Search 74,315 tutors
FIND TUTORS
Ask a question
0 0

The width of a rectangle is 31 centimeters. Find all possible values for the length of the rectangle if the perimeter is al least 692 centimerts

Tutors, please sign in to answer this question.

3 Answers

 P = 2L + 2W
 
P ≥ 692
 
 L+ W ≥ 346
 
  L+31≥ 346
 
 L ≥315
 
 
    This is simply an inequality problem.( ≥ ) should be used for at least ( minimum value),
 
   
In a rectangle opposite sides are equal
 
perimeter = 2(length + Width)
 
2(l + W) must greater than or equal to 692
L + W must be greater than or equal to (692/2)346
L + 31 must be greater than or equal to 346
there fore L must be greater than or equal to (346 - 31) 315
:)
 
 
Good morning! Recall that perimeter of a rectangle is the sum of the lengths of all sides, so:
 
P= L + L + W + W, or P + 2L + 2W
 
If the width is 31 and total perimeter is at least 692, then the minimum would be:
 
692 = 2L + 2(31), or 692 = 2L + 62
 
Subtract 62 from both sides:
 
630 = 2L
 
Divide both sides by 2:
 
315 = L
 
Remember that the perimeter is at least 692; therefore the length would be at least 315 cm
 
Hope this helps
 
Mike

Woodbridge algebra tutors