Surface Area Net 33Eb96

1. **State the problem:** We need to find the surface area of the three-dimensional figure represented by the net. 2. **Analyze the net:** The net consists of: - One long horizontal rectangle (base) - One vertical rectangle centered on the base (height) - Two congruent triangles attached to the top and bottom of the vertical rectangle 3. **Identify dimensions:** Each square on the grid paper is 1 in.$^2$. - Count the length and width of the horizontal rectangle. - Count the length and width of the vertical rectangle. - Count the base and height of the triangles. 4. **Calculate areas:** - Area of horizontal rectangle = length $\times$ width - Area of vertical rectangle = length $\times$ width - Area of one triangle = $\frac{1}{2} \times$ base $\times$ height - Since there are two congruent triangles, total triangle area = 2 $\times$ area of one triangle 5. **Sum all areas:** $$\text{Surface Area} = \text{Area of horizontal rectangle} + \text{Area of vertical rectangle} + 2 \times \text{Area of one triangle}$$ 6. **Example calculation (assuming dimensions from the grid):** - Horizontal rectangle: length = 6 in., width = 2 in. - Vertical rectangle: length = 2 in., width = 2 in. - Triangles: base = 2 in., height = 2 in. Calculate each area: - Horizontal rectangle area = $6 \times 2 = 12$ in.$^2$ - Vertical rectangle area = $2 \times 2 = 4$ in.$^2$ - One triangle area = $\frac{1}{2} \times 2 \times 2 = 2$ in.$^2$ - Two triangles area = $2 \times 2 = 4$ in.$^2$ 7. **Total surface area:** $$12 + 4 + 4 = 20 \text{ in.}^2$$ **Final answer:** $$\boxed{20 \text{ in.}^2}$$