We have the similar triangles. Therefore the lengths of the sides are in proportion:
[tex]\dfrac{x}{22}=\dfrac{42}{11+22}\\\\\dfrac{x}{22}=\dfrac{42}{33}\qquad\text{cross multiply}\\\\33x=(22)(42)\qquad\text{divide both sides by 33}\\\\x=\dfrac{(22\!\!\!\!\!\diagup^2)(42)}{33\!\!\!\!\!\diagup_3}\\\\x=\dfrac{(2)(42\!\!\!\!\!\diagup^{14})}{\not3_1}\\\\x=(2)(14)\\\\\boxed{x=28}[/tex]
