Let 'x' be the length of the corresponding side in the small triangle. Since both triangles are similar, we have the following equation:
[tex]\begin{gathered} \frac{\text{area of big triangle}}{area\text{ of small triangle}}=(\frac{36}{x})^2 \\ \Rightarrow\frac{81}{49}=\frac{1296}{x^2} \end{gathered}[/tex]solving for 'x', we get:
[tex]\begin{gathered} \frac{81}{49}=\frac{1296}{x^2} \\ \Rightarrow81\cdot x^2=1296\cdot49 \\ \Rightarrow81x^2=63504 \\ \Rightarrow x^2=\frac{63504}{81}=784 \\ \Rightarrow x=\sqrt[]{784}=28 \\ x=28 \end{gathered}[/tex]therefore, the length of the corresponding side in the small triangle is 28 cm.
Now, we can find the ratio of the areas using the first equation. Let A be the area of the big triangle, and le t a be the area of the small triangle, then:
[tex]\begin{gathered} \frac{A}{a}=(\frac{36}{28})^2=1.65_{} \\ \Rightarrow A=1.65\cdot a \end{gathered}[/tex]notice that the area of the big triangle is 1.65 times the area of the small triangle, thus, the ratio of the areas is:
[tex]\frac{81}{49}=\frac{1296}{784}[/tex]Finally, we have the following for the scale factor o the two corresponding sides:
[tex]\begin{gathered} 28\cdot k=36 \\ \Rightarrow k=\frac{36}{28}=\frac{9}{7} \\ k=\frac{9}{7} \end{gathered}[/tex]therefore, the scale factor of the two sides is k = 9/7
