Question

ma puteti ajuta cu ex 1? va rog

Answer 1

fie DP⊥AB

    CW⊥AB

=>DP si CW inaltimi

DC=PW

AD=AM=>ΔDAM ->isoscel

fie AN⊥DM=>AN inaltime ,bisectoare,mediana,mediatoare in ΔDAM

BC=DM =6 cm=>ΔDAM ->echilateral

DN=NM=3 cm

PM=AP=3 cm

fieMQ⊥DC=>MQ->inaltime

MQ=DP=CW

DQ=PM=3 cm

In ΔDPM,m(∡P)=90°=>sinM=DP/DM=>DP=sin60°*6=√3/2*6=3√3 cm

MQ=DP=CW=3√3 cm

InΔCWB , m(∡W)=90°=>cosB=WB/CB=>WB=cos60°*6=6*1/2=3 cm

MW=QC=MB-WB=12-3=9 cm

DC=DQ+QC=3+9=12 cm

AB=AP+PM+MW+WB=3+3+9+3=18 cm

A=(DC+AB)*DP/2=(12+18)*3√3/2=30*3√3/2=15*3√3=45√3 cm²

In ΔCWM , m(∡W)=90°=>(T.P):CM=√(CW²+MW²)=√[(3√3)²+9²)=√108=6√3 cm

In ΔCQM, m(∡Q)=90°=>sin QMC=QC/MC=9/6√3=3/2√3=√3/2=>m(∡QMC)=60°

In ΔDQM , m(∡Q)=90°=>sinQMD=DQ/DM=3/6=1/2=>m(∡QMD)=30°

m(∡QMC)+m(∡QMD)=60°+30°=90°=>DM⊥MC