puoi svolgere questo problema

Area quadrilatero = Area quadrato - (Area dei quattro triangoli);

Area quadrato = l^2;

Area APD = AP * PD / 2;

AP = y;  y = PD + 2 , allora PD = y - 2;

Area APD = y * (y - 2) /2 = y^2 /2 - 2y/2 = [y^2 / 2] - y;

Area AQB = AQ * BQ / 2;

AQ = l - y; BQ = x;

Area  AQB = (l - y) * x / 2 = l x/2 - x y/2;

Area BCR = CR * BR / 2;

CR = x;  BR = l - x;

Area BCR = x * (l - x) / 2 = l x / 2 - x^2 / 2;

Area  CDS = CS * DS / 2;

CS = l - x;  DS = l - (y - 2)  = l - y + 2;

Area  CDS = (l - x) * (l - y + 2) / 2 = l^2 / 2 - l y / 2 + l - l x /2 + x y/2 - x ;

Area Quadrilatero = l^2 - [y^2 / 2 - y] - [l x/2 - x y/2] - [l x / 2 - x^2 / 2] - 

  - [l^2 / 2 - l y / 2 + l - l x /2 + x y/2 - x] =

 = l^2 - y^2 / 2 + y - l x/2 + x y/2 - l x / 2 + x^2 / 2 - l^2 / 2 + l y / 2 - l +    + l x /2 - x y/2 + x = 

= l^2 - y^2 / 2 + y - l x / 2 + x^2 / 2 - l^2 / 2 + l y/2 - l + x =

= l^2 /2 - y^2 / 2 + y - l x / 2 + x^2 / 2 + + l y/2 - l + x ;

l = 10;  x = 8; y = 7;

Area ABCD = 10^2 / 2 - 7^2 / 2 + 7 - 10 * 8 / 2 + 8^2 / 2 + 10 * 7 /2 - 10 + 8 =

= 50 - 24,5 + 7 - 40 + 32 + 35 - 10 + 8 =

= 57,5.

Ciao @siria_cusini