HL congruence applies to right triangles. It says that if the hypotenuse of one right triangle is congruent to the hypotenuse of another right triangle and a leg is congruent to the corresponding leg, then the two triangles are congruent. The right angle in each triangle fixes the orientation, so matching the hypotenuse and one leg provides enough information to determine the entire triangle, making all other parts align as well. This is why the correct description is using the hypotenuse and a leg. The other options don’t guarantee congruence on their own: just matching two legs or two hypotenuses, or only the acute angles, isn’t enough to fix all side lengths and angles.

The exterior angle of a triangle equals the sum of the measures of the two interior angles not adjacent to it. This works because the exterior angle forms a linear pair with the adjacent interior angle, so together they add to 180 degrees; and the sum of all three interior angles is also 180 degrees. Therefore, the exterior angle must equal the sum of the other two interior angles—the remote interior angles. For example, if the two nonadjacent interior angles measure 30 and 40 degrees, the exterior angle would be 70 degrees. This aligns with the theorem and shows why the statement is correct. The adjacent interior angle alone isn’t equal to the exterior angle (they’re supplementary), and the exterior angle isn’t the sum of all interior angles or twice any interior angle.

The area of a circle depends on the square of its radius. If you think of filling the circle with many thin rings from the center out to radius r, each ring at radius x has area roughly its circumference 2πx times a small thickness dx, so dA = 2πx dx. Adding up all those rings from 0 to r gives ∫0^r 2πx dx = πr^2. Therefore the area is πr^2. Why the other forms don’t fit: 2πr is the circumference, not area. πr is missing a factor of r, so it’s not area. 2r^2 has the incorrect scaling and lacks π.

The midpoint of a segment in the coordinate plane is found by averaging the x-coordinates and the y-coordinates of the endpoints. For A(1,3) and B(5,7), the x-coordinates average to (1+5)/2 = 3, and the y-coordinates average to (3+7)/2 = 5. So the midpoint is (3,5). You can also see this by noting that x increases by 4 and y increases by 4 from A to B, so halfway along adds 2 to each coordinate: (1+2, 3+2) = (3,5).