Chapter 12.5, Problem 7PSA

To determine

To calculate: The volume of water the well could hold.

Answer to Problem 7PSA

The volume of water the well could hold is $1451f{t}^3$ .

Explanation of Solution

Given information:

Diameter of well = 6 m,

Slant height of cone = 5 m,

Height of well = 50 m.

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:

  

Volume of cylinder $=B\times h$

Volume of cylinder $=9\pi \times 50$

Volume of cylinder $=450\pi$

Side AB can be calculated by applying Pythagoras Theorem.

In right angled triangle ABC , we get

  $\begin{array}{l}{\left(AC\right)}^2={\left(AB\right)}^2+{\left(BC\right)}^2\\ BC=\frac{r}{2}=\frac{6}{2}=3\\ {\left(5\right)}^2={\left(AB\right)}^2+{\left(3\right)}^2\\ {\left(AB\right)}^2=25-9\\ {\left(AB\right)}^2=16\\ AB=\sqrt{16}\\ AB=4\\ h=AB=4\end{array}$

Volume of cone:

  $\begin{array}{l}{V}_{Cone}=\frac{1}{3}\times B\times h\\ V_{Cone}=\frac{1}{3}\times 9\pi \times 4\\ V_{Cone}=12\pi \end{array}$

Total Volume = Volume of cylinder + Volume of cone

Total Volume $=450\pi +12\pi$

Total Volume $=462\pi$

Total Volume $\approx 1451f{t}^3$