1. Problem 2 states: The measured distance from A to B is 165.20 m using a tape that is 0.01 m too short for every 50 m. 2. Since the tape is short, the actual distance is longer than measured. We find the correcting factor for the tape: $$\text{Tape length corrected} = 50 + 0.01 = 50.01 \text{ m}\n\text{Correction factor} = \frac{50.01}{50} = 1.0002$$ 3. Multiply the measured distance by the correction factor: $$\text{Correct distance} = 165.20 \times 1.0002 = 165.23304 \text{ m}$$ 4. The correct distance from A to B is approximately 165.23 m. 5. Problem 3 states: A rectangular lot has an area of two hectares (20,000 m²). Its length is twice its width. 6. Let the width be $w$ meters, then the length is $2w$ meters. Area formula: $$\text{Area} = \text{length} \times \text{width} = 2w \times w = 2w^2$$ 7. Given the area is 20,000 m²: $$2w^2 = 20000$$ 8. Solve for $w$: $$w^2 = \frac{20000}{2} = 10000$$ $$w = \sqrt{10000} = 100 \text{ m}$$ 9. Length is twice width: $$\text{Length} = 2 \times 100 = 200 \text{ m}$$ Final Answers: - Correct distance from A to B: $165.23$ m - Width of lot: $100$ m - Length of lot: $200$ m