[tex]\text{Width} ~= w ~ cm \\\\ \text{Length}~,l = (20+ w) ~ cm \\\\\text{Area of rectangle} = 1000 cm^2\\\\\\\text{Now,}\\\\l \cdot w = 1000\\\\\implies (20 + w) w= 1000\\\\\implies w^2 +20w -1000=0\\\\\\\implies w = \dfrac{-20 \pm \sqrt{20^2 - 4 \cdot 1 \cdot (-1000)}}{2 \cdot 1}\\\\\\\implies w = \dfrac{-20 \pm \sqrt{4400}}{2}= \dfrac{-20\pm 20\sqrt{11}}2 = -10 \pm 10\sqrt{11}\\\\\text{Width cannot be negative so,} \\\\w= -10+10\sqrt{11}=10\left(\sqrt{11} - 1 \right) ~ cm\\\\[/tex]
[tex]\text{And length,}[/tex]
[tex]l= 20+ 10\left(\sqrt{11} - 1} \right) = 20 + 10\sqrt{11} - 10 = 10 + 10\sqrt{11} = 10 \left(1+ \sqrt{11} \right)~ cm[/tex]
