Question

The length of the rectangle is 2 inches more than the width. if the diagonal is the square root of 20 inches find the area of the rectangle.

Answer 1

To start we need to assign variables for what were trying to find. Lets say the rectangle has width w. Now we know that the rectangle has a length 20 more than twice its width. So in algebra this would be l = 20+2w. Now we know that the diagonal is 2 more than the length. In algebra this would be d = 2+l. Substituting our length equation into the diagonal expression we get d = 2 + 20 + 2w = 22+2w.

So our equations are l = 20+2w and d = 22+2w. We also know that the angles of rectangle are right angles. So the length, width, and diagonal form a right triangle and using the Pythagorean theorem we find that:

(20+2w)^2 + w^2 = (22+2w)^2.

Now we can solve for the width.

(20+2w)^2 + w^2 = (22+2w)^2.
400 + 80w + 4w^2 + w^2 = 484 + 44w + w^2

Answer 2

ill leave u with the rest im not giving the answer away

Answer 3

@kg1975, thanks this is a review assignment so I already have a list with the answers. I just didn't know what to do with the diagonal.