Question

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Answer 1

Fie DF⊥AB=>DF->inaltime=>DF=DA

In ΔACB =>AB²=BC²+AC²<=>25²=20²+15²<=>625=400+225<=>625=625 (A)=>(Reciproca teoremei lui Pitagora):ΔACB->dreptunghic , m(∡C)=90°=>AC⊥BC

In ΔACB , m(∡C)=90°=>sinB=AC/AB=20/25=4/5

In ΔACB , m(∡C)=90°=>cosB=BC/AB=15/25=3/5

In ΔCFB , m(∡F)=90°=>sinB=CF/BC=>CF=BCsinB=15*4/5=3*4=12 cm=>CF=DA=12 cm

In ΔCFB , m(∡F)=90°=>cosB=FB/BC=>FB=BCcosB=15*3/5=3*3=9 cm=>FB=9 cm

AB=AF+FB=>AF=AB-FB=25-9=16 cm=>DC=AF=16 cm

A=(AB+DC)*AD/2=(25+16)*12/2=41*6=246 cm²=>A=246 cm²

tgB=sinB/cosB=4/5/3/5=4/3

In ΔPAB , m(∡A)=90°=>tgB=PA/AB=>PA=ABtgB=25*4/3=100/3 cm=>PA=100/3 cm

A_PAB=PA*AB/2=100/3 *25/2=250/6=125/3 cm²=>A_PAB=125/3 cm²