+0  
 
0
716
7
avatar+154 

Hi, 

 

Here are a few measurement problems I would appreciate some help with:

 

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

 

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

 

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?

b) The area of metal used to cover the curved surfaces of the model was 2750 cm^2. What area of metal covered the curved suraces of the actual rocket, in square metres?

c) The nose cone of the rocket had a volume of 96 m^3. What was the volume of the nose cone of the model, in cubic centimetres?

 

Thanks!

Jedithious  Jun 15, 2014

Best Answer 

 #3
avatar+27228 
+10

Here are my solutions:

ratios 

Alan  Jun 15, 2014
 #1
avatar+92674 
+10

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"

 

CPhill  Jun 15, 2014
 #2
avatar+94105 
+10

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

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

-------------------------------------------------------------

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. 

Melody  Jun 15, 2014
 #3
avatar+27228 
+10
Best Answer

Here are my solutions:

ratios 

Alan  Jun 15, 2014
 #4
avatar+94105 
0

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

Thank you Alan.

Melody  Jun 15, 2014
 #5
avatar+92674 
0

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

 

CPhill  Jun 16, 2014
 #6
avatar+154 
+5

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

Jedithious  Jun 17, 2014
 #7
avatar+94105 
0

Thank you Jedithious.

We always appreciate your excellent manners too!

Melody  Jun 17, 2014

6 Online Users

avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.