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

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

   =  $[4 +/- \sqrt{(16 + 960) }]/2$        

  = $[4 +/- \sqrt{(976) }]/2$  

  = $[2 +/- 4\sqrt{(61) }]/2$  

  = $1 +/- 2\sqrt{(61) }$

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