A rectangle has a width that is 3 more inches than the length. The area of the rectangle is 40 square inches. Find the dimensions of the rectangle.

Answer 1

Hi, the problem tells us that the width is 3in longer than the width:

(1) W=3+L

And that the area is 40in^2, meaning that the width times the length is equal to 40in^2.

(2) 40in^2=W*L

We have two equations and two unknowns therefore we can solve the problem. Plug in the first equation into the second equation and solve for L.

40=(3+L)*L

0=L^2+3L-40

We can solve this using the quadratic formula (take the positive answer):

L=5in.

Now that we know the value of L, we can find the value of W by plugging it into the first equation.

W=3+5=8in.

Hope this helps.

Answer 2

Consider x as the length

So, x+3 would be the width

The area of a rectangle is length times width, so the area would be:

x(x+3)=40

x²+3x-40=0

Then you would factor this, and get a positive answer and a negative answer. You would use the positive answer for x as the length, and then add 3 to get the width.