SAT Question 54df8076: Right Triangles And Trigonometry answer and explanation with Desmos solution

Since the perimeter of a triangle is the sum of the lengths of its sides, and the given triangle is equilateral, the length of each side is $\frac{852}{3}$, or $284$, centimeters (cm). Right triangle $AMO$ can be formed, where $M$ is the midpoint of one of the triangle's sides, $A$ is one of this side's endpoints, and $O$ is the center of the circle. It follows that $AM$ is $\frac{284}{2}$, or $142$, cm. Additionally, triangle $AMO$ has angles measuring $30^{\circ }$, $60^{\circ }$, and $90^{\circ }$, where the measure of angle $OMA$ is $90^{\circ }$ and the measure of angle $OAM$ is $30^{\circ }$. It follows that the length of side $MO$ is half the length of hypotenuse $AO$, and the length of side $AM$ is $\sqrt{3}$ times the length of side $MO$. It's given that $AO=w\sqrt{3}$ cm. Therefore, $MO=\frac{w\sqrt{3}}{2}$ cm and $AM=\frac{w\sqrt{3}\sqrt{3}}{2}$ cm, which is equivalent to $AM=\frac{3w}{2}$ cm. Since $AM=142$ cm, it follows that $\frac{3w}{2}=142$. Multiplying both sides of this equation by $2$ yields $3w=284$. Dividing both sides of this equation by $3$ yields $w=\frac{284}{3}$. Note that 284/3, 94.66, and 94.67 are examples of ways to enter a correct answer.