SOLUTION: The length of a rectangle is 20 inches greater than twice the width. If the diagonal is 2 inches more than the length, find the dimensions of the rectangle. I can't seem to find

Question 227047: The length of a rectangle is 20 inches greater than twice the width. If the diagonal is 2 inches more than the length, find the dimensions of the rectangle.
I can't seem to find this formula and I have a midterm today and want to make sure I can solve it properly. I have 2h+22 and 2h+20 and also h by itself.
Answer by algebrapro18(249) (Show Source): You can put this solution on YOUR website!
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

I will leave the rest to you but just a hint, you will have to use the quadratic formula. You will eventually get it down to which if you plug into your calculator should come out to 1.9226 or -10.9226.

Since we know that a measure of a side can't be negative we know that the width has to be 1.9226. So now we go back and find the length and the diagonal. I will leave this for you to do. You should get L = 23.8452 and D = 25.8452.
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