Question:
In triangle PQR, \( PQ = 12 \) cm, \( PR = 9 \) cm, and \( \angle Q + \angle R = 120^\circ \). Find the length of QR.
In triangle PQR, \( PQ = 12 \) cm, \( PR = 9 \) cm, and \( \angle Q + \angle R = 120^\circ \). Find the length of QR.
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When you know two sides and the included angle, always use the cosine rule. Use angle sum property to find the third angle if needed.
Updated On: Jul 28, 2025
- \( \dfrac{15}{\sqrt{2}} \) cm
- \( 3\sqrt{13} \) cm
- \( 5\sqrt{5} \) cm
- \( 5\sqrt{17} \) cm
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The Correct Option is B
Solution and Explanation
Given:
- \( PQ = 12 \) cm
- \( PR = 9 \) cm
- \( \angle Q + \angle R = 120^\circ \Rightarrow \angle P = 60^\circ \)
We will apply the Cosine Rule: \[ QR^2 = PQ^2 + PR^2 - 2 PQ PR \cos(\angle P) \] Substitute the values: \[ QR^2 = 12^2 + 9^2 - 2 12 9 \cos(60^\circ) = 144 + 81 - 216 \frac{1}{2} = 225 - 108 = 117 \Rightarrow QR = \sqrt{117} = \sqrt{9 13} = 3\sqrt{13} \] Final Answer: \( \boxed{3\sqrt{13} \text{ cm}} \)
- \( PR = 9 \) cm
- \( \angle Q + \angle R = 120^\circ \Rightarrow \angle P = 60^\circ \)
We will apply the Cosine Rule: \[ QR^2 = PQ^2 + PR^2 - 2 PQ PR \cos(\angle P) \] Substitute the values: \[ QR^2 = 12^2 + 9^2 - 2 12 9 \cos(60^\circ) = 144 + 81 - 216 \frac{1}{2} = 225 - 108 = 117 \Rightarrow QR = \sqrt{117} = \sqrt{9 13} = 3\sqrt{13} \] Final Answer: \( \boxed{3\sqrt{13} \text{ cm}} \)
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