En trapezoide abcd

Solución: AE es perpendicular a BC, AE = 4, BE =√(AB^2-AE^2)=3. Salida

AB = CD = 5, trapezoide trapezoide isósceles ∠ABE = ∠C, AD = BC-2BE = 7

AD + DP = X, PD = X-7; = CD - PD = 5 - (X-7) = 12-X

PF es perpendicular a BC

∵∠ABE = ∠C, ∠AEB = ∠PFC = 90° .

∴⊿AEB∽⊿PFC, AE/PF = AB/PC 4/PF = 5/(12-X), PF = (48-4X)/5.

ABCD trapezoidal = (AD + BC) * AE / 2 = (700 +13) * 4/2 = 40

S⊿BCP = BC * PF / 2 = ( 312-26X)/5.

Por lo tanto: Y = S trapezoide ABCD-S⊿BCP = 40 - (312-26X)/5

Es: Y = (26/5)X-112/5 . (7≤X≤12)