The width of a rectangle is 4 less than twice its length. if the area of the rectangle is 153 cm2, what is the length of the diagonal? give your answer to 2 decimal places. if anyone could explain this i would appreciate it, all the answers i kept getting on similar questions were a few numbers off and i don't know why.

  diagonal ≈ 18.43 cm

Step-by-step explanation:

Let L represent the length of the rectangle. Then the width is ...

  w = 2L -4 . . . . . . 4 less than twice the length

The area is ...

  A = wL = (2L -4)L = 2L² -4L

The area is said to be 153 cm², so we have ...

  2L² -4L = 153

  2L² -4L -153 = 0 . . . . . . subtract 153 to put into standard form

We can find the solution to this using the quadratic formula. It tells us the solution to ax²+bx+c=0 is given by ...

  x = (-b±√(b²-4ac))/(2a)

We have a=2, b=-4, c=-153, so our solution for L is ...

  L = (-(-4) ±√((-4)²-4(2)(-153)))/(2(2)) = (4±√1240)/4

Only the positive solution is of interest, so L = 1+√77.5.

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Now we know the rectangle is 1+√77.5 long and -2+2√77.5 wide. The diagonal (d) is the hypotenuse of a right triangle with these leg lengths. Its measure can be found from ...

  d² = w² +L² = (-2+2√77.5)² +(1+√77.5)²

It can work well to simply evaluate this using a calculator, or it can be simplified first.

  d² = 4 -8√77.5 +4·77.5 + 1 +2√77.5 +77.5 = 392.5 -6√77.5

Taking the square root gives the diagonal length:

  d = √(392.5 -6√77.5) ≈ 18.43 . . . . cm