Question

pe planul patratului ABCD,AB=12 cm.se construieste perpendiculara DE,DE=5cm.calculati distanta de la E la BC,E la AB,E la AC.

Answer 1

d (E, AB)=EA ( teorema celor 3 perpendiculare, DA⊥AB⊂(ABC))
d((E,BC)=EC(( teorema celor 3 perpendiculare, DC⊥BC⊂(ABC))
EA=EC=√(12²+5²)=13


fie AC∩BD={O}
DO⊥AC⊂(ABC)
ED⊥(ABC)⇒(T3p)EO⊥AC⇔d(E,AC)=EO
 DO=AB√2/2=12√2/2=6√2
EO= √(5²+(6√2)²)= √(25+72)=√97cm