The two triangles are similar triangles. As such, the following is true:
[tex](\frac{\text{side length of smaller triangle}}{\text{side length of bigger triangle}})^2=\frac{Area\text{ of smaller triangle}}{\text{Area of bigger triangle}}[/tex]Thus, we first have to compute the area of the bigger triangle, as follows:
[tex]\text{Area =}\frac{1}{2}\times base\times height[/tex]Since, for the bigger triangle, base = 81cm, and height = 36cm, we have:
[tex]\begin{gathered} \text{Area =}\frac{1}{2}\times base\times height \\ \Rightarrow\text{Area =}\frac{1}{2}\times81\times36 \\ \Rightarrow\text{Area =}\frac{2916}{2}=1458 \\ \Rightarrow Area=1458cm^2 \end{gathered}[/tex]Now, we find x and y, as follows:
[tex]\begin{gathered} (\frac{x}{36})^2=\frac{54}{1458} \\ \Rightarrow(\frac{x}{36})^2=\frac{1}{27} \\ \Rightarrow\frac{x}{36}=\sqrt[]{\frac{1}{27}}=\frac{1}{\sqrt[]{27}} \\ \Rightarrow\frac{x}{36}=\frac{1}{\sqrt[]{9\times3}}=\frac{1}{\sqrt[]{9}\times\sqrt[]{3}}=\frac{1}{3\times\sqrt[]{3}} \\ \Rightarrow x=36\times\frac{1}{3\times\sqrt[]{3}}=\frac{36}{3\sqrt[]{3}}=\frac{12}{\sqrt[]{3}} \\ \Rightarrow x=\frac{12}{\sqrt[]{3}}\times\frac{\sqrt[]{3}}{\sqrt[]{3}}=\frac{12\sqrt[]{3}}{\sqrt[]{9}}=\frac{12\sqrt[]{3}}{3}=4\sqrt[]{3} \\ \Rightarrow x=4\sqrt[]{3}\text{ cm} \end{gathered}[/tex]Now, y can be obtained similarly:
[tex]\begin{gathered} (\frac{y}{81})^2=\frac{54}{1458} \\ \Rightarrow(\frac{y}{81})^2=\frac{1}{27} \\ \Rightarrow\frac{y}{81}=\sqrt[]{\frac{1}{27}}=\frac{1}{\sqrt[]{27}} \\ \Rightarrow\frac{y}{81}=\frac{1}{\sqrt[]{9\times3}}=\frac{1}{\sqrt[]{9}\times\sqrt[]{3}}=\frac{1}{3\times\sqrt[]{3}} \\ \Rightarrow y=81\times\frac{1}{3\times\sqrt[]{3}}=\frac{81}{3\sqrt[]{3}}=\frac{27}{\sqrt[]{3}} \\ \Rightarrow y=\frac{27}{\sqrt[]{3}}\times\frac{\sqrt[]{3}}{\sqrt[]{3}}=\frac{27\sqrt[]{3}}{\sqrt[]{9}}=\frac{27\sqrt[]{3}}{3}=9\sqrt[]{3} \\ \Rightarrow y=9\sqrt[]{3}\text{ cm} \end{gathered}[/tex]
