CAT 2003 Slot 1 — QA Question 37

Consider the figure below,  

As AB || CD and ∠ABD = ∠CDB = ∠PQD = 90°

∴ ∠BAP = ∠CDP and ∠ABP = ∠DCP

∆CPD ~ ∆BPA                                                 …(AAA test)

$\therefore \frac{CP}{PB}=\frac{x}{3x}=\frac{1}{3}$

If CP = y,

PB = 3y

Now, ∆CBD ~ ∆PBQ                         …(AA test)

$\frac{CD}{PQ}=\frac{CB}{PB}=\frac{y+3y}{3y}=\frac{4}{3}$= 1 : 0.75

Hence, option (b).