Question #755fb

From https://www.khanacademy.org/math/geometry/hs-geo-solids/hs-geo-solids-intro/v/volume-of-a-sphere the Volume of a Sphere is:

#V = 4/3pir^3#

Therefore the volume of a hemisphere is 1/2 of this or:

#4/(3*2)pir^3 -> 2/3pir^3#

So to determine #r# we need to equation this formula to #250/3pi# and solve for #r#:

#2/3pir^3 = 250/3pi#

#3/pi * 2/3pir^3 = 3/pi * 250/3pi#

#cancel(3)/cancel(pi) * 2/cancel(3)cancel(pi)r^3 = cancel(3)/cancel(pi) * 250/cancel(3)cancel(pi)#

#2r^3 = 250#

#(2r^3)/2 = 250/2#

#r^3 = 125#

#r = 5#

https://www.math.hmc.edu/funfacts/ffiles/20004.2-3.shtml the surface area of a sphere is:

#A = 4pir^2#

Therefore the surface are of a hemisphere is 1/2 of this or:

#4pir^2/2 -> 2pir^2#

We know the radius of the square is #5# so we can substitute this into the formula and solve for #H#, the surface area of a hemisphere:

#H = 2pi5^2#

#H = 2pi*25#

#H = 50pi#

The area of a sphere is #4pir^2# so half of it is #4/2pir^2=2pir^2# and we need to evaluate the radius #r#;

The volume of the entire sphere is #V=4/3pir^3#. In your case the volume of the hemisphere should be half or: #V/2=2/3pir^3#
where the volume of the hemisphere is given as #250/3pi# (I am not sure if it is divided or multiplied by #pi# so I used multiplied) so we get:

#250/cancel(3)cancel(pi)=2/cancel(3)cancel(pi)r^3#

#r^3=250/2# so that #r=root(3)(250/2)=5#
We can use this value into the expression for half the surface of the sphere BUT....do we need also to consider the base of the hemisphere?

Considering the base as well we get:

#S=2pir^2+pir^2=3pir^2=3pi(5^2)=75pi#