Resources for Finding the Hypotenuse
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Finding the Hypotenuse Theory
![\begin{multicols}{2} \textbf{Example 1}\\ Find the exact length of the hypotenuse.\\ \begin{center} \begin{tikzpicture}[scale=0.8] \coordinate (O) at (0,0); \coordinate (A) at (0,6); \coordinate (B) at (-3,0); \draw[line width=1pt] (O)--(B)node[midway ,below] {5cm}--(A)--(O) node[midway,right] {12cm}; \pic [draw=black,line width=1pt,angle radius=0.4cm,angle eccentricity=1.6,""] {right angle=B--O--A}; \end{tikzpicture} \end{center} \textbf{Example 1 solution}\\ Let \(x=\) length of hypotenuse\\ \(\begin{aligned} \therefore x^2 & =12^2+5^2 \\ x^2 & =144+25 \\ x^2 & =169 \\ x & =\sqrt{169} \\ & =\sqrt{13 \times 13} \\ \therefore x & =13 \text{~cm} . \end{aligned}\) \columnbreak \textbf{Example 2}\\ Find the length of the hypotenuse correct to 1 decimal place\\ \begin{center} \begin{tikzpicture}[scale=0.85] \coordinate (O) at (0,0); \coordinate (A) at (0,3); \coordinate (B) at (4,0); \draw[line width=1pt] (O)--(B)node[midway ,below] {6cm}--(A)--(O) node[midway,left] {5cm}; \pic [draw=black,line width=1pt,angle radius=0.4cm,angle eccentricity=1.6,""] {right angle=B--O--A}; \end{tikzpicture} \end{center} \textbf{Example 2 solution}\\ Let \(x=\) length of hypotenuse\\ \(\begin{aligned} \therefore x^2 & =5^2+6^2 \\ x^2 & =25+36 \\ x^2 & =61 \\ x & =\sqrt{61} \\ & =7.8 \text{~cm} . \end{aligned}\) \end{multicols}](/media/k2jhvqpt/9230.png)
NSW Y8 Maths Pythagoras Finding the Hypotenuse
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Finding the Hypotenuse - Video - Finding the Hypotenuse - Example 1
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Finding the Hypotenuse - Video - Finding the Hypotenuse - Example 2
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