In an equilateral triangle ABC, D is the midpoint of side BC. If the length of BC is 8 cm, then the height of the triangle is:

$4\sqrt{3}$ cm

⇒ The median of an equilateral triangle bisects the base and makes \({90}^\circ\) so,

So, BD = DC = \(\frac{B}{2}\) = 4 cm.

Now,

\( {AD }^{2 } \) + \( {DC }^{2 } \) = \( {AC }^{2 } \)

⇒ \( {AD }^{2 } \) + \( {4 }^{2 } \) = \( {8}^{2 } \)

⇒ \( {AD }^{2 } \) + 16 = 64

⇒ \( {AD }^{2 } \) = 64 -16

⇒ \( {AD }^{2 } \) = 48

⇒ AD = 4\(\sqrt {3 }\) cm

Therefore, the height is 4\(\sqrt {3 }\) cm