2026 AMC 10A Problem 6

The triangle cuts the circle into four pieces: the triangle itself, and three curved pieces, one beyond each side of the triangle. We need the two smaller curved pieces.

1. Spot the right triangle. All three sides are even, so divide by $2$: $8, 15, 17$. That is a famous Pythagorean triple, since $8^2 + 15^2 = 64 + 225 = 289 = 17^2$. So the triangle has a right angle, with legs $16$ and $30$ and longest side $34$.

1. The longest side is a diameter. An angle with its corner on a circle is half of the arc it opens onto. A $90^\circ$ angle therefore opens onto a $180^\circ$ arc, which is half the circle. So the side across from the right angle, the $34$ side, joins two opposite points of the circle: it is a diameter. The radius is $34 \div 2 = 17$.

3. Split the circle along that diameter. The circle's area is $\pi \cdot 17^2 = 289\pi$, so each half has area $\tfrac{289\pi}{2}$.
- One half lies on the far side of the $34$ side, away from the triangle. It is a single curved piece. That is region $C$, and it is the largest.
- The other half contains the whole triangle plus the two curved pieces beyond the legs $16$ and $30$. Those two pieces are $A$ and $B$.

4. Subtract the triangle. The triangle is a right triangle, so its area is half the product of its legs:
$$\tfrac12 \times 16 \times 30 = 240.$$
So
$$A + B = \frac{289\pi}{2} - 240.$$

1. Check that $C$ really is the largest. $C$ is a whole half circle, while $A$ and $B$ together fit inside the other half circle with room left for the triangle. So $C$ is bigger than $A + B$, and certainly bigger than either one.

The answer is $\boxed{\textbf{(A)}\ \frac{289\pi}{2}-240}$.

A right triangle drawn in a circle always has its longest side as a diameter, and a diameter cuts the circle into two equal halves. That turns a problem about curved regions into "half a circle minus a triangle". Whenever side lengths are given, test the Pythagorean theorem first (dividing out a common factor makes the triple easier to recognize).

Find all three regions and pick the two smaller ones. Everything outside the triangle has area $289\pi - 240$. The region beyond the diameter is exactly half the circle, $\tfrac{289\pi}{2}$. The other two regions share the rest:
$$A + B = (289\pi - 240) - \frac{289\pi}{2} = \frac{289\pi}{2} - 240.$$
As a size check, $\tfrac{289\pi}{2} \approx 454$, so $A + B \approx 214$, which is less than $C \approx 454$, as it should be.