Chapter 9.8, Problem 13PSB

a.

To determine

To find:The length of a diagonal $\overline{AG}$ .

Answer to Problem 13PSB

The length of a diagonal $\overline{AG}$ is $5\sqrt{13}$ .

Explanation of Solution

Given information:

Side $\overline{CH}=17,$

Side $\overline{GH}=8,$

Side $\overline{FG}=6.$

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 CG , EG and AG can be calculated by applying Pythagoras Theorem.

In right angle triangle CGH , we get

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

In right angled triangle EHG , we get

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

In right angled triangle AEG , we get

  $\begin{array}{l}{\left(15\right)}^2+{\left(10\right)}^2={\left(AG\right)}^2\\ 225+100={\left(AG\right)}^2\\ $ \therefore {(AG)}^2=325$\therefore AG=\sqrt{325}\\ \therefore AG=5\sqrt{13}\end{array}$

b.

To determine

To calculate: The length of a diagonal $\overline{FG}$ .

Answer to Problem 13PSB

The length of a diagonal $\overline{FG}$ is $9\sqrt{11}$ .

Explanation of Solution

Given information:

Side $\overline{AG}=50,$

Side $\overline{AF}=40,$

Side $\overline{EF}=3.$

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,

Calculation:

  

Side EG and FG can be calculated by applying Pythagoras Theorem.

In right angle triangleAEG , we get

  $\begin{array}{l}{\left(40\right)}^2+{\left(EG\right)}^2={\left(50\right)}^2\\ {1600}^{}+{\left(EG\right)}^2=2500\\ {\left(EG\right)}^2=900\\ \therefore EG=\sqrt{900}\\ \therefore EG=30\end{array}$

In right angle triangle EFG , we get

  $\begin{array}{l}{\left(3\right)}^2+{\left(FG\right)}^2={\left(30\right)}^2\\ 9+{\left(FG\right)}^2=900\\ $ \therefore {(FG)}^2=891$\therefore FG=\sqrt{891}\\ \therefore FG=9\sqrt{11}\end{array}$