Consider the triangle BDC.
Determine the length of side BD by using pythagoras theorem.
[tex]\begin{gathered} (BD)^2=(x)^2-(3)^2 \\ =x^2-9 \end{gathered}[/tex]Consider triangle ABC.
Determine the length of side AB by using pythagoras theorem.
[tex]\begin{gathered} (AB)^2=(27)^2-x^2 \\ =729-x^2 \end{gathered}[/tex]Consider the triangle ABD.
Determine the length of side BD by using the pythagoras theorem.
[tex]\begin{gathered} (BD)^2)=(AB)^2-(AD)^2 \\ =729-x^2-(24)^2 \\ =153-x^2 \end{gathered}[/tex]So,
[tex]\begin{gathered} 153-x^2=x^2-9 \\ 153+9=2x^2 \\ x^2=\frac{162}{2} \\ x=\sqrt[]{81} \\ =\pm9 \end{gathered}[/tex]The value of length can never be negative. So x = 9.
Thus length of side BC is 9 units.
