question_answer

In \[\Delta ABC\], points P and Q are on CA and CB, respectively such that CA = 16 cm, CP = 10 cm, CB = 30 cm and CQ = 25 cm. Then,

A) \[PQ||AB\]

B) \[PQ\ne AB\]

C) \[\frac{QB}{CQ}=\frac{PA}{CP}\]

D) None of the above

Correct Answer: B

Solution :

Given, \[CQ=25\,\,cm\], \[CB=30\,\,cm\],

CP = 10 cm and CA = 16 cm

Here, \[\frac{CQ}{CB}=\frac{25}{30}=\frac{5}{6}\] and \[\frac{CP}{CA}=\frac{10}{16}=\frac{5}{8}\Rightarrow \,\,\,\frac{CQ}{CB}\ne \frac{CP}{CA}\]

\[\Rightarrow \,\,\,\frac{CB}{CQ}\ne \frac{CA}{CP}\Rightarrow \,\,\frac{CB}{CQ}-1\ne \frac{CA}{CP}-1\]

\[\Rightarrow \,\,\,\frac{CB-CQ}{CQ}\ne \frac{CA-CP}{CP}\]

\[\Rightarrow \,\,\,\frac{QB}{CQ}\ne \frac{PA}{CP}\]or  \[\frac{CQ}{QB}\ne \frac{CP}{PA}\]

Hence, by converse of basic proportionality theorem, PQ is not parallel to AB.