1. The problem asks to find the cost of a triangular lot with sides 140 ft, 65 ft, and 111 ft, given the price is 3 per square foot. 2. To find the cost, we first need the area of the triangle. We use Heron's formula for the area of a triangle with sides $a$, $b$, and $c$: $$s = \frac{a+b+c}{2}$$ $$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$$ 3. Calculate the semi-perimeter $s$: $$s = \frac{140 + 65 + 111}{2} = \frac{316}{2} = 158$$ 4. Calculate the area: $$\text{Area} = \sqrt{158(158-140)(158-65)(158-111)} = \sqrt{158 \times 18 \times 93 \times 47}$$ 5. Compute the product inside the square root: $$158 \times 18 = 2844$$ $$2844 \times 93 = 264492$$ $$264492 \times 47 = 12491124$$ 6. Take the square root: $$\text{Area} = \sqrt{12491124} \approx 3533.63 \text{ square feet}$$ 7. Calculate the cost: $$\text{Cost} = \text{Area} \times 3 = 3533.63 \times 3 = 10600.89$$ **Final answer:** The lot costs approximately 10600.89.