Chapter 12.3, Problem 10PSB

To determine

To calculate: The total area of the solid which contains slant height of cone as 5.

Answer to Problem 10PSB

The total area of solid is $93\pi$ .

Explanation of Solution

Given information:

Diameter d = 6,

Slant height of cone = 5.

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,

  $a^2+b^2\text{= }{c}^2$

Area of cylinder $=C\times h$

C = Circumference of cylinder and h = height of cylinder

Area of cone $=\frac{1}{2}\times C\times l$

C = Circumference of cone and l = slant height of cone

Area of hemisphere $=\frac{1}{2}\times 4\pi r^2$

r = radius of hemisphere

Calculation:

  

  $d=6,r=3$

Side AB can be calculated by applying Pythagoras Theorem.

In right angled triangle ABC , we get

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

Height of cone $AB=4$

Slant height of cone = 5.

Height of cylinder $=17-4-3=10$

Area of cylinder $=C\times h$

  $C=2\pi \times 3=6\pi$

  $h=10$

Area of cylinder $=6\pi \times 10$

Area of cylinder $=60\pi$

Area of cone $=\frac{1}{2}\times C\times l$

  $C=2\pi \times 3=6\pi$

  $l=5$

Area of cone $=\frac{1}{2}\times C\times l$

Area of cone $=\frac{1}{2}\times 6\pi \times 5$

Area of cone $=15\pi$

Area of hemisphere $=\frac{1}{2}\times 4\pi r^2$

Area of hemisphere $=2\pi {\left(3\right)}^2$

Area of hemisphere $=18\pi$

Total Area = Area of cylinder + Area of cone + Area of hemisphere

Total Area $=60\pi +15\pi +18\pi$

Total Area $=93\pi$