SAT Question 58c26db8: Right Triangles And Trigonometry answer and explanation with Desmos solution

Choice C is correct. The perimeter of a triangle is the sum of the lengths of its sides. Since the given triangle is an isosceles right triangle, the length of each leg is the same and the length of the hypotenuse is equal to $\sqrt{2}$ times the length of a leg. Let $x$ represent the length, in inches, of a leg of this isosceles right triangle. Therefore, the perimeter, in inches, of the triangle is $x+x+x\sqrt{2}$, or $2x+x\sqrt{2}$, which is equivalent to $x(2+\sqrt{2})$. It's given that the perimeter of this triangle is $18+18\sqrt{2}$ inches. Thus, $x\left(2+\sqrt{2}\right)=18+18\sqrt{2}$. Dividing both sides of this equation by $2+\sqrt{2}$ yields $x=\frac{18+18\sqrt{2}}{2+\sqrt{2}}$. Multiplying the right-hand side of this equation by $\frac{2-\sqrt{2}}{2-\sqrt{2}}$ yields $x=\frac{36+36\sqrt{2}-18\sqrt{2}-36}{2}$, or $x=9\sqrt{2}$. It follows that the length, in inches, of a leg of this isosceles right triangle is $9\sqrt{2}$. Therefore, the length, in inches, of the hypotenuse of this isosceles right triangle is $\left(9\sqrt{2}\right)\left(\sqrt{2}\right)$, or $18$.