Please help..

Asked by nitishkrnehu09 | 9th Dec, 2017, 12:59: PM

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 | 12th Dec, 2017, 05:41: PM