[tex]\huge\bold{Given:}[/tex]
Length of the base = 9
Length of the hypotenuse = 16 [tex]\huge\bold{To\:find:}[/tex]
✎ The length of the perpendicular ''[tex]x[/tex]".
[tex]\large\mathfrak{{\pmb{\underline{\orange{Solution}}{\orange{:}}}}}[/tex]
The length of the perpendicular [tex]x[/tex] is[tex]\boxed{±13.229}[/tex].
[tex]\large\mathfrak{{\pmb{\underline{\red{Step-by-step\:explanation}}{\orange{:}}}}}[/tex]
Using Pythagoras theorem, we have
[tex]({perpendicular})^{2} + ({base})^{2} = ({hypotenuse})^{2} \\ ⇢ {x}^{2} + {(9})^{2} = ({16})^{2} \\ ⇢ {x}^{2} + 81 = 256 \\ ⇢ {x}^{2} = 256 - 81 \\ ⇢x = \sqrt{175} \\ ⇢x = ±13.229[/tex]
[tex]\sf\blue{Therefore,\:the\:length\:of\:the\:perpendicular \:"x"\:is\:±13.229.}[/tex]
[tex]\huge\bold{To\:verify :}[/tex]
[tex]( {13.229})^{2} + ({9})^{2} = ({16})^{2} \\ ⇢ 175 + 81 = 256 \\ ⇢ 256 = 256 \\⇢ L.H.S.=R. H. S[/tex]
Hence verified. ✔
[tex]\circ \: \: { \underline{ \boxed{ \sf{ \color{green}{Happy\:learning.}}}}}∘[/tex]
