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2014-05-14T16:10:21+03:00
ABCD dreptunghi
AC , BD diag.=12

ungh.COB si DOA=60 grade⇒ung.DOC si AOB=120

AO=OC=AC/2=6
DO=OB=DB/2=6
 in ΔCOB
CO≡OB
ung. COB=60  ⇒ΔCOB=echilateral⇒CB=AD=6

ΔABC dr inB
⇒T.P  CB ^{2} +AB ^{2} =AC ^{2}
  36+AB ^{2} =144
AB=6 \sqrt{3}

P=2(L+l)
P=2(6 \sqrt{3} +6)
P=12 \sqrt{3} +12
P=12( \sqrt{3} +1)