1. **Problem:** Given that the measured distance AB is 165.20 m using a 50 m tape that is 0.01 m too short, we want to find the correct distance AB. 2. **Step 1: Understand the meaning of tape length error** The tape is shorter than the standard by 0.01 m per 50 m, so every measurement on the tape underestimates the actual length. 3. **Step 2: Calculate the correction factor** The actual tape length is $50 + 0.01 = 50.01$ m but the tape reads 50 m. Correction factor $= \frac{50.01}{50} = 1.0002$ 4. **Step 3: Calculate the correct distance** Correct distance $= 165.20 \times 1.0002 = 165.23304$ m (approximately). --- 5. **Problem:** A rectangular lot has a correct area of 2 hectares (20000 m²), length is twice the width. The sides were measured with a 50 m tape that is 0.02 m too long. Find error in area in sq. meters. 6. **Step 1: Define variables** Let width $= w$, length $= 2w$. Area $= l \times w = 2w \times w = 2w^2 = 20000$ m² Solve for $w$: $$2w^2=20000 \Rightarrow w^2=10000 \Rightarrow w=100 \text{ m}$$ Length $l=200$ m. 7. **Step 2: Effect of tape being too long** The tape length is nominally 50 m but actually $50.02$ m. Correction factor for measured distances: $$\text{Correction factor} = \frac{50}{50.02} = 0.9996$$ Because when using the tape, measured lengths are longer than actual, so the actual measured side lengths are shorter. 8. **Step 3: Calculate the measured sides** Measured width = $w_{meas} = w / 0.9996 = 100.04$ m Measured length = $l_{meas} = 200 / 0.9996 = 200.08$ m 9. **Step 4: Calculate the measured area using the incorrect tape length** $$A_{meas} = l_{meas} \times w_{meas} = 200.08 \times 100.04 = 20016.004 \text{ m}^2$$ 10. **Step 5: Calculate the error in area** $$\text{Error} = A_{meas} - A_{correct} = 20016.004 - 20000 = 16.004 \text{ m}^2$$ --- 11. **Problem:** Distance BC measured at 146.5 m using a 100 m tape which is 0.015 m too short. Find the correct distance. 12. **Step 1: Correction factor** Actual tape length = $100 + 0.015 = 100.015$ m Correction factor = $\frac{100.015}{100} = 1.00015$ 13. **Step 2: Calculate correct distance** $$\text{Correct distance} = 146.5 \times 1.00015 = 146.521975 \text{ m}$$ --- 14. **Problem:** The correct distance AB is 213.50 m. Tape length 100 m is 0.025 m too long. Find the measured distance. 15. **Step 1: Calculate tape correction factor** Actual tape length = $100 - 0.025 = 99.975$ m (because tape is too long, each measurement is shorter) Correction factor = $\frac{99.975}{100} = 0.99975$ 16. **Step 2: Calculate measured distance** $$\text{Measured distance} = 213.50 \times 0.99975 = 213.446625 \text{ m}$$ --- 17. **Problem:** Correct distance CD is 195 m. Tape 50 m is 0.015 m too short. Find measured distance. 18. **Step 1: Correction factor** Actual tape length = $50 + 0.015 = 50.015$ m Correction factor = $\frac{50.015}{50} = 1.0003$ 19. **Step 2: Calculate measured distance** $$\text{Measured distance} = 195 \times \frac{50}{50.015} = 195 \times 0.9997 = 194.94 \text{ m}$$ **Final concise answers:** 2. Correct distance AB $\approx 165.23$ m 3. Error in area $\approx 16$ m² 4. Correct distance BC $\approx 146.52$ m 5. Measured distance AB $\approx 213.45$ m 6. Measured distance CD $\approx 194.94$ m