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Mathematics 17 Online
OpenStudy (realsiege):

i got a question abt surface area (in description!)

OpenStudy (realsiege):

if i want to calculate the surface area of a cylinder mug, but there's only one base (the other base is open), do i have to edit the formula so it's (2πrh) + (πr^2) instead of (2πr^2). also how do i do it with a cone? thanks!

OpenStudy (amyhashimoto77):

Yes, you are correct.

OpenStudy (amyhashimoto77):

And what do you mean, " how do i do it with a cone"?

OpenStudy (realsiege):

i'm measuring something like an ice cream cone. if the cone has no base how do i find the surface area?

Nnesha (nnesha):

do you have height and radius of the cone ?

OpenStudy (realsiege):

h=15cm r=5cm

Nnesha (nnesha):

alright first you have to find slant height amyhashimoto77 is already helping u ( don't want to jumpppp in :=))

OpenStudy (realsiege):

so: √5^2+15^2 is it 40?

Nnesha (nnesha):

but anyz lets start first use Pythagorean theorem to find slant height |dw:1445946753923:dw| \[\rm I^2= h^2 +r^2\]

OpenStudy (realsiege):

oh sh*t im wrong √5^2+15^2 = 20?

Nnesha (nnesha):

that looks incorrect .

Nnesha (nnesha):

how did you get 20 ?

OpenStudy (realsiege):

lemme find a calculator lol

OpenStudy (realsiege):

it's 5√10

Nnesha (nnesha):

ah so u used fancy calculator yes that's correct 5^2 = 25 15^2= 225 so we get \[\sqrt{25+225} = \sqrt{250}\] 250 isn't a perfect square so we have to find factors of 250 and divide inot two number (one of them should be the perfect square root ) so \[\sqrt{250} \rightarrow \sqrt{25 \times 10} \rightarrow 5\sqrt{10}\]

Nnesha (nnesha):

|dw:1445947275978:dw| to find total surface area use the formula \[\rm SA=\pi r^2 +\pi rl\]

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