Congruence Of Triangles

∠Q

Step 1: Identify the longest side - comparing PQ = 7 cm, QR = 5 cm, PR = 9 cm, we see PR is the longest. Step 2: Apply the theorem that the largest angle is opposite to the longest side. Step 3: The angle opposite to side PR is ∠Q (angle at vertex Q). Step 4: Therefore, ∠Q is the largest angle. This follows from the property that in any triangle, the greater angle lies opposite to the longer side.

RHS

Step 1: Identify that both triangles are right-angled at B and E respectively. Step 2: Note that AC and DF are the hypotenuses (sides opposite to right angles). Step 3: Given that AC = DF (hypotenuses are equal) and AB = DE (one pair of sides are equal). Step 4: This matches the RHS (Right angle-Hypotenuse-Side) criterion which states that if hypotenuse and one side of a right triangle equal the hypotenuse and corresponding side of another right triangle, the triangles are congruent.

BC

Step 1: Identify the angles - ∠A = 30° (smallest), ∠B = 60°, ∠C = 90° (largest). Step 2: Apply the theorem that the smallest angle has the shortest side opposite to it. Step 3: The side opposite to ∠A (30°) is BC. Step 4: Therefore, BC is the shortest side. This is a fundamental property: in any triangle, the shortest side lies opposite to the smallest angle.

45°

Step 1: Understand that if △ABC ≅ △PQR, then corresponding angles are equal. Step 2: The given correspondence shows A↔P, meaning vertex A corresponds to vertex P. Step 3: Since corresponding angles of congruent triangles are equal, ∠A = ∠P. Step 4: Given ∠A = 45°, therefore ∠P = 45°. This is a direct application of the property that corresponding parts of congruent triangles are equal.