Answer:
[tex]\large\boxed{x=2\sqrt{21}\ and\ y=4\sqrt3}[/tex]
Step-by-step explanation:
Look at the picture.
ΔADC and ΔCDB are similar. Therefore the sides are in proportion:
[tex]\dfrac{AD}{CD}=\dfrac{CD}{DB}[/tex]
We have
[tex]AD=14-8=6\\CD=y\\DB=8[/tex]
Substitute:
[tex]\dfrac{6}{y}=\dfrac{y}{8}[/tex] cross multiply
[tex]y^2=(6)(8)[/tex]
[tex]y^2=48\to y=\sqrt{48}\\\\y=\sqrt{16\cdot3}\\\\y=\sqrt{16}\ cdot\sqrt3\\\\\boxed{y=4\sqrt3}[/tex]
For x use the Pythagorean theorem:
[tex]x^2=6^2+(4\sqrt3)^2\\\\x^2=36+48\\\\x^2=84\to x=\sqrt{84}\\\\x=\sqrt{4\cdot21}\\\\x=\sqrt4\cdot\sqrt{21}\\\\\boxed{x=2\sqrt{21}}[/tex]
