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 In today's lesson we are going to begin by reviewing \textbf{Pythagoras' Theorem} which you should have been introduced to in Year 8, and   
 then delve deeper. We will examine how ``\emph{irrational}" numbers pop out, what the significance of such numbers are, and what the conver 
 of this theorem states.                                                                                                                     
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 We will also learn the arithmetic of surds: i.e. what can be done with $ \sqrt{8} + 2\sqrt{2} $ and $ \sqrt{3} \div \sqrt{6} $.             
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 Next week, we shall tackle more complex and less trivial problems of \emph{rationalising the denominator} and \emph{3D trigonometry}!       
                                                                                                                                             
 \begin{fillin}                                                                                                                              
 Pythagoras' Theorem states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle)  
 equal to the sum of the squares of the lengths of the other two sides. This can be written as $a^2 + b^2 = c^2$.                            
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 \begin{solutionordottedlines}                                                                                                               
 For the expression $ \sqrt{8} + 2\sqrt{2} $, we can simplify $\sqrt{8}$ to $2\sqrt{2}$ because $8$ is $4 \times 2$ and $\sqrt{4}$ is $2$. S 
 the expression becomes $2\sqrt{2} + 2\sqrt{2} = 4\sqrt{2}$.                                                                                 
                                                                                                                                             
 For the expression $ \sqrt{3} \div \sqrt{6} $, we can simplify by dividing both the numerator and the denominator by $\sqrt{3}$, which give 
 us $\sqrt{3} \div \sqrt{6} = \sqrt{3} \div (\sqrt{3} \times \sqrt{2}) = 1/\sqrt{2}$. To rationalize the denominator, we multiply both the   
 numerator and the denominator by $\sqrt{2}$, which gives us $\sqrt{2}/2$.                                                                   
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