Question

AM NEVOIE URGENT DE PROBLEMA 18.
MULTUMESC!!!​

Answer 1

Răspuns:

a)Notam AC∩BD={o}

AO=BO | 1

m<cob=90 | 2

m,aoc=180 | 3

din 1, 2, 3 =>tr. AOB-dr. isoscel in O

tr. AOB dr. in O=>din TP ca AB^2=2AO^2(AO=BO)=>AO^2=BO^2=16^2:2=>AO=8√2=BO

-||-=> DO^2=CO^2=12^2:2=√144:2=6√2=>AC=BD=14√2

,iar ABCD ortodiagonal=> Aria ABCD = diagonala1 x diagonala2:2=14√2 x 14√2:2=392:2=196cm^2,dar A ABCD = (AB+DC)x h:2=28 x h:2=196 | x 2=> 28h=392 | :28 => h=14cm

b)Aria am aflat-o la punctul a)=196cm^2

c)tr. DOA-dr. in O=>din TP ca AD^2=AO^2+Do^2=(8√2)^2+(6√2)^2=200=> AD=√200=10√2, iar AD=BC=10√2=> perimetrul=12+16+20√2=28+20√2 cm=4(7+5√2) cm

Explicație pas cu pas: