Suppose point P is on AB, and there are four points on all sides in the counterclockwise direction (except the position of the point, everything else is the same).

The area of PQRS is equal to the area of rectangular ABCD minus four triangles DRS, RCQ, QDB and SAD.

The area of DRS =DR*DS/2=x(8-x)/2.

RCQ=CR*CQ/2=( 10-x)x/2

QPB=BQ*PB/2=(8-x)x/2

SAP=SA*AP/2==( 10-x)x/2

Area of ABCD = 10*8=80.

therefore

The area of PQRS = 80-[x (8-x)/2+(10-x) x/2+x (8-x)/2+(10-x) x/2]

=80-[x(8-x)+( 10-x)x]

=80-( 18x-2x square)

=2x squared-18x+80

Because it is known that the area of PQRS is the square of A=2X-18X+80.

therefore

2X squared-18X+80=2x squared-18X+80

For any x, the equation holds.

Because x is the length

So x & gt=0.

And because 8-x, 10-x is the length, >; =0

So x < =8

Therefore {x | 0