Explicație pas cu pas:
A (0,4); B (-3,0); C ( 4,0 )
a)
b)
AO = 4 u.m.
BO = 3 u.m.
CO = 4 u.m.
BC = BO + CO = 3 + 4 = 7 u.m.
T.P.:
AB² = AO² + BO² = 4² + 3² = 16 + 9 = 25 = 5²
=> AB = 5 u.m.
AC² = AO² + CO² = 4² + 4² = 16 + 16 = 32
=> AC = 4√2 u.m.
P(ABC) = AB+BC+ AC = 5+7+4√2 = 12+4√2
=> P(ABC) = 4(3+√2) u.m.
c)
M mijlocul segmentului AC
[tex]x_{M} = \frac{x_{A} + x_{C}}{2} = \frac{0 + 4}{2} = \frac{4}{2} = 2\\ y_{M} = \frac{y_{A} + y_{C}}{2} = \frac{4 + 0}{2} = \frac{4}{2} = 2 \\ \implies M(2;2)[/tex]
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Explicație pas cu pas:
A (0,4); B (-3,0); C ( 4,0 )
a)
b)
AO = 4 u.m.
BO = 3 u.m.
CO = 4 u.m.
BC = BO + CO = 3 + 4 = 7 u.m.
T.P.:
AB² = AO² + BO² = 4² + 3² = 16 + 9 = 25 = 5²
=> AB = 5 u.m.
AC² = AO² + CO² = 4² + 4² = 16 + 16 = 32
=> AC = 4√2 u.m.
P(ABC) = AB+BC+ AC = 5+7+4√2 = 12+4√2
=> P(ABC) = 4(3+√2) u.m.
c)
M mijlocul segmentului AC
[tex]x_{M} = \frac{x_{A} + x_{C}}{2} = \frac{0 + 4}{2} = \frac{4}{2} = 2\\ y_{M} = \frac{y_{A} + y_{C}}{2} = \frac{4 + 0}{2} = \frac{4}{2} = 2 \\ \implies M(2;2)[/tex]