show that of all the rectangles inscribed in a given fixed circle the square has the maximum area
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Area of Any rectangle (of which square is a special case) can be represented as are of two triangles in which the diagonal divides it. Here, since the area is right-angled, we can apply Pythagoras theorem.
a^2 + b^2 = c^2 ( a and b are the sides and c is the diagonal.)
Diagonals are nothing but chord in the circle. Since diameter is the longest chord, c^2 will be maximum when the diagonal passes through the centre of circle. This is the case when it is a square.
Since c^2 is max, c will be max. So a^2 + b^2 will be max. Therefore a and b will be max, therefore a x b will be max. Hence max area.
a^2 + b^2 = c^2 ( a and b are the sides and c is the diagonal.)
Diagonals are nothing but chord in the circle. Since diameter is the longest chord, c^2 will be maximum when the diagonal passes through the centre of circle. This is the case when it is a square.
Since c^2 is max, c will be max. So a^2 + b^2 will be max. Therefore a and b will be max, therefore a x b will be max. Hence max area.
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