Answer:
[tex] \sqrt{55} [/tex]
Step-by-step explanation:
∆ BCD ~ ∆ DCA
[tex] \frac{bc}{dc} = \frac{dc}{ac} [/tex]
Plug the values:
[tex] \frac{5}{y} = \frac{y}{6 + 5} [/tex]
[tex] \frac{5}{y} = \frac{ y}{11} [/tex]
Apply cross product property
[tex]y \times y = 11 \times 5[/tex]
Calculate the product
[tex] {y}^{2} = 55[/tex]
[tex]y = \sqrt{55} [/tex]
Hope this helps...
Good luck on your assignment..
