Giải thích các bước giải:

Ta có:

$MA=MB=\dfrac12AB=\dfrac32a$

$DM=\sqrt{AD^2+AM^2}=\sqrt{(3a)^2+(\dfrac32a)^2}=\dfrac{3a\sqrt5}2$

$CM=\sqrt{CB^2+BM^2}=\sqrt{(3a)^2+(\dfrac32a)^2}=\dfrac{3a\sqrt5}2$

$CD=3a$

Gọi $E$ là trung điểm $CM$

$\to DE=\sqrt{\dfrac{DM^2+DC^2}2-\dfrac{CM^2}4}$

$\to DE=\sqrt{\dfrac{(\dfrac{3a\sqrt5}2)^2+(3a)^2}2-\dfrac{(\dfrac{3a\sqrt5}2)^2}4}$

$\to DE=\dfrac{3a\sqrt{13}}4$

$\to |\vec{DC}+\vec{DM}|=|2\vec{DE}|=2DE=\dfrac{3a\sqrt{13}}2$