Question 642251

given:  
Two sides of a triangle have the {{{same}}}{{{ length}}}. 

let's sides be {{{a}}} and {{{b}}}....=>...{{{a=b}}}

The third side {{{c}}} measures {{{3m}}}{{{ less}}} than {{{twice}}} the common length. 

{{{c+3m=2a}}}...=>...{{{(c+3m)/2=a}}}

The {{{perimeter}}} of the triangle is {{{13m}}}.

{{{P=13m}}}

solution:

 {{{P=a+b+c}}}....since {{{a=b}}} and {{{P=13m}}}, we have

{{{13m=a+a+c}}}

{{{13m=2a+c}}}...since {{{c+3m=2a}}}, we have

{{{13m=c+3m+c}}}

{{{13m=2c+3m}}}...solve for {{{c}}}

{{{13m-3m=2c}}}

{{{10m=2c}}}

{{{10m/2=c}}}

{{{highlight(5m=c)}}}

now find {{{(c+3m)/2=a}}}

{{{(5m+3m)/2=a}}}

{{{8m/2=a}}}

{{{highlight(4m=a)}}}

now find {{{b}}}....since...{{{a=b}}}..=>...{{{highlight(4m=b)}}}