➝ Hypotenuse of triangle ( a ) = 21.63 mm
➝ Hypotenuse of triangle ( b ) = 150 mm
➝ Hypotenuse of triangle ( c ) = 111.80 mm
[tex] \quad\rule{300pt}{1.5pt}\quad[/tex]
We have to find the length of hypotenuse in the given 3 triangles, which can be done by using Pythagoras theorem.
" In a right angled triangle, the square of hypotenuse side is equal to the sum of square of other two sides "
[tex] \qquad \bull \:{\pmb{\mathfrak{ h^2 = b^2 + p^2}}}[/tex]
And we have to convert the answer to the units indicated in red i.e, in mm.
Since 1cm = 10 mm, we will convert the given values of length of side in mm before putting the values in the formula
[tex] :\implies\qquad \sf{ h^2 = b^2 + p^2}[/tex]
[tex] :\implies\qquad \sf{ h =\sqrt{b^2 + p^2}}[/tex]
[tex] :\implies\qquad \sf{h=\sqrt{ (12)^2 + (18)^2 }}[/tex]
[tex] :\implies\qquad \sf{ h= \sqrt{144 + 324}}[/tex]
[tex] :\implies\qquad \sf{ h = \sqrt{468}}[/tex]
[tex] :\implies\qquad\underline{\underline{\pmb{ \sf{ h = 21.63 mm}}}}[/tex]
[tex] :\implies\qquad \sf{ h^2 = b^2 + p^2}[/tex]
[tex] :\implies\qquad \sf{h =\sqrt{b^2 + p^2} }[/tex]
[tex] :\implies\qquad \sf{ h = \sqrt{(90)^2+(120)^2}}[/tex]
[tex] :\implies\qquad \sf{ h=\sqrt{8100+14400}}[/tex]
[tex] :\implies\qquad \sf{ h =\sqrt{22500}}[/tex]
[tex] :\implies\qquad\underline{\underline{\pmb{ \sf{h = 150mm}}} }[/tex]
[tex] :\implies\qquad \sf{h^2 = b^2 + p^2 }[/tex]
[tex] :\implies\qquad \sf{ h=\sqrt{b^2 + p^2}}[/tex]
[tex] :\implies\qquad \sf{ h =\sqrt{(100)^2)+(50)^2}}[/tex]
[tex] :\implies\qquad \sf{ h=\sqrt{10000+2500}}[/tex]
[tex] :\implies\qquad \sf{ h =\sqrt{12500}}[/tex]
[tex] :\implies\qquad \underline{\underline{\pmb{\sf{h = 111.80mm}}} }[/tex]
Hypotenuse of triangle ( a ) = 21.63 mm
Hypotenuse of triangle ( b ) = 150 mm
Hypotenuse of triangle ( c ) = 111.80 mm
find the length of hypotenuse in the given 3 triangles, which can be done by using Pythagoras theorem.
[tex]h^2 = b^2 + p^2[/tex]
And we have to convert the answer to the units indicated in red i.e, in mm.
Since 1cm = 10 mm, we will convert the given values of length of side in mm before putting the values in the formula
[tex]For \: \: triangle ( a )
\qquad \sf{ h^2 = b^2 + p^2}[/tex]
[tex]\qquad \sf{ h =\sqrt{b^2 + p^2}}[/tex][tex]\qquad \sf{h=\sqrt{ (12)^2 + (18)^2 }}[/tex][tex]\qquad\sf{h=\sqrt{ (12)^2 + (18)^2 }} \\ \\ \qquad \sf{ h= \sqrt{144 + 324}} \\ \\ \qquad \sf{ h = \sqrt{468}}
\\ \\\qquad\underline{\underline{\pmb{ \sf{ h = 21.63 mm}}}} \\ \\ For \: \: triangle ( b ) \qquad \sf{ h^2 = b^2 + p^2} \\ \\ \qquad \sf{h =\sqrt{b^2 + p^2} } \\ \\ \qquad \sf{ h = \sqrt{(90)^2+(120)^2}} \\ \\ \qquad \sf{ h=\sqrt{8100+14400}} \\ \\
\qquad \sf{ h =\sqrt{22500}} \\ \\\qquad\underline{\underline{\pmb{ \sf{h = 150mm}}} } \\ \\ For \: \: triangle ( c ) \qquad \sf{h^2 = b^2 + p^2 } \\ \\ \qquad \sf{ h=\sqrt{b^2 + p^2}} \\ \\\qquad\sf{ h=\sqrt{(100)^2)+(50)^2}} \\ \\\qquad\sf{ h=\sqrt{10000+2500}} \\ \\ \qquad \sf{ h =\sqrt{12500}} \\ \\
\qquad\underline{\underline{\pmb{\sf{h = 111.80mm}}} } [/tex]