SAT Question 25da87f8: Right Triangles And Trigonometry answer and explanation with Desmos solution

It is given that the triangle has angle measures of $30^{\circ }$, $60^{\circ }$, and $90^{\circ }$, and so the triangle is a special right triangle. The side measures of this type of special triangle are in the ratio $x:x\sqrt{3}:2x$. If $x$ is the measure of the shortest leg, then the measure of the other leg is $x\sqrt{3}$ and the measure of the hypotenuse is $2x$. The perimeter of the triangle is given to be $18+6\sqrt{3}$, and so the equation for the perimeter can be written as $2x+x+x\sqrt{3}=18+6\sqrt{3}$. Combining like terms and factoring out a common factor of $x$ on the left-hand side of the equation gives $(3+\sqrt{3})\times x=18+6\sqrt{3}$. Rewriting the right-hand side of the equation by factoring out $6$ gives $(3+\sqrt{3})\times x=6(3+\sqrt{3})$. Dividing both sides of the equation by the common factor $(3+\sqrt{3})$ gives $x=6$. The longest side of the right triangle, the hypotenuse, has a length of $2x$, or $2(6)$, which is $12$.