Regular hexagon ABCDEF is inscribed in a circle. If the shortest diago

Question

https://gmatclub.com/forum/download/file.php?id=54213 Regular hexagon ABCDEF is inscribed in a circle. If the shortest diagonal of the hexagon is 3 cm, what is the area of the shaded region? A. \frac{1}{6}(2π − 2√3) B. \frac{1}{6}(2π − 3√3) C. \frac{1}{6}(3π − \frac{9√3}{2}) D. \frac{1}{6}(3π − 4√3) E. \frac{1}{6}(6π − \frac{15√3}{2}) Are You Up For the Challenge: 700 Level Questions Geometry-Qn-15-1.png

gkumar08 · Answer

Ans will be C

For a regular hexagon the shortest diagonal is √3a where a is its side.
So √3a=3 so a=√3
And if we take any diagonal other than shortest diagonal its length is 2a.
The circumradius will be a.
Now area of circle= area of regular hexagon + 6*area of shaded region
Area of circle = π*√3*√3=3π
Area of a regular hexagon= n*(a^2/4)*cot (π/n)
So area of regular hexagon=(6*3/4)*cot(π/6)=9√3/2
So area of shaded region= 1/6(3π-9√3/2)

So ans is C

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nick1816 · Answer

ABC is an isosceles triangle. Hence, BG bisects the the AC.

AG=GC = \frac{3}{2}

Triangle ABG is 30-60-90 triangle; hence, AB : AG = 2 : √3

AB = \frac{2}{√3} * \frac{3}{2} = √3

Area of shaded region = Area of Sector (EOF) - Area of triangle (EOF)

Area of Sector (EOF) =  \frac{60}{360} *     =  \frac{ π}{2}

Area of triangle (EOF) =  \frac{√3}{4} * (√3)^2  =  \frac{3√3}{4}

Area of shaded region =  \frac{ π}{2}  -  \frac{3√3}{4}  =  \frac{1}{6}(3π − \frac{9√3}{2})

kris19 · Answer

Regular hexagon has 9 diagonal, of which the shortest one connects A and C, and A and E.
If we take diagonal AE, then it is given that AE = 3 cm ------(1)
Angle   AFE = (6-2)*\frac{180}{6}  = 120 and FE = FA (isosceles triangle)
If we draw a perpendicular line from F to diagonal AE, touching at G, then angle AFG = 60.
From (1),  AG = \frac{AE}{2}= \frac{3}{2} . 
From triangle AFG,  AF = \sqrt{3}   and  GF = \sqrt{3}/2, ED = AF = \sqrt{3}  (regular hexagon)
From right triangle AED, hypotenuse AD (dia of the circle) =  \sqrt{(AE^2 + ED^2)} = 2*\sqrt{3}  
So, the area of the circle =  \pi*(\sqrt{3})^2 = 3*\pi 
Area of the hexagon = area of rectangle ABDE + 2*(area of triangle AFE) =  AE*ED + 2*(1/2*FG*AE) = 3*\sqrt{3} + 2*(1/2*\sqrt{3}/2*3) 
=  3*\sqrt{3} + \frac{3*\sqrt{3}}{2} 
Now, the difference between area of the circle and area of the hexagon =  3*\pi - (3*\sqrt{3}+\frac{3*\sqrt{3}}{2}) = 3*\pi - \frac{9*\sqrt{3}}{2} 
The shaded area = 1/6*(difference in area)
=  \frac{1}{6}(3*\pi-\frac{9*\sqrt{3}}{2}) 
Answer (C)

lacktutor · Answer

Please Find The Attached File

∆ AED is a right-angled triangle.
 3^{2} +R^{2}= (2R)^{2}

3R^{2} = 9

--> R= √3

The shaded region =  \frac{1}{6} (πR^{2}- 6(\frac{√3}{4} R^{2})= \frac{1}{6} (3π -\frac{9√3}{2} )

Answer (C).

Regular hexagon ABCDEF is inscribed in a circle. If the shortest diago

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Regular hexagon ABCDEF is inscribed in a circle. If the shortest diago

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