Question

A conic kettle has a cover which height is 30% of its total height. The height is 2 cm less than the diameter of the base, which has an area of 380 squared cm. Which is the volume capacity of the kettle?

Answer 1

In a diagram,

Therefore, we need to find the volume of a truncated cone.

Let's define the quantities below,

`\begin{gathered} h_t\rightarrow\text{ height of the truncate cone} \\ h\rightarrow\text{ height of the whole cone} \\ r\rightarrow\text{ radius of the cone} \\ A_b\rightarrow\text{ area of the base of the cone} \\ V\rightarrow\text{ volume} \end{gathered}`

Therefore, according to the question, we can identify the relations below

a) h_t=0.30h

`\begin{gathered} h_t=0.30*h \\ h=2r-2 \\ A_b=380 \end{gathered}`

Thus,

`\begin{gathered} \Rightarrow A_b=380 \\ and \\ A_b=\pi r^2 \\ \Rightarrow\pi r^2=380 \\ \Rightarrow r=\sqrt{(380)/(\pi)} \end{gathered}`

Finding h_t,

`\begin{gathered} r=\sqrt{(380)/(\pi)} \\ and \\ h_t=0.30h=0.30(2r-2)=0.6r-0.6 \\ \Rightarrow h_t=5.99885...\approx6 \\ \end{gathered}`

Thus, the volume of the first case is

`\begin{gathered} \Rightarrow V=(1)/(3)\pi r^2h=(1)/(3)*380*6=760 \\ \Rightarrow V=760 \end{gathered}`

The volume when h_t=0.30h is approximately 760cm^3.

b) h_t=0.7h

Notice that A_b=380, so the result of r obtained above applies in this case too.

Obtaining h_t,

`h_t=0.7h=0.7(2r-2)=1.4r-1.4=13.9973\approx14`

Thus,

`\Rightarrow V=(1)/(3)*380*14=1773.3333`

The volume when h_t=0.7 is approximately 1773.333... cm^3.