Question 1124892

let's sides be {{{a}}},{{{b}}}, and {{{c}}} where {{{c}}} is longest

if one side of a triangle is {{{one-third}}} the longest side, we have

{{{a=c/3}}}....eq.1


if the third side is {{{13}}} feet less than the longest side, we have

{{{b=c-13}}}.......eq.2

 
if the perimeter is {{{85}}}, we have

{{{a+b+c=85}}}....eq.3 .............substitute {{{a}}} and {{{b}}}


{{{c/3+c-13+c=85}}}...solve for {{{c}}}

{{{c+3c-13*3+3c=85*3}}}

{{{7c-39=255}}}

{{{7c=255+39}}}

{{{7c=294}}}

{{{c=294/7}}}

{{{c=42}}}

now find {{{a}}} and {{{b}}}


{{{a=c/3}}}->{{{a=42/3}}}->{{{a=14}}}

{{{b=c-13}}}->{{{b=42-13}}}->{{{b=29}}}


so, your sides are: {{{highlight(a=14)}}},{{{highlight(b=29)}}}, and {{{highlight(c=42)}}}


check the perimeter:

{{{a+b+c=85}}}

{{{14+29+42=85}}}

{{{85=85}}}