Dr Gulshan S. answered 02/05/17
Tutor
4.9
(52)
PhD In Physics and experience of teaching IB Physics and Math High sc
Let X= width of path around
Inner area = 23x16 sq ft
Outer area = (23+2X) x (16 +2X)
Difference= (outer area - inner area) = 490 sq ft Given
Which gives
4X2 +78 X -490 = 0
Solving for X
X = -49/2 ft and X = 5 ft
-ve value can be ignored
So width of path = 5 ft