Solve For Z
A. 12
B.
[tex]4 \sqrt{7} [/tex]
C.
[tex]3 \sqrt{7} [/tex]
D. 51
![Solve For ZA 12B tex4 sqrt7 texC tex3 sqrt7 texD 51 class=](https://us-static.z-dn.net/files/da3/727a6ebf9a0b33707633ded15f3d8ee2.jpg)
Triangle ABC and triangle BCD are similar, therefore the sides are in proportion:
[tex]\dfrac{z}{9+7}=\dfrac{7}{z}[/tex]
cross multiply
[tex]z^2=16\cdot7\to z=\sqrt{16\cdot7}\\\\z=\sqrt{16}\cdot\sqrt7\\\\z=4\sqrt7[/tex]
Answer: B. 4√7