Question

În atasament Subiectul 3 problema 1

Answer 1

In ΔABC , m(∡A)=90°=>(Teorema lui Pitagora):BC=√(AB²+AC²)=√(12²+16²)=√(144+256)=√400=20 cm

NC=AB-BM-MN=20-6-4=10 cm

P=AB+BC+AC=12+16+20=48 cm

fie MW⊥AB=>MW inaltime in ΔAMB d(M,AB)=MW

In ΔABC , m(∡A)=90°=>sinB=AC/BC=16/20=4/5

In ΔABC , m(∡A)=90°=>sinC=AB/BC=12/20=3/5

fie NP⊥AC=>NP inaltime in ΔANC

In ΔNPC , m(∡P)=90°=>sinC=NP/NC=>NP=NCsinC=10*3/5=6 cm

In ΔMWB, m(∡W)=90°=>sinB=MW/BM=>MW=sinB*BM=6*4/5=24/5 cm

d(M,AB)=MW=24/5 cm

At=AB*AC/2=12*16/2=96 cm²

A_violete=AB*MW/2=12*24/5/2=144/5 cm²

A-narcise=AC*NP/2=16*6/2=48 cm²

96 cm²...........................100%

144/5 cm².....................x%

x=144/5*100/100/96=0,3=30 %  violete

96 cm²...............100%

48 cm².................y%

y=48*100/100/96=0,,5=50% narcise

100%-50%-30%=20% zambile