a sixth grade teacher asked her students to draw rectangles with positive integer length and a perimeter of 42.the difference between the largest and smallest possible ares of the rectangles that the students could have come up with is?

• sum of length and breadth will be 21 units . 42/2=21
  area will be max when lxb = 11x10=110 sq units
  area will be min when lxb = 20x1=20 sq units.
  
  .the difference between the largest and smallest possible ares of the rectangles that the students could have come up with= 110-20=90 sq units
  
• 11 years agoHelpfull: Yes(47) No(3)
• agree with mayur
• 11 years agoHelpfull: Yes(4) No(3)
• 2(l+b)=42
  l+b=21
  l=20b=1
  area=20(lowest)
  if l=10
  b=11
  area=110(highest)
• 10 years agoHelpfull: Yes(3) No(0)
• smallest area will be for sides 1 and 20, area=20 sq. units. and largest area will be for sides 10,11. area=110 sq. units.
• 10 years agoHelpfull: Yes(2) No(1)
• 2(a+b)=42 then
  a+b =21
  for lowest a=1and b=20 or a=20 and b=1
  
  then area is 20
  for highest max vale of both a anb b then
  a=10 or 11 and b= 11 or 10 then area is 110
• 9 years agoHelpfull: Yes(1) No(0)