A big rectangular plot of area 4320 m$^2$ is divided into 3 square-shaped smaller plots by fencing parallel to the smaller side of the plot. However some area of land was still left as a square could not be formed. So, 3 more square-shaped plots were formed by fencing parallel to the longer side of the original plot such that no area of the plot was left surplus. What are the dimensions of the original plot?

### Concept & Geometric Partitioning Translate the geometric partitioning into algebraic equations. When a rectangle is divided by lines parallel to a side to form squares, the side of those squares equals the length of the side being paralleled against. $$ \text{Area} = L \times B $$ ### Step-by-Step Solution 1. **Define the dimensions:** - Let the original rectangular plot have length $L$ and breadth $B$ (where $L > B$). - Given Area = $L \times B = 4320 \text{ m}^2$. 2. **Analyze the first partition:** - Fencing parallel to the smaller side ($B$) forms 3 squares. Each square has a side equal to $B$. - The total length consumed by these 3 squares is $3B$. - The remaining length of the rectangle is $(L - 3B)$. - The leftover plot is a rectangle of dimensions $(L - 3B)$ and $B$. 3. **Analyze the second partition:** - The leftover rectangle is divided into 3 more squares by fencing parallel to the *longer* side of the original plot (which means the fence cuts across $B$). - This implies the 3 new squares are formed along the side $B$, so their side length must be $(L - 3B)$. - Because 3 such squares perfectly span the breadth $B$, we have: $3 \times (L - 3B) = B$. 4. **Solve the equations:** - $3L - 9B = B \Rightarrow 3L = 10B \Rightarrow L = \frac{10}{3}B$. - Substitute $L$ into the area equation: $(\frac{10}{3}B) \times B = 4320$. - $\frac{10}{3}B^2 = 4320 \Rightarrow B^2 = 432 \times 3 = 1296$. - $B = \sqrt{1296} = 36 \text{ m}$. - $L = \frac{10}{3} \times 36 = 120 \text{ m}$. ### Exam Strategy & Shortcut Use the options to test the area and partitioning condition. Check option (c) $120 \times 36$: Area is $120 \times 36 = 4320$. Three squares of $36\times36$ take up $108$ m of length. Remaining length = $12$ m. The remaining $36 \times 12$ piece can perfectly form three $12\times12$ squares. It completely satisfies all constraints immediately. ### Common Pitfall Misinterpreting "parallel to the longer side" can cause students to set up the wrong equivalence for the second set of squares, leading to impossible algebra. ### Final Answer Therefore, the correct answer is **120 m $\times$ 36 m**.