Precalc question? Determining the dimensions of a rectangle given the area?? PLEASE HELP!!?

Question

I am really stuck on this and I have not used the Pythagorean Theorem in ages. Could you please explain how you solve this as you go? Thank you for the help! I really appreciate it!

 

The width of a rectangle is 9 less than twice its length. If the area of the rectangle is 21 cm2cm2, what is the length of the diagonal?

The length of the diagonal is cm.

Give your answer to 2 decimal places.

Note: Since you are looking for the diagonal you will want to use the pythagorean theorem, a2+b2=c2

Answer 1

Let x = length of rectangle, so w = 2x - 9 and area = x(2x - 9) = 21 (given in cm²). Making a diagram of the rectangle and its diagonal will be useful, labeled as above.  And you can call the diagonal's length c at this time, too.

 

This gives you the equation  2x² - 9x = 21 or    2x² - 9x - 21 = 0  and you'll have to solve using the quadratic formula.

 

The rest is easy.  Solving for (positive) x enables you to know the length & width; these are put into the Pythagorean formula as right triangle leg lengths a & b (c²= a² + b²), c being length of hypotenuse which is in fact the diagonal that you desire.

 

Calculate carefully!