Respuesta :
Let 'x' denote the longer side of the rectangle and let 'y' denote the shorter side of the rectangle.
From the problem, we know that:
x > y
xy > 10
With some rearranging, we get that:
x > y > 10/x
From the problem, we know that:
x > y
xy > 10
With some rearranging, we get that:
x > y > 10/x
Answer:
Step-by-step explanation:
Given that area of a rectangle is greater than 10.
Normally length would be bigger than width
Let l = w+d where d is the difference between length and width
Then area[tex]=w(w+d) >10\\w^2+wd-10>0[/tex]
This is a quadratic equation and solution would be
[tex]w=\frac{-d±\sqrt{d^2+40} }{2}[/tex]
width cannot be negative
w>0
If square then l=w = \sqrt 10 = 3.162
Shortest side of the rectangle is
0<w<3.162
then only we have w shorter than length
