A farmer wishes to grow a 100 m² rectangular vegetable garden. Since he has with him only 30m barbed wire , he fences three sides of the rectangular garden letting compound wall of his house act as the fourth side fence. Find the dimensions of his garden.

Let x metres be the length of the side opposite to unfenced side, then length of each of two others sides = $\dfrac{1}{2}(30 -x).$

According to question,

$\Rightarrow x \times \dfrac{1}{2}(30 - x) = 100 \\[1em] \Rightarrow 15x - \dfrac{x^2}{2} = 100 \\[1em] \Rightarrow \dfrac{30x - x^2}{2} = 100 \text{ (On taking L.C.M.) }\\[1em] \Rightarrow 30x - x^2 = 200 \text { (On cross multiplying) } \\[1em] \Rightarrow x^2 - 30x + 200 = 0 \\[1em] \Rightarrow x^2 - 20x - 10x + 200 = 0 \\[1em] \Rightarrow x(x - 20) - 10(x - 20) = 0 \\[1em] \Rightarrow (x - 10)(x - 20) = 0 \\[1em] \Rightarrow x - 10 = 0 \text{ or } x - 20 = 0 \\[1em] x = 10 \text{ or } x = 20$

∴ if x = 10, $\dfrac{1}{2}(30 - x) :$

$= \dfrac{1}{2}(30 - 10) \\[1em] = \dfrac{1}{2} \times 20 \\[1em] = 10$

∴ if x = 20, $\dfrac{1}{2}(30 - x) :$

$= \dfrac{1}{2}(30 - 20) \\[1em] = \dfrac{1}{2} \times 10 \\[1em] = 5$

Hence, the dimensions of garden are 10m x 10m or 20m x 5m.