Question

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Answer 1

A=l²=6²=36 cm²

P=4l=4*6=24 cm

DM=MC=DC/2=6/2=3 cm

In ΔADM , m(∡D)=90°=> (Teorema lui Pitagora):AM²=AD²+DM²=6²+3²=36+9=45=>AM=√45=3√5 cm

MP=MA/2=3√5/2 cm

P=4*MP=4*3√5/2=6√5 cm

MN∩PQ={O}

MN=AB=6 cm

MO=MN/2=6/2=3 cm

In Δ MOP, m(∡O)=90°=> (Teorema lui Pitagora):PO²=MP²-MO²=(3√5/2 )²-3²=45/4-9=9/4 => PO=√(9/4)=3/2 cm

PQ=2PO=2*3/2=3 cm

A_MPNQ=PO*MN/2=3*6/2=9 cm²

A_hasurat=2*A_MPNQ=9*2=18 cm² (1)

A_nehasurat=A-A_hasurat=36-18=18 cm² (2)

Din (1) si (2)=>A_hasurat=A_nehasurat=18 cm²

DN=AM=3√5 cm

AN=DM=3 cm

In Δ ADN , m(∡A)=90°=> tg D=AN/AD=3/6=1/2=>tg B=30°

AM=MB=>ΔAMB-> isoscel , MN-> inaltime , bisectoare,mediana,mediatoare

In ΔMNA , m(∡N)=90°=>sin AMN=AN/AM=3/3√5=√5/5

sin AMB=2*sin AMN=2√5/5

Teorema lui Pitagora:Suma patratelor lungimilor catetelor este egala cu patratul lungimii ipotenuzei