Find Area of the shaded region | Square on Median | Medium Hard | PRMO RMO INMO NMTC NTSE IBPS

[@hmm9803]

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[@ZeeshanAhmad-pb6pr]

Nice mam

[@englishedgebys.k.vidyarthi2319]

maam please have a complete course on inmo

[@kaliprasadguru1921]

Mam , BAC is a right angled triangle .with A as 90° . Hence median AM is half of hypotenuse = 10 .Area of Triangle AMC = half of area of Triangle BAC= 1/2 X 1/2 X12X16 = 48 . Why going in for . Heron formula . Regards .

[@AshishYadav-gc4ol]

Some problems are really good

[@AshishYadav-gc4ol]

I'm enjoying

[@shivamshaw_07]

Nice video but has low views 😒

[@dainiusb1114]

BM=MA=MC=R=10. Draw perpendicular from M to AF, mark H. MH^2=10^2-8^2. tg(MAF)=MH/HA=FM/MA. S=10*10-0.5*10*MF.