[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\:15\:feet.}[/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} \\⇢ ({12 \: ft})^{2} + ( {9 \: ft})^{2} = {c}^{2} \\ ⇢ 144 \: {ft}^{2} + 81 \: {ft}^{2} = {c}^{2} \\ ⇢ 225 \: {ft}^{2} = {c}^{2} \\ ⇢ \sqrt{225 \: {ft}^{2} } = c \\ ⇢ \sqrt{15 \times 15 \: {ft}^{2} } = c \\ ⇢ 15 \: ft = c[/tex]
[tex]\sf\blue{Therefore,\:the\:length\:of\:the\:hypotenuse\:is\:15\:feet.}[/tex]
[tex]\huge\bold{To\:verify :}[/tex]
[tex]( {12 \: ft})^{2} + ( {9 \: ft})^{2} = ( {15 \: ft})^{2} \\⇝144 \: {ft}^{2} + 81 \: {ft}^{2} = 225 \: {ft}^{2} \\ ⇝225 \: {ft}^{2} = 225 \: {ft}^{2} \\ ⇝L.H.S.=R. H. S[/tex]
Hence verified.
[tex]\circ \: \: { \underline{ \boxed{ \sf{ \color{green}{Happy\:learning.}}}}}∘[/tex]
