Question

A rectangle has a length that is 18 feet more than the width. The area of the

Answer 1

Answer:

For a rectangle of length L and width W, the area is:

A = L*W

In this case, we know that;

The length is 18ft more than the width.

L = W + 18ft

The area is 175 ft^2

Then:

A = 175 ft^2

Then we have a system of two equations:

L = W + 18ft

A = L*W = 175 ft^2

To solve this, the first step would be to replace the first equation into the second one:

(18ft + W)*W = 175ft^2

Now we can solve this for W.

18ft*W + W^2 - 175ft^2 = 0.

This is the quadratic equation, i will solve it just to be complete.

We have a quadratic equation, using the Bhaskara formula we can find the two solutions as:

$w = \frac{-18ft +- \sqrt{(18ft)^2 - 4*1*(-175ft^2)} }{2*1} = \frac{-18ft +- 32ft}{2ft}$

Then the two solutions are:

W = (-18ft - 32ft)/2 = -25ft    (This option can be discarded, because a                negative width has no sense)

W = (-18ft + 32ft)/2 = 7ft

Then the width is 7ft long, and the lenght will be:

L = W + 18ft = 7ft + 18ft = 25ft