AJIT AskiitiansExpert-IITD
Last Activity: 14 Years ago
Dear Shubham , 
Please refer to the figure above.
area of parallelogram AFDE = base * length of altitude
= AF *AE sint = hksint
Area of triangle ECD = 1/2*(EC)*ED = 1/2*(m-k)*h*sint
Area of triangle FDB = 1/2*(FB)*FD = 1/2*(l-h)*k*sint
Summing the area to area of triangle ABC
(1/2)*lmsint =hksint +(1/2)*(m-k)*h*sint +(1/2)*(l-h)*k*sint
(1/2)*lm = hk +(1/2)*(m-k)*h+(1/2)*(l-h)*k
using equations we get , h = (lm - lk/m)
therefore area of AFDE = (lm - lk/m)*k sint
to maximize this area differentiate this area w.r.t to k
d(area)/dt = 0 will give k = m/2 and so h = l /2 .
therefore the points chosen are the mid points of the triangle.
therefore area of parallelogram is half the area of triangle .
therefore first find the area of traingle by matrix method and then half it ,
you will get the result .
GOOD luck !!!
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