Chapter 12.5, Problem 3PSA

a.

To determine

To calculate: The volume of a pyramid which has a square base and height of 12.

Answer to Problem 3PSA

The volume of a pyramid is $400$ .

Explanation of Solution

Given information:

Base of pyramid = 10,

Height of pyramid h = 12,

Formula used:

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

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

Area of square: $A=s^2$

s = side of square

Calculation:

  

  $\begin{array}{l}B=A_{Square}={\left(10\right)}^2\\ B=A_{Square}=100\end{array}$

  $\begin{array}{l}{V}_{Pyramid}=\frac{1}{3}\times B\times h\\ V_{Pyramid}=\frac{1}{3}\times 100\times 12\\ V_{Pyramid}=400\end{array}$

b.

To determine

To find: The total area of a pyramid which has a square base of side 10.

Answer to Problem 3PSA

The total area of a pyramid is $360$ .

Explanation of Solution

Given information:

Base of pyramid = 10,

Height of pyramid h = 12,

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 triangle: $A=\frac{1}{2}\times b\times h$

b = base of triangle

h = height of triangle Area of square: $A=s^2$

s = side of square

Calculation:

  

Side AC can be calculated by applying Pythagoras Theorem.

In right angled triangle ABC , we get

  $\begin{array}{l}{\left(AC\right)}^2={\left(AB\right)}^2+{\left(BC\right)}^2\\ {\left(AC\right)}^2={\left(12\right)}^2+{\left(5\right)}^2\\ BC=\frac{10}{2}=5\\ {\left(AC\right)}^2=144+25\\ {\left(AC\right)}^2=169\\ AC=\sqrt{169}\\ AC=13\end{array}$

  $\begin{array}{l}B=A_{Square}={\left(10\right)}^2\\ B=A_{Square}=100\end{array}$

Lateral area = Area of triangle $=\frac{1}{2}\times b\times h$

Lateral area $=\frac{1}{2}\times 10\times 13$

Lateral area $=65$

There are four triangles.

Total Area $=4\times$ Lateral area + B

Total Area $=4\times$ 65 + 100

Total Area $=260+100$

Total Area $=360$