Question

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?

Answer 1

`348cm^2`

Explanation:

Lets first consider the cross-section of the cone.

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:

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`