Consider the traingle as per the attachment
In [tex]{\triangle}[/tex]ADB. By Pythagoras theorem:
[tex]{:\implies \quad \sf x^{2}+4^{2}=AB^{2}}[/tex]
[tex]{:\implies \quad \sf AB^{2}=x^{2}+16\quad ...(i)}[/tex]
Now, In [tex]{\triangle}[/tex]ADC. By Pythagoras theorem:
[tex]{:\implies \quad \sf x^{2}+9^{2}=AC^{2}}[/tex]
[tex]{:\implies \quad \sf AC^{2}=x^{2}+81\quad ...(ii)}[/tex]
Now, In [tex]{\triangle}[/tex]BAC. By Pythagoras theorem:
[tex]{:\implies \quad \sf AB^{2}+AC^{2}=BC^{2}}[/tex]
By (i) and (ii)
[tex]{:\implies \quad \sf x^{2}+16+x^{2}+81=(BD+DC)^{2}}[/tex]
[tex]{:\implies \quad \sf 2x^{2}+97=(4+9)^{2}}[/tex]
[tex]{:\implies \quad \sf 2x^{2}=169-97}[/tex]
[tex]{:\implies \quad \sf 2x^{2}=72}[/tex]
[tex]{:\implies \quad \sf x^{2}=36}[/tex]
[tex]{:\implies \quad \sf x=\sqrt{36}=\boxed{\bf{6}}}[/tex]
Hence, Option b) is correct
Pythagoras theorem:- It states that, in a right angled triangle the square of hypotenuse is equal to the sum of the square of base and square of perpendicular
