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01 May 2020, 11:13
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C
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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
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01 May 2020, 13:08
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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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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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01 May 2020, 13:25
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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) =\( \frac{60}{360} * [π *(√3)^2] \) = \(\frac{ π}{2}\)
Area of triangle (EOF) = \(\frac{√3}{4} * (√3)^2 \)= \(\frac{3√3}{4}\)
Area of shaded region = \(\frac{ π}{2}\) - \(\frac{3√3}{4}\) = \(\frac{1}{6}(3π − \frac{9√3}{2})\)
\(AG=GC = \frac{3}{2}\)
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) =\( \frac{60}{360} * [π *(√3)^2] \) = \(\frac{ π}{2}\)
Area of triangle (EOF) = \(\frac{√3}{4} * (√3)^2 \)= \(\frac{3√3}{4}\)
Area of shaded region = \(\frac{ π}{2}\) - \(\frac{3√3}{4}\) = \(\frac{1}{6}(3π − \frac{9√3}{2})\)
Bunuel
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