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Given that, 

  • $\text{AB} = 6$ cm
  • $\text{BC} = 8$ cm
  • $\angle \text{ABC} = 90^\circ$

In $\triangle \text{ABC},$ we can apply the Pythagorean theorem.

$(\text{AC})^{2} = (\text{AB})^{2}+(\text{BC})^{2}$

$\Rightarrow (\text{AC})^{2} = 6^{2}+8^{2}$

$\Rightarrow (\text{AC})^{2} = 36+64$

$\Rightarrow (\text{AC})^{2} = 100$

$\Rightarrow \boxed{\text{AC} = 10\;\text{cm}}$

$\text{D}$ is the mid point of $\text{AC}$, thus $\text{AD = DC} = \frac{10}{2}=5$ cm

Also, $(\text{BD})^{2} = (\text{AD})\times(\text{DC})$ 

$\Rightarrow (\text{BD})^{2} = 5 \times 5 = 25$

$\Rightarrow \boxed{\text{BD} = 5\;\text{cm}}$

$\therefore$ The length of $\text{BD}$ is $5\;\text{cm}.$

Correct Answer $: 5$

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