\begin{theorembox}                                                                                                                          
     \section*{Theorem}                                                                                                                      
     For a right-angled triangle, the following relationship holds between the three sides: \[a^2 + b^2 = c^2\]                              
     Where $c$ is the length of the hypotenuse, and $a$ and $b$, the lengths of the remaining two sides.                                     
 \end{theorembox}                                                                                                                            
                                                                                                                                             
 \begin{boxdef}                                                                                                                              
     \subsubsection*{Theorem:}                                                                                                               
     \begin{solutionordottedlines}[2cm]                                                                                                      
         A statement that has been proven on the basis of previously established statements.                                                 
     \end{solutionordottedlines}                                                                                                             
 \end{boxdef}                                                                                                                                
                                                                                                                                             
 \begin{boxdef}                                                                                                                              
     \subsubsection*{Hypotenuse}                                                                                                             
     \begin{solutionordottedlines}[2cm]                                                                                                      
         The longest side of a right-angled triangle, opposite the right angle.                                                              
     \end{solutionordottedlines}                                                                                                             
 \end{boxdef}                                                                                                                                
                                                                                                                                             
 This is probably the first \textbf{Theorem} that you have ever seen and it is also one of the most mathematically significant ones. It must 
 be emphasised that a Theorem means nothing without a \textbf{Proof}.                                                                        
 \\12-year old Einstein, President James Garfield, and Euclid all found unique proofs to $a^2 + b^2 = c^2$. Below we outline Pythagoras'     
 original proof by rearrangement so that you can visually grasp the geometry of this fact.                                                   
                                                                                                                                             
 \begin{center}                                                                                                                              
 \begin{tikzpicture}                                                                                                                         
     % First Square                                                                                                                          
     \draw (0,0) -- (5,0) -- (5,5) -- (0,5) -- cycle;                                                                                        
                                                                                                                                             
     % Triangles for first square                                                                                                            
     \draw (0,3.5) -- (1.5,0);                                                                                                               
     \draw (1.5,0) -- (5,1.5);                                                                                                               
     \draw (5,1.5) -- (3.5,5);                                                                                                               
     \draw (3.5,5) -- (0,3.5);                                                                                                               
                                                                                                                                             
     % Labels for first square                                                                                                               
     \node at (2.5,2.5) {$c^2$};                                                                                                             
     \node at (0,3.5/2) [left] {$a$};                                                                                                        
     \node at (0,3.5+1.5/2) [left] {$b$};                                                                                                    
     \node at (3.5/2,5) [above] {$a$};                                                                                                       
     \node at (3.5+1.5/2,5) [above] {$b$};                                                                                                   
     \node at (5,3.5+1.5/2) [right] {$a$};                                                                                                   
     \node at (5,1.5/2) [right] {$b$};                                                                                                       
     \node at (3.5+1.5/2,0) [below] {$a$};                                                                                                   
     \node at (1.5/2,0) [below] {$b$};                                                                                                       
                                                                                                                                             
     % Second Square                                                                                                                         
     \draw (7,0) rectangle (12,5);                                                                                                           
     \draw (7,0) rectangle (7+3.5,3.5);                                                                                                      
     \draw (7+3.5,3.5) rectangle (12,5);                                                                                                     
                                                                                                                                             
     \draw (7,3.5) -- (7+3.5,5);                                                                                                             
     \draw (7+3.5,0) -- (12,3.5);                                                                                                            
                                                                                                                                             
     \node at (7+3.5/2, 3.5/2) {$a^2$};                                                                                                      
     \node at (7+3.5+1.5/2,3.5+1.5/2) {$b^2$};                                                                                               
     \node at (7+3.5/2,5-1.5) [above] {$a$};                                                                                                 
     \node at (7+3.5,3.5+1.5/2) [left] {$b$};                                                                                                
     \node at (7+3.5+1.5/2,3.5) [below] {$b$};                                                                                               
     \node at (12-1.5,3.5/2) [left] {$a$};                                                                                                   
                                                                                                                                             
 \end{tikzpicture}                                                                                                                           
 \end{center}                                                                                                                                
                                                                                                                                             
 \subsection*{Explanation}                                                                                                                   
 If you don't understand it immediately, do not panic. It takes a moment to click. What is happening here is that the 4 triangles in the fir 
 figure are the same as the 4 triangles in the second figure. So the free space of figure 1 must be the same as the free space in figure 2,  
 and thus $ c^2 = a^2 + b^2$.                                                                                                                
                                                                                                                                             
 Now that we all believe \emph{The Pythagorean Theorem}, let us use this mathematical result to discover things about our world:             
                                                                                                                                             
 \begin{examplebox}                                                                                                                          
     \section*{Examples}                                                                                                                     
         \begin{questions}                                                                                                                   
             \Question[1]\,                                                                                                                  
                 \begin{center}                                                                                                              
                     \begin{tikzpicture}[scale=1.5]                                                                                          
                         \draw (0,0) -- (4,0) node[midway,below] {$40$};                                                                     
                         \draw (4,0) -- (0,0.9) node[midway,sloped,above] {$x$};                                                             
                         \draw (0,0.9) -- (0,0) node[midway,left] {$9$};                                                                     
                         \draw (0,0) rectangle (0.2, 0.2);                                                                                   
                     \end{tikzpicture}                                                                                                       
                 \end{center}                                                                                                                
                 \begin{solutionordottedlines}[1in]                                                                                          
                     $x = \sqrt{40^2 + 9^2} = \sqrt{1600 + 81} = \sqrt{1681} = 41$                                                           
                 \end{solutionordottedlines}                                                                                                 
             \Question[1]\,                                                                                                                  
                 \begin{center}                                                                                                              
                     \begin{tikzpicture}[scale=1.5]                                                                                          
                         \draw (0,0) -- (1.6,0) node[midway,below] {$x$};                                                                    
                         \draw (1.6,0) -- (1.6,3) node[midway, right] {$30$};                                                                
                         \draw (1.6,3)--(0,0) node[midway,above,sloped] {$34$};                                                              
                         \draw (1.6,0) rectangle (1.4, 0.2);                                                                                 
                     \end{tikzpicture}                                                                                                       
                 \end{center}                                                                                                                
                 \begin{solutionordottedlines}[1in]                                                                                          
                     $x = \sqrt{34^2 - 30^2} = \sqrt{1156 - 900} = \sqrt{256} = 16$                                                          
                 \end{solutionordottedlines}                                                                                                 
             \Question[1]\,                                                                                                                  
                 \begin{center}                                                                                                              
                     \begin{tikzpicture}[scale=1.5]                                                                                          
                         \draw (0,0) -- (3,0) node[midway,below] {$30$};                                                                     
                         \draw (3,0) -- (0,1.2) node[midway, above] {$x$};                                                                   
                         \draw (0,1.2)--(0,0) node[midway,left] {$12$};                                                                      
                         \draw (0,0) rectangle (0.2, 0.2);                                                                                   
                     \end{tikzpicture}                                                                                                       
                 \end{center}                                                                                                                
                 \begin{solutionordottedlines}[1in]                                                                                          
                     $x = \sqrt{30^2 + 12^2} = \sqrt{900 + 144} = \sqrt{1044} \approx 32.31$                                                 
                 \end{solutionordottedlines}                                                                                                 
         \end{questions}                                                                                                                     
 \end{examplebox}                                                                                                                            
                                                                                                                                             
 What is the difference between the third example and the previous 2?                                                                        
 \begin{solutionordottedlines}[1in]                                                                                                          
     The third example involves finding the hypotenuse $x$ given the two shorter sides, whereas the previous two examples involve finding on 
 of the shorter sides given the hypotenuse and the other shorter side.                                                                       
 \end{solutionordottedlines}                                                                                                                 
                                                                                                                                             
