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in Olympiad by (65.3k points)

Consider the areas of the four triangles obtained by drawing the diagonals AC and BD of a trapezium ABCD. The product of these areas, taken two at time, are computed. If among the six products so obtained, two products are 1296 and 576, determine the square root of the maximum possible area of the trapezium of the nearest integer.

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by (70.9k points)
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Case-I

∴ Square root of maximum possible area is 13

by (10 points)
When area of two triangles is one
And other two is 576 and 1296 the area of trapezium will be maximum
576 ×1 is will be the product of one pair and 1296×1 will be the product of another pair.
So 576+1296+1+1=1874 will be the maximum area and answer will be root1874
by (45.9k points)
It is not possible because
area of Δ AOD = area of Δ BOC in trapezium
ABCD centered at 0.
and area of ΔDOC > area of ΔAOD > area of ΔAOB
that's why two triangle having area 1 is not possible.

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