Question

The length of a rectangle is 4 m less than the diagonal and the width is 5 m less than the diagonal. If the area is 82 m^2, how long is the diagonal in meters? Round your answers to the nearest tenth.​

Answer 1

I hate rounding.

Let's call the diagonal x.  It's the hypotenuse of the right triangle whose legs are the rectangle sides.

According to the problem we have a length x-4 and a width x-5 and an area

82 = (x-4)(x-5)

82 = x^2 - 9x + 20

0 = x^2 - 9x - 62

That one doesn't seem to factor so we go to the quadratic formula

$x = \frac 1 2(9 \pm \sqrt{9^2-4(62)}) = \frac 1 2(9 \pm \sqrt{329})$

Only the positive value makes any sense for this problem, so we conclude

$x = \frac 1 2(9 \pm \sqrt{329})$

That's the exact answer.  Did I mention I hate rounding?  That's about

x = 13.6 meters

Answer: 13.6

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It's not clear to me this problem is consistent.  By the Pythagorean Theorem the diagonal satisfies

$x^2 = (x-4)^2 + (x-5)^2$

which works out to

$x=9 \pm 2\sqrt{10}$

That's not consistent with the first answer; this problem really has no solution.  Tell your teacher to get better material.