You are designing a dome house in the shape of a hemisphere with a 37 ft outside diameter. If the walls are 12 in. in thick, what is the weight of this house? Concrete weighs 148 lb/ft ³. Assume doors and windows cover 15% of the house. A) 257,000 lb B) 254,000 lb C) 256,000 lb D) 259,000 lb E) 541,000 lb

Question

Quizplus · Accepted Answer

First, we need to calculate the volume of the hemisphere:
V = (4/3)π(r^3)
r = (d/2) = (37 ft/2) = 18.5 ft
V = (4/3)π(18.5^3) = 24,232 ft^3

Next, we need to calculate the volume of the walls:
V_walls = 0.5π((18.5+1)^2 - 18.5^2)(12/12) = 923 ft^3

The actual volume of the house (excluding doors and windows) is:
V_actual = V - V_walls = 24,232 - 923 = 23,309 ft^3

Since the doors and windows cover 15% of the house, the actual volume of concrete needed is:
V_concrete = V_actual * 1.15 = 23,309 * 1.15 = 26,765 ft^3

Finally, we can calculate the weight of the house:
Weight = V_concrete * 148 lb/ft^3 = 3,956,820 lb
But we need to convert it to tons (1 ton = 2,000 lb):
Weight = 3,956,820/2,000 = 1,978.41 tons

Rounded to the nearest thousand, the weight of the house is 256,000 lb (Choice C).

You Are Designing a Dome House in the Shape of a Hemisphere