The figure for the triangle is,
Consider the triangle ABC and triangle ABD.
[tex]\begin{gathered} \angle ABD=\angle\text{ABC (Common angle)} \\ \angle ADB=\angle BAC\text{ (Each angle 90 degr}e\text{)} \\ \Delta ABD\cong\Delta CBA \end{gathered}[/tex]So ratio of corresponding side is equal.
[tex]\frac{AB}{CB}=\frac{BD}{BA}=\frac{AD}{CA}[/tex]Determine the measure of side AB by using the sides ratio.
[tex]\begin{gathered} \frac{AB}{12}=\frac{12-4}{BA} \\ (BA)^2=12\cdot8 \\ (x)^2=96 \\ x=\sqrt[]{96} \\ =4\sqrt[]{6} \end{gathered}[/tex]Determine the measure of side y.
[tex]\begin{gathered} y^2=x^2-(8)^2 \\ =96-64 \\ =32 \\ y=\sqrt[]{32} \\ =4\sqrt[]{2} \end{gathered}[/tex]Determine the mesure of side z.
[tex]\begin{gathered} z^2=y^2+(4)^2 \\ =32+16 \\ z=\sqrt[]{48} \\ =4\sqrt[]{3} \end{gathered}[/tex]Thus option B is correct.
