In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. If $AB = 2\sqrt{10}$ and $AM = 5$,find $CM$.

  • A
    $5$
  • B
    $10$
  • C
    $15$
  • D
    $3$

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