Given a regular hexagon ABCDEF whose area is 42sq m,than find the area of a pentagon made by joining the verticesABCEF.
35,49/root3,

• ans is 35
  since regular hexagon comprise of 6 equilateral triangles of equal areas,area of each triangle 7 sqm
  area of pentagon abcef = area of hexagon abcdef - area of triangle edc i.e 7
  so ans comes out to be 35
• 10 years agoHelpfull: Yes(36) No(12)
• area of hexagon= 42sqm
  so regular hexagon has 6 equal triangles.
  so area of triangle cde= 42/6= 7
  now area of pentagonal=42-7=35sqm
• 10 years agoHelpfull: Yes(9) No(5)
• ANS= 49/(ROOT 3)
  Formula used, area of regular polygon= ((n*s^2)/4)*cot(180/n)
  where, n= no. of sides and s= length of side
• 10 years agoHelpfull: Yes(7) No(5)
• Area of Hexagon= 6*[root(3)/4] a^2 where a=side of hexagon
  so, 42=6*[root(3)/4]a^2
  a^2=28/root(3)
  thus Area of pentagon= 1.72 * a^2=1.72*28/root(3)=49/root(3)
• 10 years agoHelpfull: Yes(7) No(2)
• dont be confused guys the correct answer will be 35
  
  the formula which u r using like area=ns^2 *cot 180/n is for irregular polygons ok. but here question is regular polygon.
  plz go through the correct propeties of polygon.
• 9 years agoHelpfull: Yes(3) No(0)
• Ans: 35 m^2
  
  Area of Regular hexagon=(3(root3)/2)*a^2= 42 m^2
  a=root(28) / (3 ^ 0.25)---------(1)
  We should cut the Triangle CDE.
  CDE is two sided equal triangle.
  Sin 60=(CE/2) / a
  CE=b=root28 * 3^0.25-------(2)
  Area of isosceles triangle(CDE)(with sides a,a,b)
  A=(b/4) * [root( (4*a^2) - b^2) ]------(3)
  Using (1) , (2) & (3) A= 7 m^2
  Required Area=42-7=35 m^2
• 8 years agoHelpfull: Yes(3) No(0)
• answer is 35
  
• 10 years agoHelpfull: Yes(2) No(6)
• since area of hexagon consist of 6 equilateral triangle, therefore the area of pentagon is
  42-(42/6)=35
• 10 years agoHelpfull: Yes(1) No(3)
• Hexagon is comprised of 6 equilateral triangles. Itz Area is given to be 42 sq m i.e each triangle is having area=7sq m.
  Hence required area of pentagon formed is 42-7=35sq m.
• 10 years agoHelpfull: Yes(1) No(1)
• Ans: 35
  
  
  
  
• 9 years agoHelpfull: Yes(1) No(0)
• answer is 31.5 because hexagon divided into 8 parts...then pentagon having 6 parts(triangles)=(42/8)6=31.5
• 10 years agoHelpfull: Yes(0) No(8)
• area of reg hexa=6*(Rt3/4)a^2=42
  a^2=4*42/6rt3
  
  area of pentagon=5*(rt3/4)a^2=35
• 10 years agoHelpfull: Yes(0) No(0)
• area of hexagonal=(square root three)(side square)(divided by four)=42
  =>square of side =56*(square root of three)(divided by 6).
  Now. area to be discarded = 7
  then Ans- is 35
• 9 years agoHelpfull: Yes(0) No(0)
• the answer is first correct because the REGULAR HEXAGON .
• 9 years agoHelpfull: Yes(0) No(0)
• Only restriction given is that base of the cone should be on one of the face of cuboid. Since there are no other restrictions, any value of the base radius is possible such as r=1, 2, 0.2, 0.01 etc because all such bases can be on one of the face of cuboid. Theoretically, infinite number of cones can be cut.
• 7 years agoHelpfull: Yes(0) No(1)