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Joining the mid points of a regular hexagon of side a units a second hexagon is formed. Again a third hexagon is formed by joining the mid points of the second hexagon. This process is repeated infinite number of times. What is the sum of areas of all the hexagons so formed?
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
Joining the mid points of a regular hexagon of side a units a second h...
Area of the outermost hexagon ABCDEF =
As shown in the figure infinite number of hexagons is drawn.
Draw perpendicular AM on side UP.
Since m∠UAP = 120° and AM is the perpendicular bisector, Δ AMU is a 30- 60-90 triangle
UP — 2UM — side of the second hexagon =
Area of the second hexagon PQRSTU =
∴ A rea o f th e second hexagon is 3/4tim es th a t o f th e first

∴ Similarly , area of the third hexa gon is 3/4 times  that of  the  seco nd and  so on, 
∴ The areas of all the inner hexagons form an infinite decreasing G.P.
Com m on ratio (r) 3/4 nd first term (a )
 ∴ Sum of the are as  of  all  the  hexagons = 
Hence, option 2.
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Joining the mid points of a regular hexagon of side a units a second hexagon is formed. Again a third hexagon is formed by joining the mid points of the second hexagon. This process is repeated infinite number of times. What is the sum of areas of all the hexagons so formed?a)b)c)d)Correct answer is option 'B'. Can you explain this answer?
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