A farmer wishes to start a 100 sq. m rectangular vegetable garden. Since he has only 30 m barbed wire, he fences three sides of the garden letting his house compound wall act as the fourth side fencing. The dimension of the garden is:

### Concept & Optimization The problem involves finding dimensions that satisfy two constraints simultaneously: a fixed area and a fixed length for three sides. $$ \text{Area} = l \times b = 100 $$ $$ \text{Fencing} = l + 2b = 30 \text{ (or } 2l + b = 30 \text{)} $$ ### Step-by-Step Solution * Let the length be $l$ and the breadth be $b$. The fencing covers three sides. * This means the fencing length is either $l + 2b = 30$ (if the wall is on the length side) or $2l + b = 30$ (if the wall is on the breadth side). * Test Option (a): $15 \times 6.67$. Area $\approx 100$. Fencing could be $15 + 2(6.67) = 28.34$ or $2(15) + 6.67 = 36.67$. Neither is 30. * Test Option (b): $20 \times 5$. Area = $20 \times 5 = 100$. Fencing could be $20 + 2(5) = 30$. This perfectly matches the $30 \text{ m}$ of wire available! * Test Option (c): $30 \times 3.33$. Area $\approx 100$. Fencing could be $30 + 2(3.33) = 36.66$. Does not match. * Test Option (d): $40 \times 2.5$. Area = 100. Fencing could be $40 + 2(2.5) = 45$ or $2(40) + 2.5 = 82.5$. Does not match. ### Exam Strategy & Shortcut For dual-constraint problems with options provided, back-solving is infinitely faster than setting up a quadratic equation ($l(15 - 0.5l) = 100$). Simply plug the dimensions into $L + 2B$ to see which one equals 30. ### Common Pitfall Attempting to solve this by assuming a square or a specific ratio first, rather than letting the two given conditions (Area = 100, Fencing = 30) guide the selection. ### Final Answer Therefore, the correct answer is **20 m $\times$ 5 m**.