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)
\[\therefore \frac{AX}{BX} = \frac{3}{\boxed{2}} \ \\ \frac{AX + BX}{BX} = \frac{3 + 2}{2} \quad \text{(by componendo)} \ \\ \frac{BA}{BX} = \frac{5}{2} \quad \dots \text{(I)} [6pt] \\ \text{Now } \triangle BCA \sim \triangle BYX ; (\boxed{\text{AA}} \text{ test of similarity}) [4pt] \\ \therefore \frac{BA}{BX} = \frac{AC}{XY} \quad \text{(corresponding sides of similar triangles)} [4pt] \\ \frac{5}{2} = \frac{AC}{9} \quad \text{from (I)} [4pt] \\ \therefore AC = \boxed{22.5}\]
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