Question 1067985
*[illustration FCX1.png].
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Set up a coordinate system with B at the origin ({{{0}}},{{{0}}}) and s is the length of the side of the square.
So then,
N is located at ({{{s/2}}},{{{s}}}) 
K is located at ({{{0}}},{{{s/2}}})
C is located at ({{{s}}},{{{0}}})
M is located at ({{{s}}},{{{s/2}}})
Find the line BN,
{{{m=(s-0)/(s/2-0)=2}}}
{{{y[BN]-0=2(x-0)}}}
{{{y[BN]=2x}}}
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Find the line KC,
{{{m=(s/2-0)/(0-s)=-1/2}}}
{{{y[KC]-s/2=-(1/2)(x-0)}}}
{{{y[KC]=-(1/2)x+s/2}}}
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Find the line AM,
{{{m=-1/2}}}
{{{y[AM]=-(1/2)x+s}}}
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P is the intersection of BN and KC,
{{{2x=-(1/2)x+s/2}}}
{{{(5/2)x=s/2}}}
{{{x[P]=s/5}}}
and
{{{y[P]=(2/5)s}}}
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Q is the intersection of BN and AM,
{{{2x=-(1/2)x+s}}}
{{{(5/2)x=s}}}
{{{x[Q]=(2/5)s}}}
and
{{{y[Q]=(4/5)s}}}
So then the distance PQ is,
{{{PQ^2=(s/5-(2/5)s)^2+((2/5)s-(4/5)s)^2}}}
{{{PQ^2=(-(1/5)s)^2+(-(2/5)s)^2}}}
{{{PQ^2=s^2/25+(4/25)s^2}}}
{{{PQ^2=(5/25)s^2}}}
{{{PQ=(sqrt(5)/5)s}}}
In this case,
{{{s^2=320}}}
{{{s=8*sqrt(5)}}}
So,
{{{PQ=(sqrt(5)/5)8*sqrt(5)}}}
{{{PQ=8}}}