Question 227047
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.<br>

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:<br>

(20+2w)^2 + w^2 = (22+2w)^2.<br>

Now we can solve for the width.<br>

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

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 {{{w = (-9 +- sqrt( 165))/(2) }}} which if you plug into your calculator should come out to 1.9226 or -10.9226.  <br>

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.