Re: In the figure above, ABCD is a square with sides equal to
[#permalink]
19 Sep 2017, 08:10
If the side of the square is equal to 1 its diagonal is \(sqrt(2)\), which coincides with the rhombus main diagonal.
To find the smaller one, I exploit the fact that DE = 1 since it is a radius of the semicircle so as AD, side of the square. Thus, the smaller diagonal is equal to the diagonal of the square minus twice DF, i.e. \(EF=sqrt(2)-2*(sqrt(2)-1)\).
Given the two diagonals we can compute the area of the rhombus as \(\frac{sqrt(2)*[sqrt(2)-2*(sqrt(2)-1]}{2}=sqrt(2)-1\). Thus, the answer is B!