Look at the picture.
The triangle ADC and the triangle CDB are similar.
Therefore corresponding sides are all in the same proportion:
[tex]\dfrac{y}{4}=\dfrac{12}{y}\ \ \ \ |cross\ multiply\\\\y^2=48\to y=\sqrt{48}\\\\y=\sqrt{16\cdot3}\\\\y=\sqrt{16}\cdot\sqrt3\\\\y=4\sqrt3[/tex]
Answer: 4√3
