From the previous question: - \( PQ = 12 \), \( PR = 9 \), \( \angle P = 60^\circ \), \( QR = 3\sqrt{13} \) Let angle bisector of \( \angle P \) meet QR at M.
By Angle Bisector Theorem: \[ \frac{QM}{MR} = \frac{PQ}{PR} = \frac{12}{9} = \frac{4}{3} \Rightarrow QM = \frac{4}{7} QR, MR = \frac{3}{7} QR \] So: \[ QM = \frac{4}{7} 3\sqrt{13} = \frac{12\sqrt{13}}{7}, MR = \frac{9\sqrt{13}}{7} \] Use Angle Bisector Length Formula: \[ PM^2 = PQ PR \left[ 1 - \left( \frac{QR^2}{(PQ + PR)^2} \right) \right] \] Substitute: \[ PM^2 = 12 9 \left[ 1 - \left( \frac{(3\sqrt{13})^2}{(12 + 9)^2} \right) \right] = 108 \left[ 1 - \left( \frac{117}{441} \right) \right] = 108 \frac{324}{441} = \frac{34992}{441} \] Simplify: \[ PM = \sqrt{\frac{34992}{441}} = \sqrt{79.333...} \approx 8.9 \] Try options: (A) \( \dfrac{28\sqrt{5}}{9} \approx 8.72 \)
(b) \( \dfrac{42\sqrt{5}}{11} \approx 8.62 \)
(c) \( \dfrac{36\sqrt{3}}{7} \approx 8.86 \)
(d) \( 4\sqrt{3} \approx 6.93 \)
Only option
(c) matches approx. So, Final Answer: \( \boxed{\dfrac{36\sqrt{3}}{7}} \)
In the given figure, the numbers associated with the rectangle, triangle, and ellipse are 1, 2, and 3, respectively. Which one among the given options is the most appropriate combination of \( P \), \( Q \), and \( R \)?
