Sol:

Let the length of one side be x metres

Let the length of the other side be y metres

Here one side is a compound wall and 30 m wire is used for fencing.

\( \Rightarrow x + y + x = 30 \) \( \Rightarrow y = 30 – 2x \)

Area of the vegetable garden =\( 100m^{2} \) \( \Rightarrow xy = 100 \) \( \Rightarrow x(30 – 2x) = 100 \) \( \Rightarrow 30x – 2x^{2} = 100 \) \( \Rightarrow 15x – x^{2} = 50 \) \( \Rightarrow x^{2} – 15x + 50 = 0 \) \( \Rightarrow x^{2} – 10x – 5x + 50 = 0 \) \( \Rightarrow x (x – 10) – 5(x – 10) = 0 \) \( \Rightarrow (x – 10x)(x + 5) = 0 \) \( \Rightarrow x = 5 and 10 \)

When x = 5

y = 30 – (2 * 5)

\( \Rightarrow y = 30 – 10 \) \( \Rightarrow y = 20 \)

When x = 10,

y = 30 – (2 * 10)

\( \Rightarrow y = 30 – 20 \) \( \Rightarrow y = 10 \)

Therefore, the dimensions of his garden will be 5m * 20m or 10m * 10m

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