Question

The length of your family's garden is 3 feet greater than the width. The area of the garden is 460 square feet. What are the dimensions of the garden?

Answer 1

Given:

The length of your family's garden is 3 feet greater than the width.

The area of the garden is 460 square feet.

To find:

The dimensions of the garden.

Solution:

Let x feet be the width of the garden. Then,

Length = $x+3$ feet

The area of a rectangle is:

$A=l\times w$

Where, l is the length and w is the width of the rectangle.

The area of the rectangular garden is:

$A=(x+3)\times x$

$A=x^2+3x$

It is given that the area of the garden is 460 square feet.

Putting $A=460$, we get

$460=x^2+3x$

$0=x^2+3x-460$

Splitting the middle term, we get

$x^2+23x-20x-460=0$

$x(x+23)-20(x+23)=0$

$(x+23)(x-20)=0$

$x=-23,20$

The width of a garden cannot be negative. So, $x=20$.

Now,

$l=x+3$

$l=20+3$

$l=23$

Therefore, the length of the garden is 23 feet and the width of the garden is 20 feet.