florencejennifer wrote:There is an error in the construction of the question, if the rotation is made at the center of the square c = c' and the difference is zero. If the rotation is made at any vertex then it makes some change. This is not a question to work through conventional methods it chould be done by approximation.
In general, since each side of a square is of length 5 units, the diagonal is of length 5√2.
The point C is center on the diagonal, when the square is rotated 450 the point C is moved to the point C' which is obviously on the side of the square as shown in the figure .
So CB = 5√2/2 and C'B = 5√2/2 , CA = 5/2 and AC' = (5√2/2) - 5/2.
By Pythagoras theorem CC' = (5/2)( √(4-2√2)).
But by plugging method we can understand that the value of CC' lies between 5/2 and 5/√2.
I think that question said: If the square of side 5 units is rotated by 45 degrees about one of its corners, how far would be its new center from the old one.
because, as jennifer pointed out, rotation about the center would hardly make any difference.
Approach:
we know that the diagonal of a square of 'a' units is a√2 and the center is at a distance of a/√2 from the square's corner.
So, if we rotate the square about one of its corners, the new center would still be at a distance of a/√2 units from the corner about which it is rotated.
(imagine making an arc of radius a/√2 units and of 45 degrees measure)
now in that arc, the chord making 45 degrees will be the distance between the two centers.
Drop a perpendicular from the rotated corner on the chord and you get
1/2 (chord distance) = Sin(45/2) * a/√2
=> chord distance/ distance between the centers = Sin 22.5 * a/√2
