a)
[tex]Formula~pentru~aria~trapezului:~\boxed{A_{trapez}=\frac{(B+b)\cdot h}{2}}[/tex]
Unde B=AB=18cm
b=CD=10cm
h=AD=?
Ducem CE ⊥ AB, E €AB
Dar AD ⊥ AB, D€AB
Si AB || CD
=> AECD dreptunghi
Cu AE=CD=10cm
Si AD=CE
EB=AB-AE=AB-CD=18-10=8(cm)
In tr. CEB, m(CEB)=90°=> (conform T.P.) ca CE²=BC²-EB²
CE²=10²-8²=100-64=36 =>
CE=6(cm)
=>AD=6cm
Revenim la formula ariei, acum ca stim toate necunoscutele, si le inlocuim:
[tex]A_{ABCD}=\frac{(18+10)\cdot 6}{2}=28\cdot 3=84(cm^2)[/tex]
Perimetrul lui ABCD=AB+BC+CD+AD<br />P=18+10+10+6=44(cm)
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b) AC, BD=?
In tr. DAB, m(DAB)=90° => (conform T.P.) BD²=AB²+AD²BD²=18²+6²=6²(3²+1)=6²·10=> BD=6√10 (cm)
In tr. ACD, m(ADC)=90°
=> (conform T.P.) AC²=AD²+CD²
AC²=6²+10²=2²(3²+5²)=2²·34
=> AC=2√34 (cm)
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c) A tr. ABC=?
[tex]A_{ABC}=\frac{AB\cdot CE}{2}=\frac{18\cdot 6}{2}=9\cdot 6=54~(cm^2)[/tex]
