question_answer

ABCD is a rectangle E, F of are the mid points of BC and AD respectively and G is any point on EF, then\[\Delta GAB\] equals

A)  \[\frac{1}{6}\](area of rectangle ABCD)

B)  \[\frac{1}{4}\](area of rectangle ABCD)

C)  \[\frac{1}{3}\](area of rectangle ABCD)

D)  \[\frac{1}{2}\](area of rectangle ABCD)

Correct Answer: B

Solution :

AB || EF|| CD, So, ABEF is a rectangle \[\therefore \Delta AGB=\frac{1}{2}\]area of rectangle ABEF \[=\frac{1}{2}\times \left( \frac{1}{2}area\,\,of\,\,rec\tan gle\,\,ABCD \right)\] \[=\frac{1}{4}\](area of rectangle ABCD)