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Let A B C D be a convex cyclic quadrilateral. Suppose P is a point in the plane of the quadrilateral such that the sum of its distances from the vertices of ABCD is the least. If PA , PB , PC , PD = 3 , 4,6,8 What is the maximum possible area of A B C D ?

Correct answer

55

Step-by-step solution

P must be point of intersection of diagonals A C and B D Let APB = , then area of PAB = 1 2 3 4 area of PAD = 1 2 3 6 ( - ) area of PDC = 1 2 8 6 area of PCB = 1 2 8 4 ( - ) area of quadrilateral ABCD is aligned & 1 2 (12+18+48+32) & =(6+9+24+16) aligned Maximum area of quadrilateral ABCD is 6+9+24+16=55 .

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