Question 3·25 Super-Hard SAT Geometry and Trigonometry Questions·Geometry and Trigonometry

Let $s$ be the length of the minor arc, the shorter path from $A$ to $B$, and let $r$ be the radius. The major arc is $14\pi$ units longer, so its length is $s+14\pi$. The two arcs make the full circumference, $2\pi r$, so $s+(s+14\pi)=2\pi r$.

Combine the $s$ terms: $2s+14\pi=2\pi r$. Subtract $14\pi$: $2s=2\pi r-14\pi$. Divide by $2$: $s=\pi(r-7)$. The minor arc has positive length, so $r>7$.

The sector takes the same fraction of the circle's area as its arc takes of the circumference. If $A$ is the minor sector's area, then $\frac{A}{\pi r^2}=\frac{s}{2\pi r}$. Multiply by the whole circle's area, $\pi r^2$: $A=\frac{s}{2\pi r}(\pi r^2)$. Cancel $\pi r$: $A=\frac{rs}{2}$.

The minor sector's area is $49\pi$, and its arc length is $\pi(r-7)$. Put both into $A=\frac{rs}{2}$: $49\pi=\frac{r\pi(r-7)}{2}$. Divide by $\pi$: $49=\frac{r(r-7)}{2}$. Don't use $49\pi$ as the whole circle's area; it belongs to only the minor sector.

Type $\frac{x_{1}(x_{1}-7)}{2}\sim49\left\{x_{1}>7\right\}$ in Desmos. Here $x_1$ stands for the radius, $\sim$ tells Desmos to solve, and the restriction keeps the minor arc's length positive. Under PARAMETERS, Desmos shows $x_1=14$. So the radius is $14$ units. Grid in 14.