SAT Question ae041e52: Right Triangles And Trigonometry answer and explanation with Desmos solution

Choice A is correct. When a square is inscribed in a circle, a diagonal of the square is a diameter of the circle. It's given that a square is inscribed in a circle and the length of a radius of the circle is $\frac{20\sqrt{2}}{2}$ inches. Therefore, the length of a diameter of the circle is $2\left(\frac{20\sqrt{2}}{2}\right)$ inches, or $20\sqrt{2}$ inches. It follows that the length of a diagonal of the square is $20\sqrt{2}$ inches. A diagonal of a square separates the square into two right triangles in which the legs are the sides of the square and the hypotenuse is a diagonal. Since a square has $4$ congruent sides, each of these two right triangles has congruent legs and a hypotenuse of length $20\sqrt{2}$ inches. Since each of these two right triangles has congruent legs, they are both $45$ - $45$ - $90$ triangles. In a $45$ - $45$ - $90$ triangle, the length of the hypotenuse is $\sqrt{2}$ times the length of a leg. Let $s$ represent the length of a leg of one of these $45$ - $45$ - $90$ triangles. It follows that $20\sqrt{2}=\sqrt{2}\left(s\right)$. Dividing both sides of this equation by $\sqrt{2}$ yields $20=s$. Therefore, the length of a leg of one of these $45$ - $45$ - $90$ triangles is $20$ inches. Since the legs of these two $45$ - $45$ - $90$ triangles are the sides of the square, it follows that the side length of the square is $20$ inches.