 What is the significance of this result? Why did it hurt the brains of the \emph{Ancient Greeks}?                                           
 \begin{solutionordottedlines}[1.5in]                                                                                                        
     Because the answer is irrational. And what does it really mean to have an irrational line length?? How could you ever accurately draw   
 something like that?!                                                                                                                       
 \end{solutionordottedlines}                                                                                                                 
                                                                                                                                             
 \begin{exercisebox}                                                                                                                         
     \section*{Exercises}                                                                                                                    
     \begin{questions}                                                                                                                       
         \question                                                                                                                           
             A door frame has height $1.7$m and width $1$m. Will a square piece of board $2$m by $2$m fit through the doorway?               
             \begin{solutionordottedlines}[1.5in]                                                                                            
                 The diagonal of the door frame is $\sqrt{1.7^2 + 1^2} = \sqrt{2.89 + 1} = \sqrt{3.89} \approx 1.97$m, which is less than th 
 side of the board ($2$m). Therefore, the board will not fit through the doorway.                                                            
             \end{solutionordottedlines}                                                                                                     
                                                                                                                                             
             \begin{multicols}{2}                                                                                                            
                 \Question[1]\,                                                                                                              
                     \begin{center}                                                                                                          
                         \drawTriangle{bottom left}{10}{6}{8}{$6$}{$8$}{$x$}{0.5}                                                            
                     \end{center}                                                                                                            
                     \begin{solutionordottedlines}[1in]                                                                                      
                         $x = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$                                                           
                     \end{solutionordottedlines}                                                                                             
                 \Question[1]\,                                                                                                              
                     \begin{center}                                                                                                          
                         \drawTriangle{top left}{13}{5}{12}{$12$}{$5$}{$x$}{0.5}                                                             
                     \end{center}                                                                                                            
                     \begin{solutionordottedlines}[1in]                                                                                      
                         $x = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13$                                                         
                     \end{solutionordottedlines}                                                                                             
                 \Question[1]\,                                                                                                              
                     \begin{center}                                                                                                          
                         \drawTriangle{bottom right}{5}{3}{4}{$3$}{$4$}{$x$}{0.75}                                                           
                     \end{center}                                                                                                            
                     \begin{solutionordottedlines}[1in]                                                                                      
                         $x = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$                                                              
                     \end{solutionordottedlines}                                                                                             
                 \Question[1]\,                                                                                                              
                     \begin{center}                                                                                                          
                         \drawTriangle{top right}{17}{8}{15}{$x$}{$15$}{$17$}{0.4}                                                           
                     \end{center}                                                                                                            
                     \begin{solutionordottedlines}[1in]                                                                                      
                         $x = \sqrt{17^2 - 15^2} = \sqrt{289 - 225} = \sqrt{64} = 8$                                                         
                     \end{solutionordottedlines}                                                                                             
             \end{multicols}                                                                                                                 
     \end{questions}                                                                                                                         
 \end{exercisebox}                                                                                                                           
                                                                                                                                             
 \subsection*{The Converse}                                                                                                                  
 In the introduction we mentioned \emph{the converse} of the \textbf{Pythagorean Theorem}. Let us recall that the theorem states: ``in a     
 \emph{right-angled} triangle $a^2 + b^2 = c^2$, thus the converse becomes, \emph{if $a^2 + b^2 = c^2$}, then the triangle must be           
 \emph{right-angled}.                                                                                                                        
                                                                                                                                             
 \begin{examplebox}                                                                                                                          
     \section*{Examples}                                                                                                                     
     Decide if the triangles with the following side lengths are \textbf{right-angled}.                                                      
     \begin{questions}                                                                                                                       
         \Question[1] 16,30,34                                                                                                               
         \begin{solutionordottedlines}[1in]                                                                                                  
             $16^2 + 30^2 = 256 + 900 = 1156$ and $34^2 = 1156$. Since $16^2 + 30^2 = 34^2$, the triangle is right-angled.                   
         \end{solutionordottedlines}                                                                                                         
         \Question[1] 10,24,26                                                                                                               
         \begin{solutionordottedlines}[1in]                                                                                                  
             $10^2 + 24^2 = 100 + 576 = 676$ and $26^2 = 676$. Since $10^2 + 24^2 = 26^2$, the triangle is right-angled.                     
         \end{solutionordottedlines}                                                                                                         
         \Question[1] 4,6,7                                                                                                                  
         \begin{solutionordottedlines}[1in]                                                                                                  
             $4^2 + 6^2 = 16 + 36 = 52$ and $7^2 = 49$. Since $4^2 + 6^2 \neq 7^2$, the triangle is not right-angled.                        
         \end{solutionordottedlines}                                                                                                         
     \end{questions}                                                                                                                         
 \end{examplebox}