To analyze whether Wyatt or Parker is correct, we need to look at the conditions provided:
- Triangle ABC has a right angle at B.
- Triangle PQR has a right angle at Q.
- \( AB \cong QR \) (one pair of congruent sides).
- \( AC \cong PR \) (a second pair of congruent sides).
Wyatt's Argument: Wyatt claims the triangles are congruent by the HL (Hypotenuse-Leg) Congruence Theorem. HL states that if you have two right triangles, and the hypotenuse and one leg of one triangle are congruent to the hypotenuse and one leg of another triangle, then the triangles are congruent.
However, in the triangles given, we do not have corresponding hypotenuses. Instead, we have one leg and the side that does not necessarily represent the hypotenuse.
Parker's Argument: Parker proposes that the triangles are congruent by the SAS (Side-Angle-Side) Congruence Theorem. SAS states that if two sides of one triangle and the included angle between them are congruent to two sides of another triangle and the included angle between them, then the triangles are congruent. However, in this scenario, we do not know the included angle between the sides \(AB\) and \(AC\) in triangle ABC or \(QR\) and \(PR\) in triangle PQR, so we cannot use SAS.
Given the information provided, neither Wyatt nor Parker has sufficient justification for their claims.
The correct choice is: Both are wrong, the triangles are not congruent.