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Mathematics 24 Online
OpenStudy (gabylovesyou):

@ganeshie8 Challenge #2 — Tennis Trouble: Now it's time to move on to the second challenge called Tennis Trouble. The second challenge is to figure out how many tennis balls fit in a specially designed container. The radio station will give away a prize pack and a pair of front row tickets to the winner of this challenge! Each tennis ball is 2.63 inches in diameter. A sketch of the specially designed container is below. How many tennis balls can fit inside the container? How many more tennis balls could fit into the container if the container’s dimensions are doubled?

OpenStudy (gabylovesyou):

|dw:1409851763186:dw|

ganeshie8 (ganeshie8):

find the volume of tennis ball find the volume of container divide

ganeshie8 (ganeshie8):

`Each tennis ball is 2.63 inches in diameter.` Diameter = 2.63, that means radius = ?

OpenStudy (gabylovesyou):

1.315

ganeshie8 (ganeshie8):

yes, whats the volume of a sphere of radius 1.315 ?

OpenStudy (gabylovesyou):

Volume of sphere = V = 4/3 * pi * r^3 so V = 4/3 * 3.14 * 1.315^3 = 9.52019

ganeshie8 (ganeshie8):

save it, find the volume of given container

ganeshie8 (ganeshie8):

Notice that its a composition of two shapes : 1) cylinder in the bottom 2) cone in the top

ganeshie8 (ganeshie8):

find the volume of each shape and add up

OpenStudy (gabylovesyou):

its a cylinder so v = pi * r^2 * h v = 3.14 * 6.5 * 36 correct ?

ganeshie8 (ganeshie8):

Correct ! find the volume of top cone also

OpenStudy (gabylovesyou):

volume of cylinder is 734.76

OpenStudy (gabylovesyou):

Volume of cone V = 1 over 3 (πr2)(h)

OpenStudy (gabylovesyou):

oh wait i did it wrong

OpenStudy (gabylovesyou):

for the cylinder the answer is 4775.94 right ?

ganeshie8 (ganeshie8):

volume of cylinder is 734.76 is correct

ganeshie8 (ganeshie8):

what are the dimensions of cone ? radius = ? height = ?

OpenStudy (gabylovesyou):

wait but i didnt do the squared part..

OpenStudy (gabylovesyou):

6.5 ^ 2

ganeshie8 (ganeshie8):

Oops! yes

ganeshie8 (ganeshie8):

4775.94 is correct volume for bottom cylinder http://www.wolframalpha.com/input/?i=3.14*6.5%5E2*36

OpenStudy (gabylovesyou):

radius of cone is 6.5 and height is 8

ganeshie8 (ganeshie8):

yep! find the volume

OpenStudy (gabylovesyou):

353.773 ?

OpenStudy (gabylovesyou):

sorry... OS is lagging -_-

OpenStudy (gabylovesyou):

@ganeshie8

OpenStudy (gabylovesyou):

@johnweldon1993

OpenStudy (johnweldon1993):

Yes that is the correct volume for the cone on top (I got a slightly different number but I assume you used 3.14 for pi) So the volume of the total composite figure will be the sum of our 2 volumes \[\large \text{Total Volume} = \text{Volume of Cylinder} + \text{Volume of Cone}\] \[\large V = 4775.94 + 353.773 = ?\]

OpenStudy (johnweldon1993):

And then, since we now know the total volume, we divide that by the volume of a single tennis ball to figure out how many can fit inside.

OpenStudy (johnweldon1993):

Okay, the volume of a cylinder as it is originally would is \[\large \pi r^2 h\] correct? if we were to double the dimensions of this cylinder (double the radius and the height) we would have \[\large \pi (2r)^2 2h\] \[\large \pi 4r^22h\] \[\large 8\pi r^2 h\] So this means if we double the dimensions of our cylinder, the volume will increase by a factor of 8

OpenStudy (johnweldon1993):

And for the cone...same idea \[\large V = \frac{\pi r^2h}{3}\] if we double the dimensions (again the radius and the height) we have \[\large V = \frac{\pi (2r)^2 2h}{3}\] \[\large V = \frac{\pi 4r^2 2h}{3}\] \[\large V = 8\frac{\pi r^2h}{3}\] so again with the cone as well, if we double the dimensions of it, we will increase the volume by a factor of 8.

OpenStudy (johnweldon1993):

So all we need to do, is multiply the volume we received for the cylinder by 8....and then multiply the volume of the cone by 8...and then add the results

OpenStudy (gabylovesyou):

ty soo much!

OpenStudy (johnweldon1993):

Anytime :)

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