Question

A rectangular garden is 2 feet longer than it is wide. If the width is doubled, 50 extra feet of fencing will be needed to keep out the rabbits. What are the dimensions of the original garden?

Answer 1

Answer:

The dimensions of the original garden is $27\times25\ ft$.

Step-by-step explanation:

Let the width of the rectangular garden be 'x'.

Now given:

A rectangular garden is 2 feet longer than it is wide.

so we can say that;

Length of the rectangle = $(x+2) \ ft$

Now to fence the garden we need to find the perimeter of the garden.

Now Perimeter of the garden is equal to twice the sum of the length times width.

framing in equation form we get;

Perimeter of the rectangle = $2(x+x+2)=2(2x+2)=4x+4$

Now given:

If the width is doubled, 50 extra feet of fencing will be needed

It  means that Perimeter of the doubled width rectangle is equal to perimeter of original garden plus 50.

framing in equation form we get;

$2(2x+x+2)=4x+4+50\\\\2(3x+2)=4x+50$

Now Applying distributive property we get;

$2\times3x+2\times2=4x+54\\\\6x+4=4x+54$

Combining the like terms we get;

$6x-4x=54-4\\\\2x=50$

Dividing both side by 2 we get;

$\frac{2x}{2}=\frac{50}{2}\\\\x=25 \ ft$

Length of the rectangular garden = $x+2=25+2=27\ ft$

Hence the dimensions of the original garden is $27\times25\ ft$.