the area of isoceles triangle EFG …

Let the length of FG be x m.

Using Heron's formula,

s = {tex}{{13 + 13 + x} \over 2} = {{26 + x} \over 2}{/tex}

Area of triangle EFG = 60 sq.m

=>          {tex}\sqrt {\left( {{{26 + x} \over 2}} \right)\left( {{{26 + x} \over 2} - 13} \right)\left( {{{26 + x} \over 2} - 13} \right)\left( {{{26 + x} \over 2} - x} \right)} {/tex} = 60

=>          {tex}\sqrt {\left( {{{26 + x} \over 2}} \right)\left( {{{26 + x - 26} \over 2}} \right)\left( {{{26 + x - 26} \over 2}} \right)\left( {{{26 + x - 2x} \over 2}} \right)} {/tex} = 60

=>         {tex}\sqrt {\left( {{{26 + x} \over 2}} \right)\left( {{x \over 2}} \right)\left( {{x \over 2}} \right)\left( {{{26 - x} \over 2}} \right)} {/tex} = 60

=>          {tex}{x \over 2}\sqrt {\left( {{{676 - {x^2}} \over 4}} \right)} = 60{/tex}

=>          {tex}{x \over 4}\sqrt {676 - {x^2}} = 60{/tex}

=>          {tex}x\sqrt {676 - {x^2}} = 240{/tex}

Squaring both sides,

=>          {tex}{x^2}\left( {676 - {x^2}} \right) = 57600{/tex}

=>          {tex}{x^4} - 676{x^2} + 57600 = 0{/tex}

=>          {tex}{x^4} - 576{x^2} - 100{x^2} + 57600 = 0{/tex}

=>          {tex}{x^2}\left( {{x^2} - 576} \right) - 100\left( {{x^2} - 576} \right) = 0{/tex}

=>          {tex}\left( {{x^2} - 576} \right)\left( {{x^2} - 100} \right) = 0{/tex}

=>          {tex}{x^2} - 576 = 0{/tex}     and     {tex}{x^2} - 100 = 0{/tex}

=>          {tex}{x^2} = 576{/tex}     and     {tex}{x^2} = 100{/tex}

=>          {tex}x=24{/tex} m      and      {tex}x = 10{/tex} m