Question 1083771

let the hypotenuse be {{{c}}},legs {{{a}}} and {{{b}}}


in a right triangle: {{{c^2=a^2+b^2}}}

if the hypotenuse {{{c}}} of a right triangle is{{{ twice }}}as long as one of the legs, let’s say {{{a}}}, we have {{{c=2a}}}->{{{a=c/2}}}

if the other leg {{{b=sqrt(8)}}}, than we have:


{{{c^2=(c/2)^2+(sqrt(8))^2}}}


{{{c^2=c^2/4+8}}}


{{{c^2-c^2/4=8}}}


{{{3c^2/4=8}}}


{{{3c^2=8*4}}}


{{{c^2=32/3}}}


{{{c=sqrt(32/3)}}}


{{{c=3.265986323710904}}}


{{{c=3.27}}}


->{{{a=c/2}}}->{{{a=3.27/2}}}->{{{highlight(a=1.64)}}}->other leg


{{{b=sqrt(8)}}}->{{{b=2.83}}}


check:

{{{c^2=a^2+b^2}}}

{{{(3.27)^2=(1.64)^2+(2.83)^2}}}

{{{10.6929=2.6896+8.0089}}}

{{{10.6929=10.6985}}}.........round to one decimal place

{{{10.7=10.7}}}