Triangle Area 3Dac4A

1. **Problem Statement:** Given a square \(PQRS\) with diagonal \(PR = 10\) cm, and \(PS\) extended to point \(T\), find the area of triangle \(\Delta QRT\). 2. **Understanding the figure and properties:** - Since \(PQRS\) is a square, all sides are equal and all angles are \(90^\circ\). - The diagonal \(PR\) of a square relates to the side length \(s\) by the formula: $$PR = s\sqrt{2}$$ 3. **Calculate the side length \(s\) of the square:** $$s = \frac{PR}{\sqrt{2}} = \frac{10}{\sqrt{2}} = 5\sqrt{2}$$ 4. **Coordinates assignment for clarity:** Assume the square is placed on the coordinate plane for easier calculation: - Let \(P = (0,0)\) - Since \(PQRS\) is a square, let \(Q = (s,0) = (5\sqrt{2},0)\) - \(R = (s,s) = (5\sqrt{2},5\sqrt{2})\) - \(S = (0,s) = (0,5\sqrt{2})\) 5. **Point \(T\) lies on the line extending \(PS\):** - \(PS\) is vertical from \(P(0,0)\) to \(S(0,5\sqrt{2})\). - Extending \(PS\) beyond \(S\) means \(T\) has coordinates \((0,y)\) with \(y > 5\sqrt{2}\). 6. **Find coordinates of \(T\):** - Since \(T\) lies on the extension of \(PS\), let \(T = (0,t)\) where \(t > 5\sqrt{2}\). 7. **Calculate area of \(\Delta QRT\):** - Points are \(Q = (5\sqrt{2},0)\), \(R = (5\sqrt{2},5\sqrt{2})\), and \(T = (0,t)\). - Area of triangle with vertices \((x_1,y_1), (x_2,y_2), (x_3,y_3)\) is: $$\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$$ Substitute: $$= \frac{1}{2} | 5\sqrt{2}(5\sqrt{2} - t) + 5\sqrt{2}(t - 0) + 0(0 - 5\sqrt{2}) |$$ Simplify inside: $$= \frac{1}{2} | 5\sqrt{2} \times 5\sqrt{2} - 5\sqrt{2} t + 5\sqrt{2} t | = \frac{1}{2} | 5\sqrt{2} \times 5\sqrt{2} |$$ Since \(-5\sqrt{2} t + 5\sqrt{2} t = 0\), Calculate \(5\sqrt{2} \times 5\sqrt{2} = 25 \times 2 = 50\), So area = \(\frac{1}{2} \times 50 = 25\). 8. **Final answer:** The area of \(\Delta QRT\) is \(25\) square centimeters. **Note:** The area is independent of the exact position of \(T\) on the extension of \(PS\) because the terms involving \(t\) cancel out.