perimeter of a regular hexagon ABCDEF is 42.find the area of ABDEF polygon


option are : a) 49
b)49 root 3
c) 36
d) 36 root3/2

asked today

• none of the answers is correct .there is some problem with the question,i also think they have asked for quad ABED ,then only 49 root 3 could be the answer
• 7 years agoHelpfull: Yes(7) No(1)
• perimeter=42,which is equal to the sum of side.
  here a=6
  so a=42/6=7
  area of hexagon=(3root3*a^2)/2
  (3root3*49)/2
• 7 years agoHelpfull: Yes(5) No(1)
• Let ABCDEF be a regular hexagon.
  Given that perimeter is 42.
  therefore,side will be 42/6 = 7.
  Area of regular hexagon = (3*root(3)a^2)/2 = (3*root(3)*49)/2
  For regular hexagon, BC = a, AD = 2a, BD = root(3)*a = 7*root(3).
  In isisceles traingle BCD, BC = CD = 7 and BD = root(3)*7
  From Pythagoras th. height will be 7/2.
  So Area of traingle BCD = base*height/2 = 49*root(3)/4.
  So area of ABDEF = ABCDEF - BCD = ((49*3*root(3)/2) - (49*root(3))/4)) = 49*root(3)*5/4
• 6 years agoHelpfull: Yes(5) No(1)
• Actually polygon ABDEF, there are 4 equilateral triangle with side 7.
  Then the option B is correct, 49 root3
• 7 years agoHelpfull: Yes(3) No(7)
• quesion was asked that to find the area of ABDE only not F.
• 7 years agoHelpfull: Yes(3) No(1)
• 36
  Req Area=Area of hexagon -Area of Triangle BCD
  Angle BCD=120
  Area of triangle =1/2*Product of two sides *Angle between them
  1/2*Root 3/2*24/Root 3=6
  Required Area =42-6=36
• 7 years agoHelpfull: Yes(1) No(0)
• the diagram ABDE would be a parllelogram
  so area will be
  1/2*d1*d2*sin(theta)=1/2*14*14*root(3)/2=49root3
• 7 years agoHelpfull: Yes(0) No(4)
• Answer: b) 49 root 3
  Explanation:
  6*a=42
  therefore a=7
  we are asked to find the area of ABDEF leaving the node C.
  hence we will have only 4 equilateral triangles instead of 6.
  therefore, required area= 4* (root3/4) *a*a
  req. area= 49 root3
  :P
• 7 years agoHelpfull: Yes(0) No(7)
• Sorry for the previous post answer would be 35 .In place of 24/Root3 there would be 28/root3.And here Area is equal to 42 not perimeter.
• 7 years agoHelpfull: Yes(0) No(0)
• Answer - 5/4* 49root(3)
  (a=7)
  Area of regular hexagon = (3 root(3))/2 * a^2 = (3 root(3))/2 * 49
  Draw perpendicular from Vertex C to line Segment BD
  Now, Using Trigonometry find BD, which is equivalent to 7root(3)
  Area of rectangle ABDE = 49root(3)
  Area of two triangle i.e. AFE & BCD = Area of Hexagon - Area of rectangle ABDE
  = 49 root(3)/2
  Area of two triangle = 49 root(3)/4
  Area of polygon ABDEF = (1+1/4) 49 root(3) = 5/4 49root(3)
• 6 years agoHelpfull: Yes(0) No(0)
• none of the options is correct.
  the area of ABDEF is 245 root 3 by 4 squints.
• 6 years agoHelpfull: Yes(0) No(0)