[tex]\huge\bold{To\:find:}[/tex]
✎ The length of the hypotenuse.
[tex]\large\mathfrak{{\pmb{\underline{\orange{Solution}}{\orange{:}}}}}[/tex]
[tex]\sf\purple{The\:length\:of\:the\:hypotenuse \:"c"\:is\:5.10\:km.}[/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} \\ ⇢ ( {5 \: km})^{2} + ( {1 \: km})^{2} = ( {c})^{2} \\ ⇢ 25 \: {km}^{2} + 1 \: {km}^{2} = {c}^{2} \\ ⇢ 26 \: {km}^{2} = {c}^{2} \\ ⇢ \sqrt{26 \: {km}^{2} } = c \\ ⇢5.0990 \: km = c \\ ⇢5.10 \: km = c[/tex]
[tex]\sf\blue{Therefore,\:the\:length\:of\:the\:hypotenuse\:is\:5.10\:km.}[/tex]
[tex]\huge\bold{To\:verify :}[/tex]
[tex]( {5 \: km})^{2} + ( {1 \: km})^{2} = ( {5.10 \: km})^{2} \\ ⇝25 \: {km}^{2} + 1 \: {km}^{2} = 26 \: {km}^{2} \\ ⇝26 \: {km}^{2} = 26\: {km}^{2} \\ ⇝L.H.S.=R. H. S[/tex]
Hence verified.
[tex]\circ \: \: { \underline{ \boxed{ \sf{ \color{green}{Happy\:learning.}}}}}∘[/tex]
