Please do q11 and q10

Hungenahalli Sitaramarao BadarinathBadarinath
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,

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