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CBSE Class 11-science Answered

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Asked by nitishkrnehu09 | 09 Dec, 2017, 12:59: PM
answered-by-expert Expert Answer
begin mathsize 16px style edge space is space fraction numerator 1 minus straight x over denominator 2 end fraction
applying space pythagoras
straight x squared equals open parentheses fraction numerator 1 minus straight x over denominator 2 end fraction close parentheses squared plus open parentheses fraction numerator 1 minus straight x over denominator 2 end fraction close parentheses squared
solving space we space get
straight x equals negative 1 plus square root of 2 space or space minus 1 minus square root of 2 space
so space straight x equals negative 1 plus square root of 2
area space of space regular space octagon space is space 2 open parentheses 1 plus square root of 2 close parentheses cross times side squared
end style
Answered by Arun | 12 Dec, 2017, 05:41: PM

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