[tex]\huge\bold{Given:}[/tex]
Length of the base = 16 km.
Length of the hypotenuse = 34 km. [tex]\huge\bold{To\:find:}[/tex]
✎ The length of the missing leg ''[tex]a[/tex]".
[tex]\large\mathfrak{{\pmb{\underline{\orange{Solution}}{\orange{:}}}}}[/tex]
The length of the missing leg "a" is[tex]\boxed{30\: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} \\ ⇢ {a}^{2} + ({16 \: km})^{2} = ({34 \: km})^{2} \\ ⇢ {a}^{2} + 256 \: {km}^{2} = 1156 \: {km}^{2} \\ ⇢ {a}^{2} = 1156 \: {km}^{2} - 256 \: {km}^{2} \\ ⇢ {a}^{2} = 900 \: {km}^{2} \\ ⇢a \: = \sqrt{900 \: {km}^{2} } \\ ⇢a = \sqrt{30 \times 30 \: {km}^{2} } \\ ⇢a = 30 \: km[/tex]
[tex]\sf\blue{Therefore,\:the\:length\:of\:the\:missing\:leg\:"a"\:is\:30\:km.}[/tex]
[tex]\huge\bold{To\:verify :}[/tex]
[tex]( {30 \: km})^{2} + ({16 \: km})^{2} =( {34 \: km})^{2} \\ ⇝900 \: {km}^{2} + 256 \: {km}^{2} = 1156 \: {km}^{2} \\⇝1156 \: {km}^{2} = 1156 \: {km}^{2} \\ ⇝L.H.S.=R. H. S[/tex]
Hence verified. ✔
[tex]\circ \: \: { \underline{ \boxed{ \sf{ \color{green}{Happy\:learning.}}}}}∘[/tex]
