[tex]\text{Square}A\sim\text{Square}B\ in\ scale\ k,\ then\ \text{Area}A:\text{Area}B=k^2\\\\a,\ b-\text{the lenght of the sides of the squares}\\\\a:b=11:9\to A_A:A_B=\left(\dfrac{11}{9}\right)^2\\\\A_A=448\ cm^2\\\\\text{Substitute}\\\\\dfrac{448}{A_B}=\left(\dfrac{11}{9}\right)^2\\\\\dfrac{448}{A_B}=\dfrac{121}{81}\qquad|\text{cross multiply}\\\\121A_B=(448)(81)\qquad|:121\\\\A_B=\dfrac{(448)(81)}{121}\\\\A_B=b^2\to b^2=\dfrac{(448)(81)}{121}\to b=\sqrt{\dfrac{(448)(81)}{121}}[/tex]
[tex]b=\sqrt{\dfrac{(64)(81)(7)}{121}}\\\\b=\dfrac{\sqrt{(64)(81)(7)}}{\sqrt{121}}\\\\b=\dfrac{\sqrt{64}\cdot\sqrt{81}\cdot\sqrt7}{11}\\\\b=\dfrac{(8)(9)\sqrt7}{11}\\\\b=\dfrac{72\sqrt7}{11}\\\\\text{The perimeter is}\ P_B=4b\to P_B=4\cdot\dfrac{72\sqrt7}{11}=\dfrac{288\sqty7}{11}\ cm[/tex]
[tex]Used:\\\\\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}[/tex]
