problema depaseste cl a 8-a
se foloseste formula disdtantei intre 2 puncte
AB= √((xB-xA)²+(yB-yA)²)=√12²+6²=6√5
analog, obtinem BC=√(4²+8²)=4√5
AC=√16²+2²=√260=
AB²+BC² =36*5+16*5=52*5=260=AC²
Deci ABC dtreptunghic in A deci aria ABC= cat1*cat2/2=6√5 *4√5/2=60cm²
c) evident aria ABCD =1 2Arie ABC=2*60=120cm²
b) ABCD dreptunghi ⇒diagonalele se injumatatesc⇔mijlocAC= mijlocBD
(x A+xC)/2=(xB+xD)/2
(-10+6)/2=(2+xd)/2
2+xD=-4
xD=-6
(yA+yC)=(yB+yD)/2
(4+2)/2=(10+yD)/2
4+2=10+yD
yD=-4
deci D (-6;-4)
