Quadrilateral $RSTU$ is circumscribed around circle $J$. The perimeter of quadrilateral $RSTU$ is 50 ft. We are given $UC = 5$ ft, $BR = 9$ ft, and $TB = 7$ ft. We need to find the value of $x$, where $x = DS$.

1. Problem Description

Quadrilateral $RSTU$ is circumscribed around circle $J$. The perimeter of quadrilateral $RSTU$ is 50 ft. We are given $UC = 5$ ft, $BR = 9$ ft, and $TB = 7$ ft. We need to find the value of $x$, where $x = DS$.

2. Solution Steps

Since the quadrilateral $RSTU$ is circumscribed around circle $J$, tangents from the same point to the circle have equal lengths. Therefore, we have the following:

$UC = UA = 5$

$TB = TA = 7$

$RC = RD = 9$

$DS = SS$

The perimeter of quadrilateral $RSTU$ is given by:

$RS + ST + TU + UR = 50$

We can express the sides of the quadrilateral in terms of the tangent segments:

$RS = RD + DS = 9 + x$

$ST = SA + AT = x + 7$

$TU = UA + UC = 5 + 7$

$UR = RC + UC = 9 + 5$

Substituting these into the perimeter equation:

$(9 + x) + (x + 7) + (5 + 7) + (9 + 5) = 50$

$9 + x + x + 7 + 5 + 7 + 9 + 5 = 50$

$2x + 42 = 50$

$2x = 50 - 42$

$2x = 8$

$x = 4$

3. Final Answer

$x = 4$