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On the portion of the straight line x+y=2 which is intercepted b/w the axes,a square is constructed ,away from the origin ,with this portion as one of its side .if p denotes perpendicular distance of a side of this square from the origin ,then determine max value of p
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|dw:1330253164848:dw| We can calculate the side of the square as two vertices are (0,2) and (2,0) therefore the side is two units. As the intercepted triangle is equilateral the perpendicular will bisect the intercepted side. Using pythagoras theorom OP = \[\sqrt{4 - 1}\] = \[\sqrt{3}\] The max value of p is if we calculate the OP" = \[\sqrt{3} + 2\]
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