Chapter 9.8, Problem 14PSB

a.

To determine

To calculate: Thelength of a side $ID$ .

Answer to Problem 14PSB

The length of a side $ID$ is $12$ .

Explanation of Solution

Given information:

In a regular square pyramid,

Side $PR=10,$

Diagonal $MD=IA=12\sqrt{2}$

Formula used:

Base diagonal of a square are $side\sqrt{2}$ .

Calculation:

  

It is given that Diagonal $MD=IA=12\sqrt{2}$ .

If base diagonals of a square are $12\sqrt{2}$ ,a side =12, so ID = 12.

b.

To determine

To find: The altitude of the pyramid.

Answer to Problem 14PSB

The altitude of the pyramid is 8.

Explanation of Solution

Given information:

In a regular square pyramid,

Side $PR=10,$

Diagonal $MD=IA=12\sqrt{2}$

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:

  

As PY is drawn perpendicular on YD , YD is half of any of the side of a square.

  $YR=\frac{1}{2}(ID)=\frac{1}{2}\ast 12=6$

Side PY can be calculated by applying Pythagoras Theorem.

In right angle triangle PYR , we get

  $\begin{array}{l}{\left(PY\right)}^2+{\left(YR\right)}^2={\left(PR\right)}^2\\ {\left(PY\right)}^2+{\left(6\right)}^2={\left(10\right)}^2\\ {\left(PY\right)}^2=100-36\\ \therefore PY=\sqrt{64}\\ \therefore PY=8\end{array}$

c.

To determine

To calculate: The length of side $RD$ .

Answer to Problem 14PSB

The length of side $RD$ is $6.$

Explanation of Solution

Given information:

In a regular square pyramid,

Side $PR=10,$

Diagonal $MD=IA=12\sqrt{2}$

Formula used:

A side is perpendicular bisector to other side, then the side is divided into two equal parts.

Calculation:

  

  $PR$ is a perpendicular bisector to $DA$ .

  $AR=RD=\frac{1}{2}\ast PR=\frac{1}{2}\ast 12=6$

d.

To determine

To find: The length of a sidePD .

Answer to Problem 14PSB

The length of a sidePD is $2\sqrt{34}$ .

Explanation of Solution

Given information:

In a regular square pyramid,

Side $PR=10,$

Diagonal $MD=IA=12\sqrt{2}$

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 PD can be calculated by applying Pythagoras Theorem.

In right angle triangle PDR , we get

  $\begin{array}{l}{\left(PD\right)}^2={\left(RD\right)}^2+{\left(PR\right)}^2\\ {\left(PD\right)}^2={\left(6\right)}^2+{\left(10\right)}^2\\ {\left(PD\right)}^2=36+100\\ \therefore PD=\sqrt{136}\\ \therefore PD=2\sqrt{34}\end{array}$