Show your work and include the units! Given: \( \triangle R B J \sim \triangle Y X W \) \( B=3 \operatorname{sen} \geqslant Q \) \( J R=10 \mathrm{ft} \) \( W Y=6 \mathrm{ft} \) \( x=5 \) Perimeter of \( \triangle R B J=30 \mathrm{ft} \) IR \( = \) \( 6 \frac{5}{30} \) \( \begin{array}{r}10 \\ \times 3 \\ \hline 30\end{array} \) \( B=3 \) \( w y=B \) Find the perimeter of \( \triangle Y X W \). Provide an answer accurate to the nearest hundredth.

To find the perimeter of triangle \( \triangle Y X W \) using the properties of similar triangles, we can use the ratio of the corresponding sides from triangles \( R B J \) and \( Y X W \). Given that \( \frac{B}{W Y} = \frac{R B J}{Y X W} \), we have: 1. The perimeter of triangle \( R B J \) is \( 30 \, \text{ft} \). 2. Since \( \triangle R B J \sim \triangle Y X W \), the ratio of their perimeters is the same as the ratio of their corresponding sides: \[ \frac{30 \, \text{ft}}{6 \, \text{ft}} = 5 \] Thus: \[ \text{Perimeter of } \triangle Y X W = \frac{1}{5} \times 30 = 6 \, \text{ft} \] Now, converting it into the required form, we can conclude: The perimeter of \( \triangle Y X W \) is \( 6.00 \, \text{ft} \) when rounded to the nearest hundredth.