a) Verific daca: AB²=AC²+BC²
(25 cm)²=(20 cm)²+(15 cm)²
625 cm²=400 cm²+225 cm²
625 cm²=625 cm² (adevarat)
Din R.T.P. (Reciproca teoremei lui Pitagora)=>ΔACB e dreptunghic, m(∡ACB)=90°=>AC⊥BC
b) Fie CE⊥AB, E∈(AB).
CE║AD, CE=AD
m(∡CEA)=90°
CEAD dreptunghi=>CD║AE si CD=AE
Atrapez=(B+b)*h/2
In ΔACB, m(∡C)=90 din T.C. (Teorema catetei)=>AC²=AE*AB=>AE=AC²/AB=20 cm*20 cm/25 cm=4 cm*4=16 cm
=>BE=AB-AE=25 cm-16 cm=9 cm
In ΔACB, m(∡C)=90° din T.I. (Teorema inaltimii)=>CE²=AE*BE=>CE=√(AE*BE)=√(16*9) cm=4*3 cm=12 cm
A trapez=(AB+CD)/2*AD=(25 cm+16 cm)/2*12 cm=41 cm*6 cm=246 cm²
c) AD∩BC={P}
CD║AB din T.F.A.=>ΔPDC~ΔPAB=>PD/PA=PC/PB=CD/AB
16 cm/25 cm=PD/(PD+AD)=PC/(PC+BC)
16/25=PD/(PD+AD)
25PD=16(PD+AD)
25PD=16PD+16AD
9PD=16AD=>PD=16AD/9=16*12 cm/9=16*4 cm/3=64 cm/3
AΔPDC=(c1*c2)/2=(PD*DC)/2=64 cm/3*16 cm*1/2=64/3*8 cm²=512 cm²/3
AΔPAB=AΔPDC+Atrapez=512 cm²/3+246 cm²=(512 cm²+738 cm²)/3=1250 cm²/3=416,(6)≅417 cm²
