Chapter 12.2, Problem 9PSB

a.

To determine

To calculate: The height PR , height PS , lateral area and total area of pyramid PABCD if PQ is 8, CD is 12 and BC is 30.

Answer to Problem 9PSB

The height PR is 17, height PS is 10, lateral area is $504$ and total area is $864$ .

Explanation of Solution

Given information:

Side PQ = 8,

Side CD = 12,

Side BC = 30.

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 rectangle: $A=l\times w$

l = length of rectangle

w= width of rectangle

Calculation:

  

In △PQR,

PQ = 8

  $QR=\frac{BC}{2}=\frac{30}{2}=15$

Side PR can be calculated by applying Pythagoras Theorem.

In right angle triangle PQR , we get

  $\begin{array}{l}{\left(PQ\right)}^2+{\left(QR\right)}^2={\left(PR\right)}^2\\ {\left(8\right)}^2+{\left(15\right)}^2={\left(PR\right)}^2\\ 64+225={\left(PR\right)}^2\\ {\left(PR\right)}^2=289\\ \left(PR\right)=\sqrt{289}\\ PR=17\end{array}$

In △PQS,

PQ = 8

  $QS=\frac{DC}{2}=\frac{12}{2}=6$

Side PS can be calculated by applying Pythagoras Theorem.

In right angle triangle PQS , we get

  $\begin{array}{l}{\left(PQ\right)}^2+{\left(QS\right)}^2={\left(PS\right)}^2\\ {\left(8\right)}^2+{\left(6\right)}^2={\left(PS\right)}^2\\ 64+36={\left(PS\right)}^2\\ {\left(PS\right)}^2=100\\ \left(PS\right)=100\\ PS=10\end{array}$

Area of triangle:

  $\begin{array}{l}A\left(△PDC\right)=\frac{1}{2}\times DC\times PR\\ A\left(△PDC\right)=\frac{1}{2}\times 12\times 17\\ A\left(△PDC\right)=102\end{array}$

  $\begin{array}{l}A\left(△PBC\right)=\frac{1}{2}\times BC\times PS\\ A\left(△PBC\right)=\frac{1}{2}\times 30\times 10\\ A\left(△PBC\right)=150\end{array}$

In a pyramid there are two $△PDC$ and two $△PBC$ .

Lateral Area $=2\times △PDC+2\times △PBC$

Lateral Area $=\left(2\times 102\right)+\left(2\times 150\right)$

Lateral Area $=504$

Area of rectangular base ABCD $=30\times 12=360$

Total Area = Lateral Area + Area of rectangular base ABCD

Total Area $=504+360$

Total Area $=864$

b.

To determine

To find: The altitude PQ if each lateral edge is 25 and rectangular dimensions of 24 by 30.

Answer to Problem 9PSB

The altitude PQ is $16$ .

Explanation of Solution

Given information:

Side PA=PB=PC=PD = 25,

Side AB=CD = 24,

Side DC = AB =30.

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,

  $a^2+b^2\text{= }{c}^2$

Calculation:

  

Side $PS=PA=17$

Side AC can be calculated by applying Pythagoras Theorem.

In right angle triangle ABC , we get

  $\begin{array}{l}{\left(AB\right)}^2+{\left(CB\right)}^2={\left(AC\right)}^2\\ {\left(30\right)}^2+{\left(24\right)}^2={\left(AC\right)}^2\\ {\left(AC\right)}^2=900+576\\ {\left(AC\right)}^2=1476\\ \left(AC\right)=\sqrt{1476}\\ AC=6\sqrt{41}\\ CQ=\frac{AC}{2}=\frac{6\sqrt{41}}{2}=3\sqrt{41}\end{array}$

Side PQ can be calculated by applying Pythagoras Theorem.

In right angle triangle PQC , we get

  $\begin{array}{l}{\left(PQ\right)}^2+{\left(QC\right)}^2={\left(PC\right)}^2\\ {\left(PQ\right)}^2+{\left(3\sqrt{41}\right)}^2={\left(25\right)}^2\\ {\left(PQ\right)}^2=625-369\\ {\left(PQ\right)}^2=256\\ \left(PQ\right)=\sqrt{256}\\ PQ=16\end{array}$