Question 906627

let's the third side be {{{a}}}  

given: each of {{{two}}}{{{ sides}}} ({{{b}}} and {{{c}}})of a triangle is {{{8in}}} longer than the third side   {{{a}}} ; that means the other two sides are same length, so
{{{b=a+8in}}} and {{{c=a+8in}}})

if the perimeter is {{{P=43in}}}, than we have


{{{P=a+b+c}}} 

{{{43in=a+b+c}}} .... substitute {{{b}}} and {{{c}}}

{{{43in=a+a+8in+a+8in}}}

{{{43in=3a+16in}}}

{{{43in-16in=3a}}}

{{{27in=3a}}}

{{{27/3in=a}}}

{{{highlight(9in=a)}}}

now find {{{b}}} and {{{c}}}

{{{b=a+8in}}} 
{{{b=9in+8in}}} 
{{{highlight(b=17in)}}} 


and 
{{{c=a+8in}}}

{{{c=9in+8in}}} 
{{{highlight(c=17in)}}} 

check:

{{{P=a+b+c}}} 

{{{43in=9in+17in+17in}}} 
{{{43in=43in}}}