Question

There is a rectangular parking lot with a length of 2x and a width of x. if the ratio of the perimeter of the parking lot to the area of the parking lot is given by $\dfrac{k}{x}$ , then find the value of k.

Answer 1

Hint: To find the value of k, we have to find the perimeter and area of the parking lot. We are given that the length of the rectangular parking lot is 2x and width is x. Hence, we will get the perimeter of the parking lot $=2\left( 2x+x \right)=6x$ and area of parking lot $=2x\times x=2{{x}^{2}}$ . Now, we have to substitute these in $\dfrac{\text{Perimeter of parking lot}}{\text{Area of parking lot}}=\dfrac{k}{x}$ and find the value of k.

Complete step-by-step answer:
We are given that the length of rectangular parking lot is 2x, that is,
$l=2x$
We also have the width of the parking lot as x. Let us denote this as
$b=x$

We are given that the ratio of perimeter to the area of the parking lot $=\dfrac{k}{x}$ .
$\Rightarrow \dfrac{\text{Perimeter of parking lot}}{\text{Area of parking lot}}=\dfrac{k}{x}...(i)$
First, let us find the perimeter of the given rectangular parking lot.
We know that the perimeter of a rectangle $=2\left( l+b \right)$ . Let’s substitute the values. We will get
Perimeter of the parking lot $=2\left( 2x+x \right)=2\left( 3x \right)=6x$
Now, let’s find the area of the parking lot.
We know that area of a rectangle $=lb$ . When we substitute the values, we will get
Area of parking lot $=2x\times x=2{{x}^{2}}$
Now, let’s substitute these in (i).
$\Rightarrow \dfrac{6x}{2{{x}^{2}}}=\dfrac{k}{x}$
We have to find the value of k. For this, let’s solve the above expression.
Let’s cancel the common terms from numerator and denominator of LHS.
$\Rightarrow \dfrac{6}{2x}=\dfrac{k}{x}$
Let us now cross multiply. We will get
$6x=2xk$
Let us cancel x from both sides. Hence, the above expression becomes
$6=2k$
Let us rewrite the above expression.
$2k=6$
Let us take 2 from LHS to RHS.
$\Rightarrow k=\dfrac{6}{2}=3$
Hence, the value of k is 3.

Note: You must know the perimeter and area of the rectangle to solve this question. You may make a mistake by writing the area of the rectangle as $l+b$ and perimeter of the rectangle as $2\left( lb \right)$ . You may also make mistakes by taking the given ratio as $\dfrac{\text{Area of parking lot}}{\text{Perimeter of parking lot}}=\dfrac{k}{x}$ . Students must be thorough with algebraic rules in order to solve the algebraic expressions.