Chapter 12.2, Problem 5PSA

a.

To determine

To calculate: The lateral area of regular pyramid PRXYZ .

Answer to Problem 5PSA

The lateral area of regular pyramid PRXYZ is $192$ .

Explanation of Solution

Given information:

PRXYZ is a regular pyramid. The midpoints of its lateral edges are joined to form a square ABCD

Side PR= 10

Side RX= 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

  

Side PU can be calculated by applying Pythagoras Theorem.

In right angle triangle PUX , we get

  $\begin{array}{l}{\left(PU\right)}^2+{\left(UX\right)}^2={\left(PX\right)}^2\\ UX=\frac{RX}{2}=\frac{12}{2}=6\\ {\left(PU\right)}^2+{\left(6\right)}^2={\left(10\right)}^2\\ {\left(PU\right)}^2+36=100\\ {\left(PU\right)}^2=100-36\\ {\left(PU\right)}^2=64\\ PU=\sqrt{64}=8\end{array}$

Area of triangle:

  $\begin{array}{l}A=\frac{1}{2}\times b\times h\\ A=\frac{1}{2}\times 12\times 8\\ A=48\end{array}$

There are four triangles

Lateral Area = 4 $\times$ Area of triangle

Lateral Area = $4\times 48$

Lateral Area $=192$

b.

To determine

To calculate: The lateral area of regular pyramid PABCD .

Answer to Problem 5PSA

The lateral area of regular pyramid PABCD is $48$ .

Explanation of Solution

Given information:

PRXYZ is a regular pyramid. The midpoints of its lateral edges are joined to form a square ABCD

Side PR= 10

Side RX= 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

  

Side PS can be calculated by applying Pythagoras Theorem.

In right angle triangle PSB , we get

  $\begin{array}{l}{\left(PS\right)}^2+{\left(SB\right)}^2={\left(PB\right)}^2\\ AB=6\\ SB=\frac{AB}{2}=\frac{6}{2}=3\\ PX=PR=10\\ PB=\frac{PX}{2}=\frac{10}{2}=5\\ {\left(PS\right)}^2+{\left(3\right)}^2={\left(5\right)}^2\\ {\left(PS\right)}^2+9=25\\ {\left(PS\right)}^2=25-9\\ {\left(PS\right)}^2=16\\ PS=\sqrt{16}=4\end{array}$

Area of triangle:

  $\begin{array}{l}A=\frac{1}{2}\times b\times h\\ A=\frac{1}{2}\times 6\times 4\\ A=12\end{array}$

There are four triangles

Lateral Area = 4 $\times$ Area of triangle

Lateral Area = $4\times 12$

Lateral Area $=48$

c.

To determine

To calculate: The area of square ABCD .

Answer to Problem 5PSA

The area of square ABCD is $36$ .

Explanation of Solution

Given information:

PRXYZ is a regular pyramid. The midpoints of its lateral edges are joined to form a square ABCD

Side PR= 10

Side RX= 12

Formula used:

Area of square: $A=s^2$

s = side of square

  

Area of square ABCD :

Side AB = 6

  $\begin{array}{l}A={\left(AB\right)}^2\\ A={\left(6\right)}^2\\ A=36\end{array}$

d.

To determine

To calculate: The area of square RXYZ .

Answer to Problem 5PSA

The area of square RXYZ is $144$ .

Explanation of Solution

Given information:

PRXYZ is a regular pyramid. The midpoints of its lateral edges are joined to form a square ABCD

Side PR= 10

Side RX= 12

Formula used:

Area of square: $A=s^2$

s = side of square

  

Area of square RXYZ :

Side RX= 12

  $\begin{array}{l}A={\left(RX\right)}^2\\ A={\left(12\right)}^2\\ A=144\end{array}$

d.

To determine

To calculate: The area ratio of square ABCD and square RXYZ .

Answer to Problem 5PSA

The area ratio is $1:4$ .

Explanation of Solution

Given information:

PRXYZ is a regular pyramid. The midpoints of its lateral edges are joined to form a square ABCD

Side PR= 10

Side RX= 12

Formula used:

Area of square: $A=s^2$

s = side of square

  

Area of square ABCD :

Side AB = 6

Area of square RXYZ :

Side RX= 12

  $\begin{array}{l}{A}_{RXYZ}={\left(RX\right)}^2\\ A_{RXYZ}={\left(12\right)}^2\\ A_{RXYZ}=144\end{array}$

The area ratio is as follows:

  $\frac{A_{ABCD}}{A_{RXYZ}}=\frac{36}{144}=\frac{1}{4}$

f.

To determine

To calculate: The area of trapezoid ABXR .

Answer to Problem 5PSA

The area of trapezoid ABXR is $36$ .

Explanation of Solution

Given information:

PRXYZ is a regular pyramid. The midpoints of its lateral edges are joined to form a square ABCD

Side PR= 10

Side RX= 12

Formula used:

Area of trapezoid: $A=\frac{1}{2}\times (b_1+b_2)\times h$

b₁ and b₂ = bases of trapezoid

h = height of trapezoid

  

Area of trapezoid ABXR :

  $\begin{array}{l}A=\frac{1}{2}\times (RX+AB)\times SU\\ A=\frac{1}{2}\times (12+6)\times 4\\ A=36\end{array}$