question_answer

	Column  - I		Column  - II
P.	In \[\Delta ABC\] and \[\Delta PQR\] \[\frac{AB}{PQ}=\frac{AC}{PR}\], \[\angle A=\angle P\] \[\Rightarrow \,\,\Delta ABC\tilde{\ }\Delta PQR\]	1.	AA similarity criterion
Q.	In \[\Delta ABC\] and \[\Delta PQR\] \[\angle A=\angle P\], \[\angle B=\angle Q\]\[\Rightarrow \,\Delta ABC\tilde{\ }\Delta PQR\]	2.	SAS similarity criterion
R.	In \[\Delta ABC\] and \[\Delta PQR\]\[\frac{AB}{PQ}=\frac{AC}{PR}=\frac{BC}{QR}\] \[\Rightarrow \,\Delta ABC\tilde{\ }\Delta PQR\]	3.	SSS similarity criterion
S.	In \[\Delta ABC\], \[DE||BC\] \[\Rightarrow \,\frac{AD}{DB}=\frac{AE}{EC}\]	4.	BPT

A) P-1,      Q-2,      R-3,      S-4

B) P-2,      Q-1,      R-3,      S-4

C) P-4,      Q-3,      R-2,      S-1

D) P-4,      Q-3,      R-2,      S-4

Correct Answer: B

Solution :

(P)

Given, \[\frac{AB}{PQ}=\frac{AC}{PR}\],

\[\angle A=\angle P\]

\[\because\] \[\angle A\] is containing the sides AB and AC and is containing the sides PQ and PR.

\[\therefore \,\,\,\Delta ABC-\Delta PQR\]

[by SAS criterion of similarity]

(Q)

Given, \[\angle A=\angle P,\,\angle B=\angle Q\]

\[\therefore \Delta ABC-\angle PQR\]

[by AA criterion of similarity]

(R)

Given,

\[\frac{AB}{PQ}=\frac{AC}{PR}=\frac{BC}{QR}\]

\[\because\] Sides of the \[\Delta ABC\] and \[\Delta PQR\]are in proportion

\[\therefore \Delta ABC-\Delta PQR\]

[by SSS criterion of similarity]

(S)

Given  \[DE\,|\,\,|\,\,BC\]

\[\therefore \,\,\,\,\frac{AD}{BD}=\frac{AE}{EC}\]                      [by BPT]