A man was of a 1260 yd2 rectangular field He these run to th

A man was of a 1260 yd^2 rectangular field. He these run to the further corner. As shown in the diagram if the length L. of the field is 3 yd longer show than three times the width W, how fat the man run? For both the exact solution and an approximate solution.

Solution

Let lenght be l and width be w

Given :l*w = 1260 ----(1)

l = 3+3w -----(2)

Solve above tow get values of l and w:

So, 1260/w = 3(1+w)

420 = w(1+w)

w^2 -w -420 =0

w = 21, -20

Neglect the -ve value ,So, w = 21 yard

l =1260/21 = 60 yard

So, man ran lenght = diagonal of the rectangle

= sqrt(l^2+w^2) = sqrt( 21^2 +60^2) = sqrt(4041) = 63.57 yard

Approximately he ran 64 yards