Question 2443
 
Before proceeding further, the student is advised to see my earlier solved problem No.2444. Now let us solve this problem too. But Child, you must also try to solve.

Let the width = x cms.
So the length = 5x + 1 cms.
 
The area, i.e x*(5x + 1) = 130 Sq.Cms.
=  5{{{x^2}}} + x = 130
so 5{{{x^2}}} + x - 130 = 0
=  5{{{x^2}}}-25x+26x-130 = 0
=  5x(x-5) + 26(x-5) = 0
=  (x-5)*(5x+26) = 0
So either (x-5) =0 or
         (5x+26) = 0.
 
So x = 5  or  x = -5.2.

As 'x' denotes the width of the rectangle, it cannot be negative.  So, 
the width = 5 cms.   Answer.

gsm