question_answer

If E, F, G and H are the mid-points of sides of a parallelogram ABCD then \[ar(EFGH)=\_\_\_\_.\]

A) \[\frac{1}{3}\,ar\,(ABCD)\]                     

B)        \[\,ar\,(ABCD)\] 

C)         \[\frac{1}{2}\,ar(ABCD)\]                      

D)        \[\frac{1}{4}\,ar(ABCD)\]           

Correct Answer: C

Solution :

ABCD is a parallelogram and E, F, G, H are the midpoints of ABCD. Construction: Join HF, Such that \[HF||AB||CD\] We know, if triangle and parallelogram lie between same base and parallel then area of triangle is half the area of parallelogram. \[\therefore \]\[ar(\Delta \Epsilon F\Eta )=\frac{1}{2}(ar\,ABFH)\]                 ?(i) and \[ar(\Delta GHF)=\frac{1}{2}(ar\,DHFC)\]                  ?(ii) Adding (i) & (ii), we get \[ar(\Delta \Epsilon FH)+ar(\Delta GHF)\] \[=\frac{1}{2}ar(ABFH)+\frac{1}{2}ar(DHFC)\] \[\Rightarrow \]\[ar(EFGH)=\frac{1}{2}ar(ABCD)\]