Given: A triangle ABC with a point D on AC such that
[tex]\begin{gathered} AD=\text{ 2 units} \\ CD=8\text{ units} \\ \angle BDC=90\degree \end{gathered}[/tex]
Required: To determine the length of side AB.
Explanation: In triangle ABC, applying the Pythagoras theorem as
[tex]AC^2=AB^2+BC^2[/tex]
Now,
[tex]\begin{gathered} AC=AD+CD \\ =10\text{ units} \end{gathered}[/tex]
Thus,
[tex]100=AB^2+BC^2\text{ ...}(1)[/tex]
Now, triangle ABD is a right-angled triangle with AB as the hypotenuse. Applying the Pythagoras theorem as-
[tex]\begin{gathered} AB^2=AD^2+BD^2 \\ AB^2-BD^2=4\text{ }(\because AD=2)...(2) \end{gathered}[/tex]
Similarly, triangle BDC is a right-angled triangle. Thus-
[tex]\begin{gathered} BC^2=BD^2+CD^2 \\ BC^2-BD^2=64\text{ }(\because CD=8)\text{ ...}(3) \end{gathered}[/tex]
Subtracting equations (2) and (3) as follows-
[tex]\begin{gathered} (AB^2-BD^2)-(BC^2-BD^2)=4-64 \\ AB^2-BC^2=-60\text{ }...(4) \end{gathered}[/tex]
Again adding equations (1) and (4) as follows-
[tex]\begin{gathered} (AB^2+BC^2)+(AB^2-BC^2)=100+(-60) \\ 2AB^2=40 \\ AB^2=20 \\ AB=\sqrt{20}\text{ units} \end{gathered}[/tex]
Further solving-
[tex]\begin{gathered} AB=2\sqrt{5}\text{ units} \\ AB=4.4721\text{ units} \\ AB\approx4.5\text{ units} \end{gathered}[/tex]
Final Answer: Options (a) and (c) are correct.
