The area of a triangle having vertices $A(3, 0)$,$B(0, 3)$,and $C(3, 3)$ is:

  • A
    $9$
  • B
    $4.5$
  • C
    $6$
  • D
    $3$

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If $G$ is the centroid of a triangle having vertices $A(3h, 3k)$,$B(-3a, 0)$,and $C(3a, 0)$,then prove that $AB^2 + BC^2 + AC^2 = 3(GA^2 + GB^2 + GC^2)$.

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