Chapter 12, Problem 8RP

a.

To determine

To calculate: The lateral area of a pyramid which has a square base of 16 and slant height of 10.

Answer to Problem 8RP

The lateral area of a pyramid is $320$ .

Explanation of Solution

Given information:

Base of square = 16,

Slant height of pyramid l = 10.

Formula used:

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

b = base of triangle

h = height of triangle

Calculation:

  

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

Area of triangle $=\frac{1}{2}\times 16\times 10$

Area of triangle $=80$

There are four triangles.

Total lateral area $=4\times$ Lateral area Total lateral area $=4\times$ 80

Total lateral area $=320$

b.

To determine

To find: The total area of a pyramid which has a square base of 16 and slant height of 10.

Answer to Problem 8RP

The total area of a pyramid is $576$ .

Explanation of Solution

Given information:

Base of square = 16,

Slant height of pyramid l = 10.

Formula used:

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:

  

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

Area of triangle $=\frac{1}{2}\times 16\times 10$

Area of triangle $=80$

There are four triangles.

Total lateral area $=4\times$ Lateral area Total lateral area $=4\times$ 80

Total lateral area $=320$

  $\begin{array}{l}B=A_{Square}={\left(16\right)}^2\\ B=A_{Square}=256\end{array}$

Total Area = Total lateral area +

c.

To determine

To calculate: The total volume of a pyramid which has a square base of 16 and slant height of 10.

Answer to Problem 8RP

The total volume of a pyramid is $512$ .

Explanation of Solution

Given information:

Base of square = 16,

Slant height of pyramid l = 10.

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 square: $A=s^2$

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

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

Calculation:

  

B = Base area of pyramid

  $\begin{array}{l}B=A_{Square}={\left(16\right)}^2\\ B=A_{Square}=256\end{array}$

Side AB can be calculated by applying Pythagoras Theorem.

In right angled triangle ABC , we get

  $\begin{array}{l}BC=\frac{16}{2}=8\\ {\left(AC\right)}^2={\left(AB\right)}^2+{\left(BC\right)}^2\\ {\left(10\right)}^2={\left(AB\right)}^2+{\left(8\right)}^2\\ {\left(AB\right)}^2=100-64\\ {\left(AB\right)}^2=36\\ AB=\sqrt{36}\\ AB=6\\ h=6\end{array}$

  $\begin{array}{l}{V}_{Pyramid}=\frac{1}{3}\times B\times h\\ V_{Pyramid}=\frac{1}{3}\times 256\times 6\\ V_{Pyramid}=512\end{array}$