Question

The length of a rectangle is 5959 inches greatergreater than twice the width. if the diagonal is 2 inches more than the​ length, find the dimensions of the rectangle.

Answer 1

Length: 2w + 59

width: w

diagonal: (2w + 59) + 2 = 2w + 61

Length² + width² = diagonal²

(2w + 59)² + (w)² = (2w + 61)²

(4w² + 118w + 3481) + w² = 4w² + 122w + 3721

5w² + 118w + 3481 = 4w² + 122w + 3721

w² + 118w + 3481 = 122w + 3721

w² - 4w + 3481 = 3721

w² - 4w - 240 = 0

a = 1, b = -4, c = -240

w =
`[-(b) +/- \sqrt{(b)^(2)  - 4(a)(c) }]/2(a)`

=
`[-(-4) +/- \sqrt{(-4)^(2)  - 4(1)(-240) }]/2(1)`

=
`[4 +/- √((16  + 960) )]/2`

=
`[4 +/- √((976) )]/2`

=
`[2 +/- 4√((61) )]/2`

=
`1 +/- 2√((61) )`

since width cannot be negative, disregard 1 - 2√61

w = 1 + 2√61 ≈ 16.62

Length: 2w + 59 = 2(1 + 2√61) + 59 = 2 + 4√61 + 59 = 61 + 4√61 ≈ 92.24

Answer: width = 16.62 in, length = 92.24 in