Question 322348
{{{drawing(300,300,-10,10,-10,10,line(-5,-5,-5,5),line(-5,5,5,5),line(5,5,5,-5),line(5,-5,-5,-5), locate(-5,-5,A),locate(-5,6,B),locate(5,6,C),locate(5,-5,D),circle(5,0,0.2),locate(5.5,0.5,P),circle(0,-5,0.2),locate(0.5,-5,Q),
line(0,-5,-5,5),line(0,-5,5,0))}}}
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Let the side of the square equal {{{s}}}.
The area of QBCP plus the area of triangle ABQ and the area of triangle PDQ equal the area of the square ABCD.
The area of the triangle is,
{{{A=(1/2)bh}}}
ABQ:{{{A1=(1/2)(s/2)(s)=s^2/4}}}
PDQ:{{{A2=(1/2)(s/2)(s/2)=s^2/8}}}
QBCP:{{{15}}}
ABCD:{{{s^2}}}
Put it all together,
{{{s^2/4+s^2/8+15=s^2}}}
{{{2s^2+s^2+120=8s^2}}}
{{{5s^2=120}}}
{{{s^2=24}}}
{{{s=2sqrt(6)}}}