Wyatt and Parker are comparing two triangles to determine if they are congruent. For triangle ABC, <B measures 90 degrees and for triangle PQR, <Q is a right angle. They also know that AB¯¯¯¯¯¯¯¯ ≅QR¯¯¯¯¯¯¯¯ and AC¯¯¯¯¯¯¯¯ ≅ PR¯¯¯¯¯¯¯¯

Question

. Wyatt says the triangles are congruent by the HL Congruence Theorem and Parker thinks that they are congruent by SAS. Who is correct? (Hint: Draw a picture)(1 point)
Responses which answer is truly correct and why

Both are wrong, the triangles are congruent by SSA
Both are wrong, the triangles are congruent by SSA

Parker is correct
Parker is correct

Wyatt is correct
Wyatt is correct

Both are wrong, the triangles are not congruent

Answer 1

To determine who is correct between Wyatt and Parker, let's analyze the information provided.

1. Right Angles: Both triangles \( ABC \) and \( PQR \) have a right angle (\( \angle B \) in triangle \( ABC \) and \( \angle Q \) in triangle \( PQR \)).
2. Sides: We have \( AB \cong QR \) and \( AC \cong PR \).

Now, let's evaluate the claims:

• Wyatt claims congruence by HL (Hypotenuse-Leg) Congruence Theorem: This theorem can be applied to right triangles. For HL to be applicable, we need to know that the hypotenuse of both triangles is congruent. Since we are not given that \( BC \cong PQ \) (the corresponding hypotenuses), Wyatt does not have sufficient information to use the HL theorem, making his conclusion incorrect.

• Parker claims congruence by SAS (Side-Angle-Side): For SAS to apply, we need to have two sides and the included angle congruent. Here, we have \( AB \cong QR \), \( AC \cong PR \), and the included angle \( \angle B \) is right (also \( \angle Q \) is right). Therefore, Parker's conclusion is valid given that he has two sides and the included right angle between them.

Thus, Parker is correct, and the triangles can be shown to be congruent by SAS (Side-Angle-Side).

Response:

Parker is correct.