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Trapezul ABCD cu AB II CD, AB=48 cm si CD=30 cm ,iar diagonalele AC=52 cm si BD= 65 cm.Stiind ca AC intersectat BD={O},calculati OA, OB, OC si OD.

Va rog mult ajutati-ma la aceasta problema!


Răspuns :

AB||CD⇒ ΔAOB asemeneaΔCOD
AO/OC=OB/OD=AB/CD

AO/OC= 48/30=8/5
AO+OC/OC=8+5/5
AC/OC=13/5⇒ 52/OC=13/5⇒ OC=52*5/13= 20 cm⇒ AO=52-20=32 cm

OB/OD=48/30=8/5
OB+OD/OD=8+5/5
BD/OD=13/5
65/OD=13/5⇒ OD= 65*5/13= 25 cm⇒ OB=65-25=40 cm