Question

A rectangular field is half as wide as it is long and is completely enclosed by $x$ yards of fencing.The area in terms of $x$ is
A.\[\dfrac{{{x}^{2}}}{2}\]
B.\[2{{x}^{2}}\]
C.\[\dfrac{2{{x}^{2}}}{9}\]
D.\[\dfrac{{{x}^{2}}}{18}\]

Answer 1

Hint: In the above question we will use the formula of perimeter of rectangle and then find the value of its length in terms of the given perimeter and further we will find the area of the rectangle in terms of the given perimeter.

Complete step by step answer:
In the above question it is said that the rectangle field is half as wide as it is long.So, let us suppose that its length is “\[l\]” then its width must be equal to “\[\dfrac{l}{2}\]”.
Also we have to know about the formula of perimeter and area for rectangle which is given below;
\[\begin{align}
  & area=L\times B \\
 & perimeter=2(L\times B) \\
 & \text{where L,B are length and width respectively}\text{.} \\
\end{align}\]
Also it is given that the rectangle field is enclosed by $x$ yards of fencing which means its perimeter is equal to $x$ yards.
So, we have L= \[l\] and B= \[\dfrac{l}{2}\].
$\begin{align}
  & \Rightarrow perimeter=2(l+\dfrac{l}{2}) \\
 & \Rightarrow x=3l \\
 & \Rightarrow \dfrac{x}{3}=l \\
\end{align}$
Here, we get the length in terms of $x$.
Now, the area of the given rectangle field is in term of $x$ are as below;
\[\begin{align}
  & \Rightarrow area=L\times B \\
 & \Rightarrow area=l\times \dfrac{l}{2} \\
 & \Rightarrow area=\dfrac{{{l}^{2}}}{2} \\
\end{align}\]
Now, we will use the above value of \[l\] in term of $x$ an we get;
$\begin{align}
  & area=\dfrac{{{({}^{x}/{}_{3})}^{2}}}{2} \\
 & \Rightarrow area=\dfrac{{{x}^{2}}}{18} \\
\end{align}$
Hence, the area in terms of $x$ will be \[\dfrac{{{x}^{2}}}{18}\] .
Therefore, the correct option of the above question is option D.

NOTE: Be careful while doing calculation and also remember the formula of perimeter and area of rectangle.One must know the formulae of Area and perimeter of the rectangle to solve this problem.