Chapter 12.5, Problem 5PSA

a.

To determine

To calculate: The volume of a right circular cone which has height of 40 and slant height of 41.

Answer to Problem 5PSA

The volume of a right circular cone is $1080\pi$ .

Explanation of Solution

Given information:

Height of cone h = 40,

Slant height of cone l = 41,

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 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:

  

Side BC can be calculated by applying Pythagoras Theorem.

In right angled triangle ABC , we get

  $\begin{array}{l}{\left(AC\right)}^2={\left(AB\right)}^2+{\left(BC\right)}^2\\ {\left(41\right)}^2={\left(40\right)}^2+{\left(BC\right)}^2\\ 1681=1600+{\left(BC\right)}^2\\ {\left(BC\right)}^2=1681-1600\\ BC=\sqrt{81}\\ BC=9\\ r=BC=9\end{array}$

  $B=A_{Circle}=\pi r^2=\pi {\left(9\right)}^2=81\pi$

  $\begin{array}{l}{V}_{Cone}=\frac{1}{3}\times B\times h\\ V_{Cone}=\frac{1}{3}\times 81\pi \times 40\\ V_{Cone}=1080\pi \end{array}$

b.

To determine

To find: The lateral area of a right circular cone which has height of 40 and slant height of 41.

Answer to Problem 5PSA

The lateral area of a right circular cone is $369\pi$ .

Explanation of Solution

Given information:

Height of cone h = 40,

Slant height of cone l = 41,

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 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:

  

Side BC can be calculated by applying Pythagoras Theorem.

In right angled triangle ABC , we get

  $\begin{array}{l}{\left(AC\right)}^2={\left(AB\right)}^2+{\left(BC\right)}^2\\ {\left(41\right)}^2={\left(40\right)}^2+{\left(BC\right)}^2\\ 1681=1600+{\left(BC\right)}^2\\ {\left(BC\right)}^2=1681-1600\\ BC=\sqrt{81}\\ BC=9\\ r=BC=9\end{array}$

Lateral Area $=\pi rl$

Lateral Area $=\pi \left(9\right)\left(41\right)$

Lateral Area $=369\pi$

c.

To determine

To calculate: The total area of a right circular cone which has height of 40 and slant height of 41.

Answer to Problem 5PSA

The total area of a right circular cone is $450\pi$ .

Explanation of Solution

Given information:

Height of cone h = 40,

Slant height of cone l = 41,

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 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:

  

Side BC can be calculated by applying Pythagoras Theorem.

In right angled triangle ABC , we get

  $\begin{array}{l}{\left(AC\right)}^2={\left(AB\right)}^2+{\left(BC\right)}^2\\ {\left(41\right)}^2={\left(40\right)}^2+{\left(BC\right)}^2\\ 1681=1600+{\left(BC\right)}^2\\ {\left(BC\right)}^2=1681-1600\\ BC=\sqrt{81}\\ BC=9\\ r=BC=9\end{array}$

  $B=A_{Circle}=\pi r^2=\pi {\left(9\right)}^2=81\pi$

Lateral Area $=\pi rl$

Lateral Area $=\pi \left(9\right)\left(41\right)$

Lateral Area $=369\pi$

Total Area = Lateral Area + Area of base

Total Area $=369\pi +81\pi$

Total Area $=450\pi$