To find the value of x we can apply Pythagoras theorem in this right angled triangle.
So let's apply!
[tex]{ \boxed{ \sf{ \mapsto {Hypotenuse}^{2} ={Base }^{2} +{Perpendicular }^{2} }}}[/tex]
[tex] \sf{ : \implies {x}^{2} = {9}^{2} + {3}^{2} }[/tex]
[tex] \sf{ : \implies {x}^{2} = 81 + 9 }[/tex]
[tex] \sf{ : \implies {x}^{2} = 90 }[/tex]
[tex] \sf{ : \implies {x} = \sqrt{2 \times 3 \times 3 \times 5} }[/tex]
[tex] \sf{ : \implies {x} = 3 \sqrt{10} }[/tex]
Hence this is the value of x.
