Question

# A farmer wishes to grow a $100{m}^{2}$ rectangular vegetable garden. Since he has with the only $30m$ barbed wire, he fences three sides of the rectangular garden letting the compound wall of his house act as the fourth side – fence. Find the dimensions of his garden.

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Solution

## Step 1: Let's find the value of each sides of the fence in the garden.Let the length of one side be x metresLet the length of the other side be y metresHere one side is a compound wall and 30 m wire is used for fencing.$\begin{array}{l}⇒x+y+x=30\end{array}$$\begin{array}{l}⇒y=30–2x\end{array}$Area of the vegetable garden =$\begin{array}{l}100{m}^{2}\end{array}$$\begin{array}{l}⇒xy=100\end{array}$$\begin{array}{l}⇒x\left(30–2x\right)=100\end{array}$$\begin{array}{l}⇒30x–2{x}^{2}=100\end{array}$$\begin{array}{l}⇒15x–{x}^{2}=50\end{array}$$\begin{array}{l}⇒{x}^{2}–15x+50=0\end{array}$$\begin{array}{l}⇒{x}^{2}–10x–5x+50=0\end{array}$$\begin{array}{l}⇒x\left(x–10\right)–5\left(x–10\right)=0\end{array}$$\begin{array}{l}⇒\left(x–10x\right)\left(x+5\right)=0\end{array}$$\begin{array}{l}⇒x=5and10\end{array}$Step 2: Let's find the dimensions of the garden.When $x=5,$$y=30-\left(2×5\right)$$\begin{array}{l}⇒y=30–10\end{array}$$\begin{array}{l}⇒y=20\end{array}$When $x=10,$$y=30-\left(2×10\right)$$\begin{array}{l}⇒y=30–20\end{array}$$\begin{array}{l}⇒y=10\end{array}$Therefore, the dimensions of his garden will be $\left(5m×20m\right)$ or $\left(10m×10m\right)$

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