0 votes
in JEE by (15 points)
recategorized by

Parallelogram \( =A F E D \). 6. If \( E, E, G \) and \( H \) are respectively the midpoints of the sides of a Parallelogram \( A B C D \), show that [wr (E FCAH) \( =\frac{1}{2} \operatorname{ar}(A B C D) \) \[ \operatorname{arc}\left(E F(G H)=\frac{1}{\theta} \operatorname{ar}(A B C D)\right. \]

Please log in or register to answer this question.

1 Answer

+1 vote
by (42.4k points)

Construction: Join E & G.

\(\because \) AB = CD (opposite sides of parallelogram)

\(\frac{AB}2 = \frac{CD}2\) 

⇒ BE = GC (\(\because \) E is mid-point of AB & G is mid-point of CD)

Also BE || GC (\(\because \) AB || CD)

\(\therefore\) EBCG is a parallelogram.

Now, since, parallelogram EBCG & triangle △EFG are on the same base EG & between same parallels EG & BC.

\(\therefore\) ar(△EFG) = \(\frac12\)ar(EBCG)-----(i)

Similarly, ar(△EHG) = \(\frac12\)ar(EGDA)----(ii)

Adding equations (i) & (ii), we get

ar(△EFG) + ar(△EHG) = \(\frac12\)ar(EBCG) + \(\frac12\)ar(EGDA)

⇒ ar(EFGH) = \(\frac12\)ar(ABCD)

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.