ABC\] be an isosceles triangle and Let\[AL\bot BC.\] \[\frac{1}{2}\times BC\times AL=200\,\,c{{m}^{2}}\] \[\frac{1}{2}\times 20\,\,cm\times AL=200\,\,c{{m}^{2}}\] \[AL=\frac{200\times 2}{20}=20\,\,cm\] Now, \[BL=\frac{1}{2}BC=\left( \frac{1}{2}\times 20 \right)cm=10\,\,cm\] and \[AL=20\,\,cm\] \[\therefore \]\[a=\sqrt{A{{L}^{2}}+B{{L}^{2}}}=\sqrt{{{20}^{2}}+{{10}^{2}}}cm\] \[=\sqrt{400+100}=\sqrt{500}=10\sqrt{5}cm\] Hence PERIMETER \[=\left( 10\sqrt{5}+10\sqrt{5}+20 \right)\]\[=20\sqrt{5}+20=20\left( \sqrt{5}+1 \right)cm\]