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anonymous
 5 years ago
You and your friends decide to make a scale model of the water tower in your town. your scale factor is 1:48. the top of the water tower has a diameter of 20feet. the surface area of the top is 314ft^2. You decide to make the top of the water tower with silver foil. How many square inches of foil will you need?? The height of the actual water tower is 32 feet. What is the surface area of your scale model? Do not include the bottom base. Find the volume of the actual water tower? Use your result of the volume of the actual water tower, to find the volume of the scale model.
anonymous
 5 years ago
You and your friends decide to make a scale model of the water tower in your town. your scale factor is 1:48. the top of the water tower has a diameter of 20feet. the surface area of the top is 314ft^2. You decide to make the top of the water tower with silver foil. How many square inches of foil will you need?? The height of the actual water tower is 32 feet. What is the surface area of your scale model? Do not include the bottom base. Find the volume of the actual water tower? Use your result of the volume of the actual water tower, to find the volume of the scale model.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This look like an example of a problem with "false context". What shape is the water tower assumed to have?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There's a couple of problems to solve here. One thing to keep in mind is that "scale factor" applies to dimensions, not to area/volume. So the process here would be: 1) Given the area or volume of the actual water tower find the radius. 2) Scale height/radius by the given factor 3) Calculate the areas using newly obtained dimensions. You'll have to pay attention to conversion between square feet to square inches too.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0would this be right.. 1 ft = 12 in, so 20 ft = 240 in Radius = diameter / 2 = 240 in / 2 = 120in for the real deal. The model has a radius of 120 in • 1/ 48 = 120in / 48 = 2.5 in A (base) = (2.5)² • π A = 6.25π A ≈ 19.635 in² I need approx 19.635 in² of foil … 32 ft = 384 in 384 in / 48 = 8 in SA = 2π r² + 2π rh = 2πr(r + h) SA = 2 π (2.5) (2.5 + 8) SA = 2 π (2.5) (10.5) SA = 5 π (10.5) SA = 52.5 π SA ≈ 164.93in² We do not want to bottom base, so 164.93in²  19.635 in² = 145.295 in² The surface area is approx. 145.295in². … V = (πr²)h V = (19.635 in² )8 in V ≈ 157.01 in² The volume is approx. 157.01 in²
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