Question 1127404
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Let x be the common length of the square and the rhombus.


Use the formula for the rhombus area 

    Rhombus area = {{{x*x*sin(alpha)}}} = {{{x^2*sin(alpha)}}},

where {{{alpha}}} is the angle between (any) two adjacent sides of the rhombus.



Comparing with the formula for the square area, you can conclude that {{{sin(alpha)}}} = {{{1/2}}};

hence,  the angle   {{{alpha}}}  is either  30° or 150°, which geometrically represent the same rhombus.



Then the longer diagonal of the rhombus is  ( the law of cosine with cos(150°) = {{{-sqrt(3)/2}}} )

    {{{sqrt(x^2 + x^2 - 2*x*x*cos(150^o))}}} = {{{sqrt(2x^2 - 2x^2*(-(sqrt(3)/2)))}}} = {{{x*sqrt(2+2*(sqrt(3)/2))}}} = {{{x*sqrt(2 + sqrt(3))}}}.


The shorter diagonal is ( the law of cosine with cos(30°) = {{{sqrt(3)/2}}} )

    {{{sqrt(x^2 + x^2 - 2*x*x*cos(30^o))}}} = {{{sqrt(2x^2 - 2x^2*(sqrt(3)/2))}}} = {{{x*sqrt(2-2*(sqrt(3)/2))}}} = {{{x*sqrt(2 - sqrt(3))}}}.



Thus the ratio of the longer diagonal length to the shorter diagonal length is  {{{sqrt(2+sqrt(3))/sqrt(2-sqrt(3))}}}.


You can rationalize this fraction further

    {{{sqrt(2+sqrt(3))/sqrt(2-sqrt(3))}}} = {{{sqrt(2+sqrt(3))/sqrt(2-sqrt(3))}}}.{{{sqrt(2+sqrt(3))/sqrt(2+sqrt(3))}}} = {{{sqrt((2+sqrt(3))^2)/sqrt(2^2 - (sqrt(3))^2)}}} = {{{(2+sqrt(3))/(4-3)}}} = {{{2+sqrt(3)}}} = 3.732 (approximately).
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Solved.



Nice solution to a nice problem.