \sqrt{18.25} + 12 = 15 + 10 \times \sqrt{18.25} + 12 = 27 + 10 \times \sqrt{18.25}$$ 10. **Approximate:** $$\sqrt{18.25} \approx 4.27$$ $$\text{Surface Area} \approx 27 + 10 \times 4.27 = 27 + 42.7 = 69.7 \text{ in}^2$$ **Final answer:** $$\boxed{69.7 \text{ square inches}}$$