Please do q11 and q10
Hungenahalli Sitaramarao Badarinath, former Retired
Answered Jun 29, 2017
Dear student,
Here given ABCD is a parallelogram where L and M are the midpoints of BC and CD respectively.
Draw LX parallel to AB, so X is midpoint of AD. Also draw MY parallel to AD, so Y is midpoint of AB.
Ar (triangle ALM) = Ar(ABCD) - ar(ABL) - ar(ADM) - ar(LMC)
ar(ABL) = (1/2) of parallelogram ABLX = (1/4) Ar(ABCD)
ar(ADM) = (1/2) of parallelogram AYMD = (1/4) Ar(ABCD)
ar(LMC)_= (1/2) of parallelogram OLCM = (1/8) Ar(ABCD)
Therefore, Ar (triangle ALM) = Ar(ABCD) - (1/4) Ar(ABCD) - (1/4) Ar(ABCD) - (1/8) Ar(ABCD) = Ar(ABCD)[1 - (1/4) - (1/4) - (1/8)] = Ar(ABCD)[8-2-2-1]/8
= (3/8) ar (ABCD)
Such that 8 (area(triangle ALM)) = 3 (area (ABCD)).
Hence proved
Please ask 11Q separately
Regards,