Measurement problem (1)

Here are my solutions:

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"

Hi Jedithious,

It is always great to see you in the forum!  

1) The ratio of the surface areas of two cubes is 64:81. What is the ratio of the edges of the cubes?

the answer will just be sqrt(64):sqrt(81) = 8:9

Now I will show you why

$$\dfrac{6l_1^2}{6l_2^2}=\frac{64}{81}\qquad so \qquad \dfrac{l_1^2}{l_2^2}=\frac{64}{81}\qquad so \qquad \dfrac{l_1}{l_2}=\frac{8}{9}$$

2) If the ratio of the volumes of two similar cylinders is 125:27, what is the ratio of their surface areas?

volume is U3,  Surface area is U2

so ratio of surface areas will be  125^(2/3):27^(2/3) = 25:9

3) A model of a rocket was built to a scale of 1:40. 

a) The length of the model was 75 cm. What was the length of the actual rocket, in metres?

75*40=3000cm=30m

b) The area of metal used to cover the curved surfaces of the model was 2750 cm². What area of metal covered the curved suraces of the actual rocket, in square metres?

1:1600 = 2750:2750*1600=4,400,000cm²=440m²     (Little edit here)

c) The nose cone of the rocket had a volume of 96 m³. What was the volume of the nose cone of the model, in cubic centimetres?

1:40³ = x:96  

$$\frac{1}{64000}=\frac{x}{96}\\\\
96\times \frac{1}{64000}=x\\\\
x=1.5\times 10^{-3}\quad m^3$$

100cm in a metre

100³cm³ in a m³

$$x=1.5\times 10^{-3}\times 10^6 = 1.5\times 10^{3} = 1500cm^3$$

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NOW I have taken shourt cuts (maybe I am not allowed to) and my answers are different from CPhill if I get time I will check the long way but I think one of the other mathematicians will validate one of our answers before then. 

Okay - I made a really stupid mistake on one - I will fix that now - and the rest of mine are okay.

Thank you Alan.

I'm an idiot, Jedithious,  because I CAN'T READ !!!   Ignore my stupid answrs to 3b  and 3c !!!!

Thank-you, CPhil, Melody, and Alan for taking the time to help out! As always, much appreciated. 

Thank you Jedithious.

We always appreciate your excellent manners too!