A cone has a height of #18 cm# and its base has a radius of #5 cm#. If the cone is horizontally cut into two segments #12 cm# from the base, what would the surface area of the bottom segment be?

1 Answer
Apr 19, 2016

#348cm^2#

Explanation:

Lets first consider the cross-section of the cone.

enter image source here

Now it is given in the question, that AD = #18cm# and DC = #5cm#

given, DE = #12cm#

Hence, AE = #(18-12)cm = 6cm#

As, #DeltaADC # is similar to #DeltaAEF #, #(EF)/(DC) = (AE)/(AD)#

#:. EF = DC*(AE)/(AD)=(5cm)*6/18 = 5/3cm#

After cutting, the lower half looks like this:

enter image source here

We have calculated the smaller circle (the circular top), to have a radius of #5/3cm#.

Now lets calculate the length of the slant.

#Delta ADC# being a Right angle triangle, we can write

#AC =sqrt(AD^2+DC^2) = sqrt(18^2+5^2) cm ~~ 18.68 cm #

The surface area of the whole cone is : #pirl = pi*5*18.68 cm^2#

Using the similarity of the triangles #DeltaAEF # and #DeltaADC #, we know that all the sides of #DeltaAEF # are less than the corresponding sides of #DeltaADC # by a factor of 3.

So the slant surface area of the upper part (the smaller cone) is : #(pi*5*18.68)/(3*3)cm^2#

Hence of the slant surface area of the lower part is: #pi*5*18.68*(8/9)cm^2#

And we have the the upper and lower circular surfaces' areas as well.

So the total area is:

#pi*(5^2/3^2)_"for upper circular surface"+ pi*5*18.68*(8/9) _"for the slant surface"+ pi*(5^2) _"for lower circular surface" ~~348cm^2#