KarunMendiratta wrote:
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ABCDEF is a regular polygon inscribed in a circle with AC = 6 units.
Quantity A |
Quantity B |
Area of shaded region |
\(\frac{3}{2}\) |
A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Explanation:Refer to the figure below;
ABCDEF is a regular Hexagon with each angle 120° and AC = 6
Drop a perpendicular (BG) in Isosceles triangle ABC with AB = BC.
BG will bisect the angle ABC and the side AC i.e. AG = CG = 3
Now, ABG and BGC are two congruent 30-60-90 triangles
In triangle BGC, GC = \(\sqrt{3}x = 3\)
i.e. \(x = \frac{3}{\sqrt{3}} = \sqrt{3}\)
So, AB = BC = CD = DE = EF = EA = \(2x = 2\sqrt{3}\)
Things to Remember: 1. A regular Hexagon gets divided into 6 equilateral triangles when all its vertices are joined from the centre (here O)
2. Area of Sector = \(\frac{θ}{360}πr^2\)
3. Area of an equilateral triangle = \(\frac{\sqrt{3}}{4}a^2\)
Therefore, DE = EO = OD = \(2\sqrt{3}\)
Also, radius of the circle O = \(2\sqrt{3}\)
Area of Shaded region = Area of sector EOD - Area of equilateral triangle EOD
= \(\frac{60}{360}π(2\sqrt{3})^2\) - \(\frac{\sqrt{3}}{4}(2\sqrt{3})^2\)
= \(\frac{π}{6}(2\sqrt{3})^2 - \frac{\sqrt{3}}{4}(2\sqrt{3})^2\)
= \((\frac{π}{6}-\frac{\sqrt{3}}{4})(2\sqrt{3})^2\)
= \(\frac{4π-6\sqrt{3}}{24}(12)\)
= \(\frac{4π-6\sqrt{3}}{2}\)
Col. A: \(\frac{4π-6\sqrt{3}}{2}\)
Col. B: \(\frac{3}{2}\)
Multiplying by \(2\) both sides;
Col. A: \(4π-6\sqrt{3}\)
Col. B: \(3\)
Col. A: \(2.17\)
Col. B: \(3\)
Hence, option B
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