1. **Problem Statement:** Given a right triangle PQR with right angle at P (\(\angle QPR = 90^\circ\)), \(PQ = 6\) cm, and point M on hypotenuse PR such that \(\angle QPM = \angle RPM\). Find: - (a) \(\angle QPM\) - (b) \(\angle PQM\) and \(\angle PRM\) - (c) Length of \(PR\) 2. **Key Concepts:** - In a right triangle, the hypotenuse is the side opposite the right angle and is the longest side. - The point M on hypotenuse PR is such that \(\angle QPM = \angle RPM\), meaning M is the midpoint of the arc or related to equal angles. - Use properties of isosceles triangles and angle sums. 3. **Step-by-step Solution:** **(a) Find \(\angle QPM\):** - Since \(\angle QPM = \angle RPM\), triangle QPM and RPM share equal angles at M. - Because \(\angle QPR = 90^\circ\), triangle PQR is right angled at P. - By the property of the circle or equal angles subtended by chord PM, \(\angle QPM = 45^\circ\). **(b) Find \(\angle PQM\) and \(\angle PRM\):** - In triangle PQM, sum of angles is 180°: \(\angle PQM + \angle QPM + \angle PMQ = 180^\circ\). - We know \(\angle QPM = 45^\circ\). - Since M lies on PR, \(\angle PMQ = 90^\circ\) (right angle at P). - So, \(\angle PQM = 180^\circ - 45^\circ - 90^\circ = 45^\circ\). - Similarly, in triangle PRM, \(\angle PRM = 45^\circ\) by symmetry. **(c) Find length of \(PR\):** - Triangle PQR is right angled at P with \(PQ = 6\) cm. - Since \(\angle QPM = \angle RPM = 45^\circ\), triangle PQR is isosceles right angled with \(PQ = PR\). - Therefore, \(PR = PQ \times \sqrt{2} = 6 \times \sqrt{2} = 6\sqrt{2}\) cm. **Final answers:** - \(\angle QPM = 45^\circ\) - \(\angle PQM = 45^\circ\), \(\angle PRM = 45^\circ\) - \(PR = 6\sqrt{2}\) cm