SAT Question ffe862a3: Right Triangles And Trigonometry answer and explanation with Desmos solution

Choice C is correct. Since the triangle is an isosceles right triangle, the two sides that form the right angle must be the same length. Let $x$ be the length, in inches, of each of those sides. The Pythagorean theorem states that in a right triangle, $a^2+b^2=c^2$, where $c$ is the length of the hypotenuse and $a$ and $b$ are the lengths of the other two sides. Substituting $x$ for $a$, $x$ for $b$, and $58$ for $c$ in this equation yields $x^2+x^2=58^2$, or $2x^2=58^2$. Dividing each side of this equation by $2$ yields $x^2=\frac{58^2}{2}$, or $x^2=\frac{2\cdot 58^2}{4}$. Taking the square root of each side of this equation yields two solutions: $x=\frac{58\sqrt{2}}{2}$ and $x=-\frac{58\sqrt{2}}{2}$. The value of $x$ must be positive because it represents a side length. Therefore, $x=\frac{58\sqrt{2}}{2}$, or $x=29\sqrt{2}$. The perimeter, in inches, of the triangle is $58+x+x$, or $58+2x$. Substituting $29\sqrt{2}$ for $x$ in this expression gives a perimeter, in inches, of $58+2\left(29\sqrt{2}\right)$, or $58+58\sqrt{2}$.