A parallelogram having an acute angle of 30 degrees whose area is equal to the perimeter and the sides are whole numbers then number of such parallelograms possible are
a) 0 b) 1 c) 2 d) 3

• let the paralleogram consist of two triangle..
  the area=(1/2)*ab*sin(30) [a and b are the sides]
  =1/4ab
  now area of parallelogram=1/2ab
  peimeter=2a+2b
  so acc. to qstn
  2a+2b=1/2ab
  
  so it can be written as
  (a-4)(b-4)=16
  Thus a−4 and b−4 must have product 16.
  we may assume that a≤b
  since a−4 and b−4 cannot both be negative.
  So a−4=1, b−4=16, or a−4=2, b−4=8, or a−4=4, b−4=4.
  So ansr is 3 possible parallelogram
• 11 years agoHelpfull: Yes(30) No(12)
• area of parllgrm=a*b sin(30) =1/2 ab
  perimeter= 2(a+b)
  1/2 ab=2(a+b) --> 4a+4b=ab -->a(b-4) =4b
  a=4b/(b-4)
  then(a,b) -->(20,5) (8,8) (5,20)
  (8,8) --> is a square is a parlelogram but no side will be 30 degrees
  so only (20,5) and (5,20) are possible
  only 2 parallelograms are possible
• 11 years agoHelpfull: Yes(9) No(9)
• nikhil please
  send me
  avinashsingh56@gmail.com
• 11 years agoHelpfull: Yes(1) No(9)
• send me too nikhil at ankita.apt@gmail.com
• 11 years agoHelpfull: Yes(1) No(4)
• area of parallelogram=1/2 ab, perimeter=2(a+b)
  so 2a+2b=1/2 ab
  so,, (a-4)(b-4)=16
  area is parallelogram is 1/2(a-4)(b-4)=8
  
  therrefore possible combinations to obtain product of sides 8 should be 1,8 and 2,4.. answer 2 parallelograms.
• 11 years agoHelpfull: Yes(1) No(4)
• square is also a parallelogram
• 11 years agoHelpfull: Yes(1) No(5)
• hey nikhil if you have more questions then please post it.Or mail me at ankitsaun.thekiller@gmail.com will put it
• 11 years agoHelpfull: Yes(0) No(3)
• Given that perimeter of the parallelogram is equal to its area.
  So we have: 2(l + b) = l * h --------------- (1)
  From the diagram we have: sin300 = (h/b) => 1/2 = h/b
  => h = b/2 --------------- (2)
  From eq (1) and (2):
  2(l + b) = l * b/2
  => (l + b)/(l*b) = 1/4
  => l * b = 4(l + b)
  If you are familiar with graph you can see that the the expression on the left hand side is the equation of a hyperbola and the expression on the right hand side matches the equation of a straight line. Till (l, b) = (8, 8) the graph of hyperbola will be below the graph of the straight line after this point the graph of hyperbola will always be above the graph of the straight line. So we can have just one solution for this.
• 9 years agoHelpfull: Yes(0) No(1)