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DEAR JAYATI ,
First of all the incentre , orthocentre and centroid of the triangle coincides. now the strategy is that centre of the circle is the centre of square and triangle as well because all are symmetrical figures.
in triangle ABC , angle A is 30 deg as the line connecting A to centre is angle bisector as well. and also bc perpendicular to the side of the trianlge will imply that it is also the median as all points coincide. now apply
r/(a/2) = tan 30 ; sqrt(3) r = a/2 , so r = a /2*sqrt(3)
so the diagonal of the square is 2r = a/sqrt(3) = a/1.73
1.414 * side = a/1.73 ( apply pytha. theorem in square ) , side = a/1.73*1.414
so area = side *side = a2/6
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