A 30-60-90 triangle is half an equlateral triangle (refer to the image below).
If we call the length of the side
, we can see that it is exactly half of the side of the equilateral triangle. So, we have
![\overline{AB} = \overline{AC} = \overline{BC} = 2\cdot \overline{BD} = 2x](https://img.qammunity.org/2019/formulas/mathematics/college/b58bhem27muez5cdafa4ycx7ifilt7q7tq.png)
Moreover, we can find the height
using the pythagorean theorem, having
.
Now, you know the longer leg to be 6. The longer leg is the height, so you have
![\overline{CD} = x√(3) = 6 \implies x = \overline{BD} = \cfrac{6}{√(3)} = \cfrac{6√(3)}{3} = 2√(3)](https://img.qammunity.org/2019/formulas/mathematics/college/q9a30bowuwwbayk22ft0z37jrmfkpbizo6.png)
So, the hypotenuse is twice that value:
![BC = 2x = 2\cdot 2√(3) = 4√(3)](https://img.qammunity.org/2019/formulas/mathematics/college/8264jntmie995czf5sbfvb9armx4sb42e0.png)