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Mathematics 9 Online
OpenStudy (anonymous):

please help!!! What is the approximate volume of the cylinder? height = 52.2 cm; diameter = 20 cm

OpenStudy (tkhunny):

Volume = \(\pi r^{2}h\) We have r = 20 cm / 2 = 10 cm -- Not much need to approximation that. We have h = 52.2 cm -- Okay, that's a little long from 50 cm. If we use 50 cm, we'll be a little short with out estimate. We have \(\pi = 3.14159265358979323846264338...\) Okay, if we go low again, that's not far from 3. ShortVolume = 3*(10^2)*50 cm^3= 150*100 cm^3 = 15000 cm^3 We have r = 20 cm / 2 = 10 cm -- Not much need to approximation that. We have h = 52.2 cm -- Okay, that's a little shy of 55 cm. If we use 55 cm, we'll be a little long with out estimate. I picked 55 cm for a reason. We have \(\pi = 3.14159265358979323846264338...\) Okay, if we go high again, that's not far from 3.2. LongVolume = 3.2*(10^2)*55 cm^3 = 3.2 * 5 * 11 * 100 cm^3 = 16*11*100 cm^3 LongVolume = 176*100 cm^3 = 17600 cm^3 Let's calculate the exact value and see if we land in the middle. ExactVolume = \(\pi (10^2)\cdot 52.2\;cm^{3}\) = 16399 I like it. Feel free to express some individuality in your estimates. You don't have to do it like me or like anyone else. Just make an estimate!

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