\begin{center}\includegraphics[width=0.5\textwidth]{img/5a}\end{center}                                                              
                                                                                                                                      
 How can we find the length of the diagonal $CD$ of this cube with side lengths of $4$cm?                                             
                                                                                                                                      
 We apply Pythagoras' Theorem!                                                                                                        
                                                                                                                                      
 \begin{theorembox}                                                                                                                   
     \centering                                                                                                                       
     \textbf{Recall that:} $c^2 = a^2 + b^2$                                                                                          
 \end{theorembox}                                                                                                                     
                                                                                                                                      
 And so here we first find the length of $s$ to be $\sqrt{32}=4\sqrt{2}$ and thus we can use this length of $s$ along with the side   
 $CB$ to form another triangle and find $d$ = $\sqrt{32+16} = \sqrt{48} = 4\sqrt{3} = \approx 6.93$.                                  
                                                                                                                                      
 \begin{solutionordottedlines}[2in]                                                                                                   
 To find the length of the diagonal $CD$, we first find the diagonal $s$ of the base square using Pythagoras' theorem: $s^2 = 4^2 + 4 
 = 32$, so $s = \sqrt{32} = 4\sqrt{2}$. Then, we use $s$ to find $CD$ by forming a right-angled triangle with $s$ and side $CB$ of th 
 cube: $CD^2 = s^2 + CB^2 = (4\sqrt{2})^2 + 4^2 = 32 + 16 = 48$, so $CD = \sqrt{48} = 4\sqrt{3} \approx 6.93$ cm.                     
 \end{solutionordottedlines}                                                                                                          
                                                                                                                                      
 \begin{examplebox}                                                                                                                   
     \subsection{Example:}                                                                                                            
     A square pyramid has height \(5 \mathrm{~cm}\) and square base with side length \(4 \mathrm{~cm}\). Find the length of the edge  
 \(V C\) in this diagram.                                                                                                             
                                                                                                                                      
     \begin{center}                                                                                                                   
     \includegraphics[width=0.3\textwidth]{img/5b}                                                                                    
     \end{center}                                                                                                                     
                                                                                                                                      
     \begin{solutionordottedlines}[2in]                                                                                               
     To find the length of the edge $VC$, we can use the Pythagorean theorem on the triangle formed by the height of the pyramid, hal 
 the base, and the slant height (edge $VC$). The half of the base is $2$ cm (half of $4$ cm). So, $VC^2 = 5^2 + 2^2 = 25 + 4 = 29$,   
 thus $VC = \sqrt{29} \approx 5.39$ cm.                                                                                               
     \end{solutionordottedlines}                                                                                                      
 \end{examplebox}                                                                                                                     
                                                                                                                                      
 \begin{exercisebox}                                                                                                                  
     \subsection*{Exercises:}                                                                                                         
     \begin{questions}                                                                                                                
         \Question[6]                                                                                                                 
             \begin{center}                                                                                                           
             \includegraphics[width=0.4\textwidth]{img/5c}                                                                            
             \end{center}                                                                                                             
             The rectangular prism in the diagram has a length of \(12 \mathrm{~cm}\), a width of \(5 \mathrm{~cm}\) and a height of  
 \(6 \mathrm{~cm}\).                                                                                                                  
             \begin{parts}                                                                                                            
                 \part Consider triangle \(E F G\). Find:                                                                             
                     \begin{subparts}                                                                                                 
                         \subpart \(E F\)                                                                                             
                             \begin{solutionordottedlines}[2cm]                                                                       
                             $EF = 12$ cm (given as the length of the prism).                                                         
                             \end{solutionordottedlines}                                                                              
                         \subpart the size of \(E F G\)                                                                               
                             \begin{solutionordottedlines}[2cm]                                                                       
                             $\angle EFG$ is a right angle since $EF$ and $FG$ are edges of the rectangular prism meeting at a right  
 angle.                                                                                                                               
                             \end{solutionordottedlines}                                                                              
                     \end{subparts}                                                                                                   
                 \part Find \(E G\).                                                                                                  
                     \begin{solutionordottedlines}[2cm]                                                                               
                     To find $EG$, we use Pythagoras' theorem on triangle $EFG$: $EG^2 = EF^2 + FG^2 = 12^2 + 6^2 = 144 + 36 = 180$,  
 $EG = \sqrt{180} = 6\sqrt{5} \approx 13.42$ cm.                                                                                      
                     \end{solutionordottedlines}                                                                                      
                 \part Find \(A G\), correct to 1 decimal place.                                                                      
                     \begin{solutionordottedlines}[2cm]                                                                               
                     To find $AG$, we use Pythagoras' theorem on triangle $AFG$: $AG^2 = AF^2 + FG^2 = 5^2 + 6^2 = 25 + 36 = 61$, so  
 $AG = \sqrt{61} \approx 7.8$ cm.                                                                                                     
                     \end{solutionordottedlines}                                                                                      
             \end{parts}                                                                                                              
         \question[] Note: \(A G\) is called the space diagonal of the rectangular prism.                                             
         \Question[3] Find the length of the longest pencil that can fit inside a cylindrical pencil case of length \(15 \mathrm{~cm} 
 and radius \(2 \mathrm{~cm}\).                                                                                                       
             \begin{solutionordottedlines}[2in]                                                                                       
             The longest pencil that can fit inside the cylindrical pencil case would be the length of the diagonal of the cylinder.  
 This can be found using Pythagoras' theorem with the length and diameter of the cylinder: $pencil^2 = length^2 + diameter^2 = 15^2 + 
 (2 \times 2)^2 = 225 + 16 = 241$, so the pencil length is $\sqrt{241} \approx 15.52$ cm.                                             
             \end{solutionordottedlines}                                                                                              
     \end{questions}                                                                                                                  
 \end{exercisebox}