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}\xa0= 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}\xa0= 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}\xa0= 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}\xa0m and {tex}x = 10{/tex}\xa0m\xa0