Question: 4. The diagram shows a piece of rectangular land PQTY owned by Mr Rhulben. Parts of the land in the shapes of triangle PQR and rectangle STUV have been given to his brother. Mr Rulben plans to fence up his land. What is the cost of fencing if each metre of fence costs RM50?

1. **State the problem:** We need to find the cost of fencing Mr Rhulben's remaining land after parts shaped as triangle PQR and rectangle STUV are removed. 2. **Identify the shapes and dimensions:** - Rectangle PQTY with sides PQ = 60 m and PV = 180 m. - Triangle PQR with PQ = 60 m and QR = 60 m. - Rectangle STUV with RS = 50 m, TU = 52 m. 3. **Calculate the perimeter of the original rectangle PQTY:** $$\text{Perimeter} = 2 \times (PQ + PV) = 2 \times (60 + 180) = 2 \times 240 = 480\text{ m}$$ 4. **Calculate the lengths of the removed parts' boundaries that are inside the original rectangle and thus reduce the fencing needed:** - Triangle PQR removes side QR = 60 m and side PR (hypotenuse). Since PQR is a right triangle with legs PQ = 60 m and QR = 60 m, hypotenuse PR is: $$PR = \sqrt{60^2 + 60^2} = \sqrt{3600 + 3600} = \sqrt{7200} = 60\sqrt{2} \approx 84.85\text{ m}$$ - Rectangle STUV removes sides ST, TU, UV, and VS. Given RS = 50 m and TU = 52 m, the rectangle STUV has sides ST = RS = 50 m and TU = 52 m. 5. **Calculate the new perimeter after removing the triangle and rectangle:** The fencing will be along the outer boundary minus the parts given away, plus the new edges created by the removal. The new perimeter is: $$\text{New perimeter} = \text{Original perimeter} - (QR + RS + TU) + (PR + UV)$$ Note: UV is the side opposite to ST in rectangle STUV, so UV = ST = 50 m. Calculate: $$\text{New perimeter} = 480 - (60 + 50 + 52) + (84.85 + 50) = 480 - 162 + 134.85 = 452.85\text{ m}$$ 6. **Calculate the cost of fencing:** Cost per metre = 50 $$\text{Total cost} = 452.85 \times 50 = 22642.5$$ **Final answer:** The cost of fencing is 22642.5.