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24 votes
24 votes
What is the diameter of a hemisphere with a volume of 272
m^{3, to the nearest tenth of a meter?

User Jtello
by
2.9k points

1 Answer

19 votes
19 votes

Answer:

D = 10.1m

Explanation:

We know that the volume of a sphere of radius R is:

V = (4/3)*pi*R^3

where pi = 3.14

A hemisphere is half a sphere, then the volume of an hemisphere is half of the volume of a sphere, then the volume of a hemisphere of radius R is:

V' = (1/2)*(4/3)*pi*R^3

Now, we know that our hemisphere has a volume:

V' = 272m^3

Then we can solve the equation:

272m^3 = (1/2)*(4/3)*pi*R^3

For R.

(we want the diameter, but remember that the diameter is equal to two times the radius, so finding the radius is a good start)

272m^3 = (1/2)*(4/3)*pi*R^3

272m^3 = (4/6)*3.14*R^3

272m^3*(6/4*3.14) = R^3

∛( 272m^3*(6/4*3.14) ) = R = 5.06 m

And the diameter is two times that, so we get:

D = 2*5.06m = 10.12 m

Rounding this to the nearest tenth of a meter (first digit after the decimal dot), we get:

D = 10.1m

User Bneil
by
2.6k points
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