question_answer

ABCD is parallelogram, G is the point on AB such that AG = 2GB, E is point on DC such that CE = 2DE and F is the point on BC such that BF = 2FC. Then, match the following:

	Column-I		Column-II
p	\[ar(ADEG)\]	(i)	\[\frac{1}{6}ar(ABCD)\]
Q	\[ar(\Delta EGB)\]	(ii)	\[ar(GBCE)\]
R	\[ar(\Delta EFC)\]	(iii)	\[ar(\Delta GBCE)\]
S	\[ar(\Delta EFC)\]	(iv)	\[\frac{1}{2}ar(\Delta EBF)\]

           

A)

P-i

Q-ii

R-iii

S-iv

               

B)       

P-iii

Q-i

R-iv

S-ii

               

C)       

P-iii

Q-ii

R-iv

S-i

               

D)

P-ii

Q-i

R-iii

S-iv

Correct Answer: B

Solution :

          We have, ABCD is a parallelogram, AG = 2 GB, CE = 2 DE and BF = 2 FC. Let AL be the height of ABCD. (P) \[ar(ADEG)=\frac{1}{2}(AG+DE)AL\] [\[\because \]ADEG is a trapezium] \[=\frac{1}{2}\left[ \frac{2}{3}AB+\frac{1}{3}CD \right]AL\] \[\frac{1}{2}\left[ \frac{2}{3}CD+\frac{1}{3}AB \right]AL\] [\[\because \]ABCD is a parallelogram] \[=\frac{1}{2}[CE+GB]AL=ar(CEGB)\] (Q) \[\therefore \]Since, \[AG=2\,GB.\](given) AB = AG + GB = GB +GB = 3GB \[\Rightarrow \]\[GB=\frac{1}{3}AB\] \[\therefore \]\[ar(\Delta \Epsilon G\Beta )=\frac{1}{2}(GB)(AL)=\frac{1}{2}\left[ \frac{1}{3}AB.AL \right]\] \[=\frac{1}{6}ar(ABCD)\] (r) \[\therefore \]Since, BF = 2 FC BC = BF + FC = 2 FC +FC = 3 FC  \[\Rightarrow \]\[FC=\frac{1}{3}BC\] \[\therefore \]\[ar(\Delta \Epsilon FC)=\frac{1}{2}(CF)\times height\] \[=\frac{1}{2}\left( \frac{1}{3}BC\times height \right)\] \[=\frac{1}{2}\left[ \frac{1}{2}\left( \frac{2}{3} \right)BC\times height \right]\] \[=\frac{1}{2}\left[ \frac{1}{2}BF\times height \right]=\frac{1}{2}[ar\,\Delta \Epsilon \Beta F]\] (s) \[\therefore \]\[ar(\Delta EGB)=\frac{1}{2}\times BG\times AL\] \[=\frac{1}{2}\left[ \frac{1}{3}AB\times AL \right]\]          \[\left[ \left( \because BG=\frac{AB}{3} \right) \right]\] \[=\frac{1}{2}\left[ \frac{1}{3}CD\times AL \right]\] [\[\because \]ABCD is a parallelogram] \[=\frac{1}{2}\left[ \frac{1}{3}\times 3DE\times h \right]\]             \[(\because \,CD=3DE)\] \[=\frac{1}{2}\left[ DE\times h \right]=ar(\Delta \Epsilon DG)\]