[tex]\huge\bold{To\:find:}[/tex]
✎ The length of the hypotenuse ''[tex]x[/tex]".
[tex]\large\mathfrak{{\pmb{\underline{\orange{Solution}}{\orange{:}}}}}[/tex]
[tex]\sf\purple{The\:length\:of\:the\:hypotenuse \:"x"\:is\:10.}[/tex]
[tex]\large\mathfrak{{\pmb{\underline{\red{Step-by-step\:explanation}}{\orange{:}}}}}[/tex]
Using Pythagoras theorem, we have
[tex] ( {hypotenuse})^{2} = ( {perpendicular})^{2} + ( {base})^{2} \\ ⇢ {x}^{2} = {6}^{2} + {8}^{2} \\ ⇢ {x}^{2} = 36 + 64 \\ ⇢ {x}^{2} = 100 \\ ⇢ x = \sqrt{100} \\ ⇢x = \sqrt{10 \times 10} \\ ⇢x = \sqrt{ ({10})^{2} } \\ ⇢x = 10[/tex]
[tex]\sf\blue{Therefore,\:the\:length\:of\:the\:hypotenuse\:is\:10.}[/tex]
[tex]\huge\bold{To\:verify :}[/tex]
[tex] {x}^{2} = {6}^{2} + {8}^{2} \\⇝ ({10})^{2} = 36 + 64 \\ ⇝10 \times 10 = 100 \\ ⇝100 = 100 \\ ⇝L.H.S.=R. H. S[/tex]
Hence verified. ✔
[tex]\circ \: \: { \underline{ \boxed{ \sf{ \color{green}{Happy\:learning.}}}}}∘[/tex]
