A rectangular garden is 18.6 m long and 12.4 m. A student measured the length as 18.2 m and the width as 12.7 m. Find the percentage error in the: i. Measured length ii. Calculated perimeter iii. Calculated area

$$
\text{Error in length} = |L_M - L_T| = |18.2 - 18.6| = |-0.4| = 0.4 \text{ m}
$$

$$
\text{Percentage Error (Length)} = \frac{0.4}{18.6} \times 100\% \approx 2.1505\%
$$
Rounded to two decimal places:
$$
\boxed{\text{2.15%}}
$$

$$
P_T = 2(L_T + W_T) = 2(18.6 + 12.4) = 2(31.0) = 62.0 \text{ m}
$$
$$
P_M = 2(L_M + W_M) = 2(18.2 + 12.7) = 2(30.9) = 61.8 \text{ m}
$$

$$
\text{Error in perimeter} = |P_M - P_T| = |61.8 - 62.0| = |-0.2| = 0.2 \text{ m}
$$

$$
\text{Percentage Error (Perimeter)} = \frac{0.2}{62.0} \times 100\% \approx 0.3226\%
$$
Rounded to two decimal places:
$$
\boxed{\text{0.32%}}
$$

$$
A_T = L_T \times W_T = 18.6 \times 12.4 = 230.64 \text{ m}^2
$$
$$
A_M = L_M \times W_M = 18.2 \times 12.7 = 231.14 \text{ m}^2
$$

$$
\text{Error in area} = |A_M - A_T| = |231.14 - 230.64| = 0.50 \text{ m}^2
$$

$$
\text{Percentage Error (Area)} = \frac{0.50}{230.64} \times 100\% \approx 0.2168\%
$$
Rounded to two decimal places:
$$
\boxed{\text{0.22%}}
$$