An equilateral triangle with sides 10 inches ,is inscribed in a square ABCD in such a way that one vertex is at A ,another vertex on BC and one on CD . Find the area of square??

thanks for asking. well, first of all draw a figure and name the equilateral triangle as AEF, such that E lies on BC and F on CD. now let angle EAB= y then obviously angle FAD= 30 – y (because angle EAF=60 and angle BAD=90)

now AB=AD

so, 10cosy=10cos(30 – y)

or y=30-y

or y=15

so area of square is (10cosy)^2=100cos^2y=50(1+cos2y)=50(1+root3/2)

=25(2+root3) sq units