math!!!help~!!!!!!!!!!!!!!!!!!!!!!!!!!!please~

Let AQ = x and AP = y. Therefore,
QB = 10-x
BR = y + 3
RC = 10 - (y+3) = 7-y
PD = 10-y
DS = x- 4
SC = 10 - (x-4) = 14 - x
Area of triangle AQP = xy/2
Area of triangle QBR = QB x BR/2 = (10-x)(y+3)/2 = (10y + 30 - xy -3x)/ 2
= 5y + 15 -xy/2 -3x/2.
Area of triangle RCS = RC x SC/2 = (7-y)(14-x)/2 = (98 - 7x -14y +xy)/2
= 49 - 7x/2 - 7y + xy/2.
Area of triangle PSD = PD x DS/2 = (10-y)((x-4)/2 = (10x -xy -40 +4y)/2
= 5x -xy/2 -20 +2y.
Therefore, sum of area of the 4 triangles
= xy/2 + 5y +15 -xy/2 -3x/2 + 49 -7x/2 -7y +xy/2 + 5x -xy/2 -20 +2y
= 15 + 49 - 20 = 44.
Therefore area of shaded region PQRS = 10 x 10 - 44 = 100 - 44 = 56.