WAEC Mathematics 2010 Paper

Concept:Total surface area of the net equals the sum of the areas of all its faces: the rectangular base plus the four triangular faces.Explanation:The rectangular base has length $8 \, \text{cm}$ and width $5 \, \text{cm}$, so its area is $8 \times 5 = 40 \, \text{cm}^2$.There are two identical triangles with base $8 \, \text{cm}$ and height $4 \, \text{cm}$.Area of one such triangle is $\frac{1}{2} \times 8 \times 4 = 16 \, \text{cm}^2$.Contribution from these two triangles is $2 \times 16 = 32 \, \text{cm}^2$.There are also two identical triangles with base $5 \, \text{cm}$ and height $4 \, \text{cm}$.Area of one such triangle is $\frac{1}{2} \times 5 \times 4 = 10 \, \text{cm}^2$.Contribution from these two triangles is $2 \times 10 = 20 \, \text{cm}^2$.Total surface area is $40 + 32 + 20 = 92 \, \text{cm}^2$.Answer:$92 \, \text{cm}^2$Option C.