question_answer

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

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

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

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

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

Correct Answer: B

Solution :

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