Respuesta :

SOLUTION

Using Pythagoras theorem,

[tex]\begin{gathered} 8^2\text{ = ( 4}\sqrt[]{3})^2+h^2 \\ 64=48+h^2 \\ h^2\text{ = 64 - 48} \\ h^2\text{ = 16} \\ \text{square root both sides , we have :} \\ h\text{ = 4} \end{gathered}[/tex]

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