From the given figure,
[tex]\begin{gathered} In\text{ }\Delta ABD,\text{ BD }\perp\text{ AC} \\ \end{gathered}[/tex]
By using right angled triangle theorem,
According to right angled triangle theorem, perpendicular drawn on the hypotenuse is equal to the square root of the product of parts in which hypotenuse is divided.
[tex]\begin{gathered} x\text{ = }\sqrt[]{10\text{ }\times\text{ 4}} \\ x\text{ = }\sqrt[]{40} \\ x\text{ = 2}\sqrt[]{10} \end{gathered}[/tex]
By using Pythagoras theorem,
[tex]\begin{gathered} AB^2=AD^2+DB^2 \\ z^2=10^2+x^2\text{ } \\ z^2=10^2\text{ + (2}\sqrt[]{10})^2 \\ \end{gathered}[/tex]
Further,
[tex]\begin{gathered} z^2=\text{ 100 + 40} \\ z^2\text{ = 140} \\ z\text{ = 2}\sqrt[]{35\text{ }}\text{ } \end{gathered}[/tex]
Also,
[tex]In\text{ }\Delta ABC,[/tex]
By using Pythagoras theorem,
According to Pythagoras theorem, the square of the hypotenuse is equal to the sum of the squares of the remaining sides.
[tex]\begin{gathered} AC^2=AB^2+BC^2 \\ 14^2=z^2+y^2_{_{_{}\text{ }_{}}} \\ z^2=14^2-y^2_{_{_{}}}\text{ } \end{gathered}[/tex]
Further,
[tex]\begin{gathered} y^2=14^2-z^2 \\ y^2\text{ = 196 - (2}\sqrt[]{35})^2 \\ y^2\text{ = 196 - 140} \\ \end{gathered}[/tex]
Therefore ,
[tex]\begin{gathered} y^2\text{ = 56} \\ y\text{ = 2}\sqrt[]{14} \end{gathered}[/tex]
Thus the required values of x , y and z are
[tex]\begin{gathered} x\text{ = 2}\sqrt[]{10}\text{ units} \\ y\text{ = 2}\sqrt[]{14}\text{ units} \\ z\text{ = 2}\sqrt[]{35\text{ }}\text{ units} \end{gathered}[/tex]
