Twice the width of a rectangle is 10 meters more the its length. If the area of the region enclosed by the rectangle is 600 square meters , then find its perimeter.

Let, the length of the rectangle is x meter 

And the width of the rectangle is y meter

According to question,

$$\begin{gathered} 2y=x+10 ............ \left(i\right) \\ And xy=600 ....... \left(ii\right) \\ From equation \left(ii\right) \\ xy=600 \\ x=\frac{600}{y}...........\left(iii\right) \\ Putting the value of x in equation \left(i\right) \\ 2y=x+10 \\ \Rightarrow 2y=\frac{600}{y}+10 \\ \Rightarrow 2y^2-10y-600=0 \\ \Rightarrow y^2-5y-300=0 \left[Dividing both sides by 2\right] \\ \Rightarrow y^2-20y+15y-300=0 \\ \Rightarrow y\left(y-20\right)+15\left(y-20\right)=0 \\ \Rightarrow \left(y-20\right)\left(y+15\right)=0 \\ \therefore \left(y-20\right)\left(y+15\right)=0 \end{gathered}$$

Since width cannot be (-)ve. So we find y=20

That means the width of the rectangle is 20 meters

The length of the rectangle is $x=\frac{600}{y}$ 

$\therefore x=\frac{600}{20}=30$

The length of the rectangle is 30 meters.

 The perimeter of the rectangle = $2\times (30+20) = 2\times 50=100$ meters. 

ans. The perimeter is 100m.