An image of an ice cream cone is shown below. The slant height of the cone itself is 4 inches, and the radius of the ice cream scoop on top is 0.75 inches. What is the approximate surface area of the entire ice cream cone with the scoop of ice cream? http://learn.flvs.net/webdav/assessment_images/educator_geometry_v14/pool_Geom_3641_0805_09/image0024e848b52.gif
Am I doing this right?? \[\pi(0.75)(4)+\pi(0.75)^{2}\]
Well Done.. Yes you are right.
Now this is where I'm stuck.. how do i solve this in terms of pi?
\[Surface \; Area (Cone) = \pi r l + \pi r^2\]
You can use: \[\pi = 3.14\]
If this is right here is where I'm stuck \[\pi(3)+\pi(0.5625)\]
Now factor out \(\pi\)..
\[\pi(3)+\pi(0.5625) \implies \pi(3 + 0.5625)\] Add the brackets first..
Okay so my answer would round up to be \[(3.6)\pi\]
Is that the whole cone plus the scoop of ice cream on top?
If you want to leave it in \(\pi\) then yes your answer is right..
No I think..
Above is semi sphere.. You have to include surface area for that too I think..
Yeah. So use surface area of a hemisphere to find that and add them together?
\[Surface \; Area(Sphere) = 4 \pi r^2\] So its half will be: \[SA = 2 \pi r^2\]
Do I still use 0.75 as my radius?
Yes..
4.7π
Total ??
Yes..
for surface area of cone, you don't need the pi*r^2 because it has the scoop of ice cream so the bottom of cone is not included
I agree..
So I get \[(1.125) \pi \] and I add that to \[(3.6)\pi \] and my answer is 4.725 or \[(4.7)\pi\]
They want the surface area of the cone and the scoop of ice cream on top.
Just subtract \(0.5625 \pi\) from the final answer..
Because in this case Surface Area will be: \[SA = \pi rl + 2 \pi r^2\]
\[SA = 3 \pi + 1.125 \pi \implies SA = 4.125 \pi\]
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