SOLUTION: The length of a rectangle is 3 more than twice the width. If the perimeter is to be at least 51 meters, what are the possible values for the width? (If the perimeter is at least 5
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Question 186840: The length of a rectangle is 3 more than twice the width. If the perimeter is to be at least 51 meters, what are the possible values for the width? (If the perimeter is at least 51 meteres, than it is greater than or equal to 51 meters.) Answer by uday1435(57) (Show Source):
You can put this solution on YOUR website! The perimeter of a rectangle = 2( l +w)
Here l = w+3, pluging in that we get p = 2( w+3 + w) = 2(2w+3)
The given condition is p >= 51
That is 2(2w + 3) ≥ 51
4w + 6 ≥ 51 adding – 6 on both sides we get
4w ≥ 51- 6
4w ≥ 45
Dividing both sides by 4 we get
w ≥ 45/4
w≥ 11.25 . So the width must be at least 11.25 meters
If you still require further clarifications please write to udayakumar.t.r@gmail.com
