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ABCDEF is a regular hexagon inscribed inside a circle. If the shortest diagonal of the hexagon is of length 3 units, what is the area of the shaded region?

 
Soham Agarwal
(@soham)
Estimable Member

ABCDEF is a regular hexagon inscribed inside a circle. If the shortest diagonal of the hexagon is of length 3 units, what is the area of the shaded region?
CAT Question - Geometry - Regular Octagon

  1. 1/6(3π − (9√3)/2)
  2. 1/6(2π − (6√3)/2)
  3. 1/6(3π − (8√3)/2)
  4. 1/6(6π − (15√3)/2)
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Topic starter Posted : 23/03/2021 8:29 pm
Rahul Singh
(@rahul)
Admin

CAT Question - Geometry - Regular Octagon

Let side of regular hexagon be a.
The shortest diagonal will be of length a√3. Why?
A regular hexagon is just 6 equilateral triangles around a point. The shortest diagonal is FD.
FD = FP + PD
△FOE is equilateral and so is △ EOD.
Diagonal FD can be broken as FP + PD, both of which are altitude of equilateral �?�s.
FP = (√3a)/2
FD = √3a = shortest diagonal
The question tells us that the shortest diagonal measures 3 cm.
√3a = 3 => a = √3
Radius of circle = √3
Area of hexagon = (√3a2 )/4 * 6
Area of circle – area of hexagon = π (√3)2 − √3/4 * (√3)2 * 6
= 3π − (9√3)/2
Area of shaded region = 1/(6 ) (area(circle) – area(hexagon))
= 1/(6 )(3π − (9√3)/2)

The question is "what is the area of the shaded region?"

Hence, the answer is 1/(6 )(3π − (9√3)/2)
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Posted : 23/03/2021 9:08 pm
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