Place the correct answer in each blank. Use pi = 3.14 and, if necessary, round your answers to the nearest hundredth. The area of the largest cross section of a sphere and the circumference of the sphere are in the ratio 4 : 1. The radius of the sphere is ______ cm. The circumference of the sphere is about ______cm. The area of the largest cross section is about ______cm^2. The volume of the sphere is about _____cm^3.
I'm gonna have to mess with this one for a bit. I will return soon.
alrighty, thanks
Are you given any other information?
Nope.
I'm here, I just had to do a good deal of thinking. I think I have the solution, but It's gonna take a bit to explain.
Ahhh, I dont have much time. Okay lets try
I'll do my best to make this swift. We know the two formulas for circumference and area of a circle are \[C = 2 \pi r\] and \[A = \pi r^2\]
And we know that the ratio given is 4:1, with C first and A second, so I restate the formulas as this: \[4(\pi r^2)=1(2 \pi r)\]
Understand so far?
I think so yeah
Ok, so I simplified the equation as follows: \[4(\pi r^2)=1(2 \pi r)\] \[2(\pi r^2)=\pi r\] \[2(r)=1\] \[r= 1/2\]
Do any of my steps confuse you?
No, I think I get it, I'm scrambling to get everything haha
I know it's a lot, so I appreciate your patience. Now that we've established the radius, we can find every other value. Do you know the formulas for C, A, and V?
\[C = 2\pi \frac{ 1 }{ 2 }\]
A= \[\pi \frac{ 1 }{ 2 }^2\]
You need not do the fancy formatting, so long as you can get the answers for me to confirm. :) Just try to solve the equations and give me your answers.
alright
V= 4/3 pi 1/2 ^2 ?
The power on r should be 3, not two.
oops, sorry V = 4/3 pi 1/2^3
Yep, now solve. (Then state your answers for Area and circumference too)
C = pi A= 0.78539816 (calculator) V= 0.523759877
I'm probably wrong
All your numbers agree with mine! Just one very important note: If this gets turned in on paper, it would be in your best interest to write down all the steps it took to solve this. An instructor can help you verify or otherwise answer this.
Alrighty, will do. Thank you
Happy to help!
If you have more time, then feel free to ask more.
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