Răspunsuri

2014-04-24T17:30:16+03:00
In triunghiu AED, mE=90, mA=30⇒mD=60, AD=8⇒th 30-60-90 ED=4
aplici Pitagora AD²=ED²+AE²
8²=4²+AE²
AE²=64-16
AE=√48
AE=4√3
AE- INALTIMEA TRAPEZULUI 
IN ΔCBA, m B=90, mBCA=45⇒m BAC=45⇒ΔCBA- dreptunghic isoscel BC=4√3=BA
in trapezul ABCD, BA//CE⇒CE=4√3
CD=4√3+4
Aria= (BA+CD)*AE/2
ARIA= (4√3+4√3+4)*4√3/2=8√3(2√3+1)=48+8√3=8(6+√3)
Perimetrul= 4√3*3+4+8=12√3+12=12(√3+1)

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