EXPLANATION
9) We can see that this is an Equilateral Triangle, so the perimeter is given by the following equation:

We have that the height is equal to 4sqrt(15), but we don't know the base, so we can get this by applying the Law of Sines over the inside right triangle:
Law of Sines:

Let's call x to the unknown leg:
![(x)/(\sin 90)=\frac{4\sqrt[]{15}}{\sin 60}](https://img.qammunity.org/2023/formulas/mathematics/college/qvgv61znwa9nic4h2ciy6rrmcodvzicdp0.png)
Isolating x:
![x=(\sin90)/(\sin60)4\sqrt[]{15}=\frac{1}{\frac{\sqrt[]{3}}{2}}4\sqrt[]{15}=\frac{2}{\sqrt[]{3}}4\sqrt[]{15}=2\cdot4\frac{\sqrt[]{3}\cdot\sqrt[]{5}}{\sqrt[]{3}}](https://img.qammunity.org/2023/formulas/mathematics/college/j8zlu7ogi4qzclfz4ix3ze708vp12ntftp.png)
Simplifying terms:
![x=8\sqrt[]{5}](https://img.qammunity.org/2023/formulas/mathematics/college/nciz4rfdthssn1vvbbwnxo6r1xxkxowwse.png)
Now, we know that both legs are equal to 8*sqrt(5) because this is an Equilateral Triangle and the base is also the same.
![\text{Base }=legs=8\sqrt[]{5}](https://img.qammunity.org/2023/formulas/mathematics/college/llkn1pngk76qittjlxe6zohla2hw20b646.png)
Finally, the Perimeter is:
![\text{Perimeter}=8\sqrt[]{5}+8\sqrt[]{5}+8\sqrt[]{5}=3\cdot(8\sqrt[]{5})=24\sqrt[]{5}](https://img.qammunity.org/2023/formulas/mathematics/college/nj9m1co1z30yf2dhiuj7m204pg14rbaf0v.png)
The Perimeter is:
![\text{Perimeter}=24\sqrt[]{5}](https://img.qammunity.org/2023/formulas/mathematics/college/87pbxozeysldz5hzq6y3tzia7l1tqub59o.png)