Chapter 12.5, Problem 12PSB

To determine

To calculate: The volume of rocket portion that is available for nonfuel items.

Answer to Problem 12PSB

The volume of rocket portion is $4423$ .

Explanation of Solution

Given information:

Diameter of rocket d = 16,

Slant height of rocket = 17,

Height of rocket h = 50.

Formula used:

The below theorem is used:

Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

  

In right angle triangle,

  $a^2+b^2\text{= }{c}^2$

Volume of cylinder $=B\times h$

B = Base area of cylinder and h = height of cylinder

Volume of cone $=\frac{1}{3}\times B\times h$

B = Base area of cone and h = height of cone

Area of a circle: $A=\pi r^2$

r = radius of circle

Calculation:

  

  $r=\frac{d}{2}=\frac{16}{2}=8$

Volume of cylinder $=B\times h$

  $B=\pi {(8)}^2=64\pi$

Volume of cylinder $=\pi {(8)}^2\times 50$

Volume of cylinder $=3200\pi$

Side AB can be calculated by applying Pythagoras Theorem.

In right angled triangle ABC , we get

  $\begin{array}{l}{\left(AC\right)}^2={\left(AB\right)}^2+{\left(BC\right)}^2\\ BC=\frac{r}{2}=\frac{16}{2}=8\\ {\left(17\right)}^2={\left(AB\right)}^2+{\left(8\right)}^2\\ {\left(AB\right)}^2=289-64\\ {\left(AB\right)}^2=225\\ AB=\sqrt{225}\\ AB=15\\ h=AB=15\end{array}$

Volume of cone:

  $B=\pi {(8)}^2=64\pi$

  $\begin{array}{l}{V}_{Cone}=\frac{1}{3}\times B\times h\\ V_{Cone}=\frac{1}{3}\times 64\pi \times 15\\ V_{Cone}=320\pi \end{array}$

Total Volume = Volume of cylinder + Volume of cone

Total Volume $=3200\pi +320\pi$

Total Volume $=3520\pi$

If $60\%$ contains fuel, $40\%$ remains.

Volume for nonfuel items:

  $40\%of3520\pi =0.4\times 3520\pi =1480\pi \approx 4423$