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ABCD is a quadrilateral whose diagonals AC and BD intersect at O. If triangles AOB and COD have areas 4 and 9 respectively, then the minimum area that ABCD can have is
ABCD is a quadrilateral whose diagonals AC and BD intersect at O. If
triangles AOB and COD have areas 4 and 9 respectively, then the minimum
area that ABCD can have is
It is given that ABCD is a quadrilateral whose diagonals AC and BD intersect at O.
Now, Triangles AOB and COD have areas 4 and 9, respectively.
We need to find out the minimum area that ABCD can have is
We know ar ABCD = ar AOB + ar COD + ar BOC + ar AOD
In any quadrilateral, when the diagonals are drawn, they divide the quadrilateral into four triangles. The product of the areas of two opposite triangles is equal to the product of the areas of the other two opposite triangles
Thus, (ar AOB)*(ar COD) = (ar BOC)*(ar AOD)
(ar BOC)*(ar AOD) = 4*9 = 36
The sum has to be minimum, and the product of ar BOC and ar AOD is 36.
This will happen when both areas are equal
Thus, ar BOC = 6 = ar AOD
Therefore, the minimum area that ABCD can have is = 4+9+6+6 = 25
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