SAT Question 568d66a7: Right Triangles And Trigonometry answer and explanation with Desmos solution

Choice B is correct. It's given that the right triangle is isosceles. In an isosceles right triangle, the two legs have equal lengths, and the length of the hypotenuse is $\sqrt{2}$ times the length of one of the legs. Let $l$ represent the length, in inches, of each leg of the isosceles right triangle. It follows that the length of the hypotenuse is $l\sqrt{2}$ inches. The perimeter of a figure is the sum of the lengths of the sides of the figure. Therefore, the perimeter of the isosceles right triangle is $l+l+l\sqrt{2}$ inches. It's given that the perimeter of the triangle is $94+94\sqrt{2}$ inches. It follows that $l+l+l\sqrt{2}=94+94\sqrt{2}$. Factoring the left-hand side of this equation yields $\left(1+1+\sqrt{2}\right)l=94+94\sqrt{2}$, or $\left(2+\sqrt{2}\right)l=94+94\sqrt{2}$. Dividing both sides of this equation by $2+\sqrt{2}$ yields $l=\frac{94+94\sqrt{2}}{2+\sqrt{2}}$. Rationalizing the denominator of the right-hand side of this equation by multiplying the right-hand side of the equation by $\frac{2-\sqrt{2}}{2-\sqrt{2}}$ yields $l=\frac{\left(94+94\sqrt{2}\right)\left(2-\sqrt{2}\right)}{\left(2+\sqrt{2}\right)\left(2-\sqrt{2}\right)}$. Applying the distributive property to the numerator and to the denominator of the right-hand side of this equation yields $l=\frac{188-94\sqrt{2}+188\sqrt{2}-94\sqrt{4}}{4-2\sqrt{2}+2\sqrt{2}-\sqrt{4}}$. This is equivalent to $l=\frac{94\sqrt{2}}{2}$, or $l=47\sqrt{2}$. Therefore, the length, in inches, of one leg of the isosceles right triangle is $47\sqrt{2}$.