Triangular Playground Fence and Sod
A school plans a triangular outdoor learning area. Two sides meet at a 73° angle and measure 42.0 m and 58.0 m. The third side borders a wetland and must be fenced along a raised boardwalk; the other two sides get standard fence. The supplier sells fence in 10 m panels and recommends ordering 5% extra for corners and waste. Sod is sold by the square meter with a 10% overage allowance. The principal has budgeted for 160 m of fence and 1,300 m² of sod. A teacher says the two given sides and the included angle are enough to find everything; a parent worries the wetland side might push the fence order over budget.
1. Find the length of the wetland side and the total perimeter. Determine whether 160 m of fence is enough after the 5% overage and 10 m panel requirement.
Read the worked answer
Use the Law of Cosines with $a=42.0$, $b=58.0$, and $C=73^\circ$:
$$c^2=42^2+58^2-2(42)(58)\cos73^\circ.$$
$$c^2=1764+3364-4872(0.29237)\approx5128-1424.4=3703.6,$$
so $c\approx60.86$ m. The perimeter is
$$P\approx42.0+58.0+60.86=160.86\ \mathrm{m}.$$
Even before overage, 160 m is slightly short. With 5% extra, the order would be $160.86(1.05)\approx168.9$ m. Because panels come in 10 m lengths, the team must buy 17 panels, or 170 m. The 160 m budget is therefore insufficient by at least 10 m of purchased fence.
2. Calculate the area of the triangular region and decide whether the 1,300 m² sod budget is adequate with the 10% overage.
Read the worked answer
Use the sine area formula with the two given sides and included angle:
$$A=\frac12(42.0)(58.0)\sin73^\circ.$$
$$A=1218(0.9563)\approx1164.8\ \mathrm{m^2}.$$
With 10% overage,
$$1164.8(1.10)\approx1281.3\ \mathrm{m^2}.$$
This is below 1,300 m², so the sod budget is adequate with about 18.7 m² to spare. The judgment is not just to compare 1,164.8 with 1,300; the overage allowance is required because sod must be cut around edges and corners. The 10% allowance is enough here, but only barely, so the team should verify the final dimensions before ordering.
3. The angle measurement could be off by ±1°. Test whether that uncertainty changes the fence or sod conclusions.
Read the worked answer
At 72°, $\cos72^\circ\approx0.3090$, so
$$c^2\approx5128-4872(0.3090)=5128-1505.4=3622.6,$$
and $c\approx60.19$ m. At 74°, $\cos74^\circ\approx0.2756$, so
$$c^2\approx5128-1342.7=3785.3,$$
and $c\approx61.53$ m. The perimeter ranges from about 160.19 m to 161.53 m, both above 160 m and both requiring 17 panels after overage. For area, $\sin72^\circ\approx0.9511$ gives about 1,158 m², and $\sin74^\circ\approx0.9613$ gives about 1,171 m². With 10% overage, the range is about 1,274–1,288 m², still under 1,300 m². The conclusions are robust to a ±1° error, so the team should focus on the fence budget, not the sod budget.