In trapezul isoscel ABCD, AB//CD,AB=6 cm,CD=12 cm,masura unghiului BCD =30 grade.Calculati:a)h=? b)lungimile laturilor neparalele c)lungimile diagonalelor d)aria ABCD
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duci BM_/_CD (baza mare)
ΔDMB mas<M=90 mas<D=30 ⇒ cos<D=MD/BD=√3/2
MD=CN =CD-DM/2=CD-AB/2=12-6/2=6/2=3
CN_/_CD
3/BD=√3/2
BD=3*2/√3 rationalizam cu√3
BD=6√3/3=2√3
BD=AC=2√3
BM²=BD²-MD²=(2√3)²-3²=12-9=3⇒ BM=√3
A=(B+b)*h/2=(6+12)*√3/2=18*√3/2=9√3
ΔDMB mas<M=90 mas<D=30 ⇒ cos<D=MD/BD=√3/2
MD=CN =CD-DM/2=CD-AB/2=12-6/2=6/2=3
CN_/_CD
3/BD=√3/2
BD=3*2/√3 rationalizam cu√3
BD=6√3/3=2√3
BD=AC=2√3
BM²=BD²-MD²=(2√3)²-3²=12-9=3⇒ BM=√3
A=(B+b)*h/2=(6+12)*√3/2=18*√3/2=9√3
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