Google Classroom
GeoGebraGeoGebra Classroom

Területek

Forrás: Let the diagonals of quadrilateral PQRS intersect at T, on the exterior, as shown.  Let ∠PTQ = α.  ∠PTS = 180° - ∠PTQ = 180° - α ... linear pair of angles  sin(180° - α) = sin(α)  area(PQRS)  = area(∆PRQ) + area(∆PRS)  = area(∆PTQ) - area(∆RTQ) + area(∆PTS) - area(∆RTS)  = 1/2(TP)(TQ)sin(α) - 1/2(TR)(TQ)sin(α) + 1/2(TP)(TS)sin(180° - α) - 1/2(TR)(TS)sin(180° - α)  = 1/2(TP)(TQ)sin(α) - 1/2(TR)(TQ)sin(α) + 1/2(TP)(TS)sin(α) - 1/2(TR)(TS)sin(α)  = 1/2[(TP)(TQ) - (TR)(TQ) + (TP)(TS) - (TR)(TS)]sin(α)  = 1/2[(TP - TR)(TQ + TS)]sin(α)  = 1/2(PR)(QS)sin(α) "Pope"