Step 1: Understanding the Concept:
There is a fundamental theorem in geometry that relates the areas of similar triangles to their corresponding sides.
Step 2: Key Formula or Approach:
If two triangles are similar, the ratio of their areas is equal to the square of the ratio of their corresponding sides.
\[ \frac{\text{Area}_1}{\text{Area}_2} = \left(\frac{\text{Side}_1}{\text{Side}_2}\right)^2 \]
Therefore, the ratio of the sides is the square root of the ratio of the areas.
\[ \frac{\text{Side}_1}{\text{Side}_2} = \sqrt{\frac{\text{Area}_1}{\text{Area}_2}} \]
Step 3: Detailed Explanation:
We are given the ratio of the areas:
\[ \frac{\text{Area}_1}{\text{Area}_2} = \frac{100}{144} \]
To find the ratio of the corresponding sides, we take the square root:
\[ \frac{\text{Side}_1}{\text{Side}_2} = \sqrt{\frac{100}{144}} = \frac{\sqrt{100}}{\sqrt{144}} = \frac{10}{12} \]
So, the ratio of the corresponding sides is 10 : 12.
Step 4: Final Answer:
The ratio of their corresponding sides is 10 : 12.
In the following figure \(\angle\)MNP = 90\(^\circ\), seg NQ \(\perp\) seg MP, MQ = 9, QP = 4, find NQ. 
Solve the following sub-questions (any four): In \( \triangle ABC \), \( DE \parallel BC \). If \( DB = 5.4 \, \text{cm} \), \( AD = 1.8 \, \text{cm} \), \( EC = 7.2 \, \text{cm} \), then find \( AE \). 
In the following figure, XY \(||\) seg AC. If 2AX = 3BX and XY = 9. Complete the activity to find the value of AC.
Activity:
2AX = 3BX (Given)
\[\therefore \frac{AX}{BX} = \frac{3}{\boxed{2}} \ \\ \frac{AX + BX}{BX} = \frac{3 + 2}{2} \quad \text{(by componendo)} \ \\ \frac{BA}{BX} = \frac{5}{2} \quad \dots \text{(I)} [6pt] \\ \text{Now } \triangle BCA \sim \triangle BYX ; (\boxed{\text{AA}} \text{ test of similarity}) [4pt] \\ \therefore \frac{BA}{BX} = \frac{AC}{XY} \quad \text{(corresponding sides of similar triangles)} [4pt] \\ \frac{5}{2} = \frac{AC}{9} \quad \text{from (I)} [4pt] \\ \therefore AC = \boxed{22.5}\]