Question 1088561
let the hypotenuse be {{{c}}}, the longer leg {{{a}}}, and the shorter leg {{{b}}}
if the hypotenuse of a right triangle is {{{5m}}} long, we have

{{{c=5m}}}

if the longer leg is {{{1m}}} longer than the shorter leg, we have

{{{a=b+1m}}}

since we have a right triangle, than

{{{c^2=a^2+b^2}}}.....plug in {{{c=5m}}} and {{{a=b+1m}}}

{{{(5m)^2=(b+1m)^2+b^2}}}......solve for {{{b}}}

{{{25m^2=b^2+2b*m+1m^2+b^2}}}

{{{25m^2=2b^2+2b*m+1m^2}}}

{{{2b^2+2b*m+1m^2-25m^2=0}}}

{{{2b^2+2b*m-24m^2=0}}}...divide by {{{2}}}

{{{b^2+b*m-12m^2=0}}}...factor

{{{b^2-3b*m+4bm-12m^2=0}}}

{{{(b^2-3b*m)+(4bm-12m^2)=0}}}

{{{b(b-3m)+4m(b-3m)=0}}}

{{{(b - 3m)(b + 4m) = 0}}}

solutions:
{{{(b - 3m)= 0}}}=>{{{b=3m}}}

since triangle side, we don't need negative solution

so, {{{a=b+1m}}}=>{{{a=4m}}}



 the length of the shorter side:{{{3m}}}
the length of the longer side:{{{4m}}}
the length of the hypotenuse:already given, {{{5m}}}