1. **Problem Statement:** Given a rectangle PQRS and a parallelogram TQRU inside it, with RV perpendicular to TQ, we need to find the relation between the areas of PQRS and TQRU, prove their equality, and find the length TU given certain dimensions. 2. **Relation between areas:** The area of rectangle PQRS is given by: $$\text{Area}_{PQRS} = PQ \times QR$$ The area of parallelogram TQRU is given by: $$\text{Area}_{TQRU} = \text{base} \times \text{height}$$ Here, base = RU and height = RV (since RV is perpendicular to TQ). 3. **Proof that areas are equal:** - Since PQRS is a rectangle, $PQ = SR$ and $QR = PS$. - Parallelogram TQRU shares side QR with the rectangle. - The height of the parallelogram is $RV$, perpendicular to $TQ$. - The base of the parallelogram is $RU$. Using the properties of the rectangle and parallelogram, the area of PQRS is: $$\text{Area}_{PQRS} = PQ \times QR$$ The area of TQRU is: $$\text{Area}_{TQRU} = RU \times RV$$ Since $PQ = RU$ (opposite sides of parallelogram and rectangle) and $QR = RV$ (height), the areas are equal: $$\text{Area}_{PQRS} = \text{Area}_{TQRU}$$ 4. **Finding TU:** Given: $$PQ = 10 \text{ cm}, RU = 16 \text{ cm}, RV = 5 \text{ cm}$$ Since RV is perpendicular to TQ, triangle RVU is right-angled at V. Using Pythagoras theorem in triangle RVU: $$TU^2 = RU^2 + RV^2$$ $$TU = \sqrt{RU^2 + RV^2} = \sqrt{16^2 + 5^2} = \sqrt{256 + 25} = \sqrt{281} \approx 16.76 \text{ cm}$$ **Final answer:** $$TU \approx 16.76 \text{ cm}$$