In \( \triangle P Q R, P Q=4 \mathrm{~cm}, P R=8 \mathrm{~cm} \) and \( R T=6 \) cm . Find (i) the area of \( \triangle P Q R \) (ii) the length of \( Q S \).

To solve the problem, we'll analyze the given information step by step. ### Given: - Triangle \( PQR \) with sides: - \( PQ = 4 \) cm - \( PR = 8 \) cm - \( RT = 6 \) cm, where \( RT \) is the altitude from vertex \( R \) to side \( PQ \). ### (i) Finding the Area of \( \triangle PQR \): The area of a triangle can be calculated using the formula: \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \] Here, we can take \( PQ \) as the base and \( RT \) as the height. \[ \text{Area} = \frac{1}{2} \times 4 \times 6 = 12 \text{ cm}^2 \] ### (ii) Finding the Length of \( QS \): Assume \( S \) is the foot of the perpendicular from \( R \) to \( PQ \). To find \( QS \), we'll use coordinate geometry. 1. **Coordinate Setup:** - Place point \( P \) at \( (0, 0) \). - Point \( Q \) is at \( (4, 0) \). - Since \( RT = 6 \) cm is the altitude, point \( R \) lies somewhere above the base \( PQ \). Let’s denote \( R \) as \( (x, 6) \). 2. **Using the Distance Formula:** \[ PR = \sqrt{x^2 + 6^2} = 8 \] \[ x^2 + 36 = 64 \Rightarrow x^2 = 28 \Rightarrow x = 2\sqrt{7} \] So, \( R \) is at \( (2\sqrt{7}, 6) \). 3. **Finding \( S \):** The foot \( S \) of the perpendicular from \( R \) to \( PQ \) is directly below \( R \) on the \( x \)-axis, so \( S \) is at \( (2\sqrt{7}, 0) \). 4. **Calculating \( QS \):** \[ QS = |Q_x - S_x| = |4 - 2\sqrt{7}| \] Since \( 2\sqrt{7} \approx 5.291 \) cm, \[ QS \approx 5.291 - 4 = 1.291 \text{ cm} \] Therefore, \[ QS = 2\sqrt{7} - 4 \text{ cm} \] ### **Final Answers:** 1. **Area of \( \triangle PQR \)**: \( 12 \) cm² 2. **Length of \( QS \)**: \( 2\sqrt{7} - 4 \) cm ### **Summary:** - **(i)** The area of \( \triangle PQR \) is **12 cm²**. - **(ii)** The length of \( QS \) is **\( 2\sqrt{7} - 4 \) cm**.