1) The edge length of the bigger cube is 9/8ths that of the smaller cube. To see this, note that the surface area of a cube is just 6s^2, where s is the side length. So, let the edge length of the smaller cube be s and let the edge length of the larger cube be (9/8)s. So the ratio of the surface area of the larger cube to that of the smaller cube is......
[6*[(9/8)(s)]^2] / [6*s^2] = (9/8)^2 = 81/64
2) The ratio of the dimensions of the larger cylinder to the smaller one is just 5/3. Thus, the larger one has a radius 5/3rds times the smaller and a height 5/3rds times the smaller. To see this note that the ratio of their volumes is just....
[pi*[(5/3)r]^2*(5/3)*h] / [pi*r^2*h] = (5/3)^3 = 125/27 .....just as we thought
So...it appears that the ratio of their volumes is just the cube of the scale factor (5/3). And volume is a "cubed" quantity. And since surface area is a "squared" quantity, the ratio of their surface areas is just (5/3)^2 = (25/9).
3)(a) If the scale was 1:40, the actual rocket was 40 times as long as the model...so 40 * 75cm = 3000cm. And since a meter = 100cm, we have 3000cm/100 = 30 m.
3)(b) We can do this one by something called "dimensional analysis"......Note that 1 square meter = (1m) *(1m) = (100cm)*(100cm). Thus, a square meter = 10000cm^2 .......so we have
2750cm^2 * [(1m^2)/10000cm^2] = .275m^2 .....Notice how the cm. just "cancel out" leaving us with m^2 !!!
3)(c) For this one, we just note that 1 cubic meter = (1m) *(1m)*(1m) = (100cm)*(100cm)*(100cm). Thus, a cubic meter = 1000000cm^2 .......so doing the same sort of thing we did on the last one...
96m^3 * [1000000cm^3/1m^3] = 96000000cm^3.....again, note how the meters "cross cancel"
