+0

# Amount of Air Method?

0
777
6
+279
 A design on the surface of a balloon is 5 cm wide when the balloon holds 71 cm of air. How much air does the balloon hold when the design is 10 cm wide? Explain the method you use to find the amount of air.

Thank You All For All the Help! Couldnt do It Without You! :D

Jun 25, 2014

#2
+101431
+10

We have

V = (4/3)*pi*r^3

71 = (4/3) *pi * r^3    Solving for r , we have

71(3/4)/pi  = r^3         Taking the cube root of both sides, we have

r = 2.5687583235794156      And if the design is now 10cm wide, the dimensions of the balloon must have doubled....so doubling the radius, we have  2 * 2.5687583235794156 = 5.1375166471588312

So

V = (4/3)*pi * (5.1375166471588312)*3   = 567.99999cm^3 or  568cm^3

Obviously, Melody's path to the answer is more elegant than mine !!!

Jun 25, 2014

#1
+101771
+10

I think if a length dimension is 2*original then the volume is 2cubed=8 times original.

$$71*8=568cm^3$$

I'm another mathematician will look at my answer.

Jun 25, 2014
#2
+101431
+10

We have

V = (4/3)*pi*r^3

71 = (4/3) *pi * r^3    Solving for r , we have

71(3/4)/pi  = r^3         Taking the cube root of both sides, we have

r = 2.5687583235794156      And if the design is now 10cm wide, the dimensions of the balloon must have doubled....so doubling the radius, we have  2 * 2.5687583235794156 = 5.1375166471588312

So

V = (4/3)*pi * (5.1375166471588312)*3   = 567.99999cm^3 or  568cm^3

Obviously, Melody's path to the answer is more elegant than mine !!!

CPhill Jun 25, 2014
#3
+101771
+5

I learned that shortcut of Alan.  Nifty isn't it.

Thanks Alan   ♬

Jun 25, 2014
#4
+101431
0

Yeah...I knew that "trick," too.....I just wanted to try a different approach..but yours is definitely preferable !!!   Just like the balloon......the amount of "hot air" on the forum tends to expand in a "cubic" fashion.....sometimes......

Jun 25, 2014
#5
+279
+5

Thanks For Help Melody and CPhill!

Jun 25, 2014
#6
+22358
+6

$$V_1=\dfrac{4}{3}\pi r_1^3$$

$$V_2=\dfrac{4}{3}\pi r_2^3 \quad | \quad r_2=2r_1$$

$$V_2=\dfrac{4}{3}\pi (2r_1)^3= \underbrace{\dfrac{4}{3}\pi r_1^3}_{V_1}*\underbrace{2^3}_8$$

$$V_2=8V_1=8*71\;cm^3=568\;cm^3$$

.
Jun 26, 2014