Ip.: ABCD-trapez isoscel
AB || CD
P=42 cm
AB=15 cm
CD=7 cm
Cl.: A=?
Dem:
P=AB+BC+CD+AD
ABCD-trapez isoscel =>BC=AD =>P= AB+CD+2AD => 2AD=P-(AB+CD)
=>AD=[P-(AB+CD)]/2
=[42-(15+7)]/2
=(42-22)/2
=20/2
=10 cm
Ducem CR ⊥ AB si DG ⊥ AB
CD=GR=7cm=>AG=BR=(AB-GR)/2=(15-7)/2=8/2=4cm
In ΔCRB ,m(∡CRB)=90° =>(prin teorema lui pitagora) BC²=CR²+BR²
10²=CR²+4²=>CR²=10²-4²=>CR²=100-16=>CR=√84=2√21
CR-inaltimea trapezului isoscel ABCD
A=[(B+b) h]/2=[(AB+CD)CR]/2=[(15+7)2√21]/2=(22·2√21)/2=22√21 cm²
Sper ca te-am ajutat si ca nu am gresit vreun calcul pe acolo =)!
