If an aluminum plate has a rectangular area of 64 square feet and a triangular area of 16 square feet, how much does it weigh, rounded up to the nearest hundred pounds?

To determine the weight of the aluminum plate, we first need to calculate the total area of the plate by adding its rectangular and triangular areas together. The rectangular area is 64 square feet and the triangular area is 16 square feet, giving a combined area of 80 square feet.

Next, we need to consider the density of aluminum, which is approximately 0.1 pounds per cubic inch. To convert the area from square feet to square inches, we multiply by 144 (since there are 144 square inches in a square foot). Thus, the total area in square inches is 80 square feet × 144 square inches/square foot, which equals 11,520 square inches.

However, to compute the weight, we also require a thickness to calculate the volume; typically, a standard thickness is assumed in these types of problems. For instance, if one assumes a thickness of 0.1 inches (a commonly used thickness for sheets), the volume can be calculated by multiplying the area by the thickness, resulting in:

Volume = Area × Thickness = 11,520 square inches × 0.1 inches = 1,152 cubic inches.

To find the weight in pounds, the volume can be multiplied by the density of