SOLUTION
In a right angle triangle, the hypotenuse is always the side opposite the right angle
Hence, from the image given,
![\text{hypotenuse =x}](https://img.qammunity.org/2023/formulas/mathematics/high-school/sk9wu2dtn0hmqga4esevsjlhrvivsh7ilo.png)
Apply pythagoras rule i.e the square of the hypotenuse side is equal to the sum of squares of the other two sides“
we have
![x^2=3^2+(2\sqrt[]{2})^2](https://img.qammunity.org/2023/formulas/mathematics/high-school/6eth4mest2q6vf56k37hrkzvzl4n0l0pmw.png)
Simplify the expression above
![\begin{gathered} x^2=9+4(2) \\ x^2=9+8 \\ x^2=17 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/vwgg0btc9xmz0mg877vzlmz024nfr979uv.png)
Take the square root of both sides
![\begin{gathered} \sqrt[]{x^2}=\sqrt[]{17} \\ \text{Then} \\ x=\sqrt[]{17}in \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/m3hx917fbb82sttv3i155tosvddw85imbs.png)
Hence
The missig side in the triangle is √17in
Answer: √17in