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18 Mar 2021, 06:00
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
37% (02:47) correct
63%
(02:49)
wrong
based on 27
sessions
History
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Time
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ABCDEF is a regular hexagon of side 6 cm. What is the area of the shaded region ?
(A) 16 cm^2
(B) 18 cm^2
(C) \(16\sqrt{3}\) cm^2
(D) \(18\sqrt{3}\) cm^2
(E) \(24\sqrt{3}\) cm^2
Attachment:
2021-03-08_14-18-12.png [ 18.03 KiB | Viewed 4206 times ]
19 Mar 2021, 01:16
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Bunuel
See the attached figure.
\(\triangle aed\) is 30-60-90, so ed:ae:ad=\(1:\sqrt{3}:2=6:6\sqrt{3}:12\)
Area of \(\triangle aed =\frac{1}{2}*6*6\sqrt{3}=18\sqrt{3}\)
\(\triangle ged\) is 30-60-90, so eg:ed:gd=\(1:\sqrt{3}:2=\frac{6}{\sqrt{3}}:6:\frac{12}{\sqrt{3}}\)
Area of \(\triangle ged =\frac{1}{2}*6*6/\sqrt{3}=\frac{18}{\sqrt{3}}=6\sqrt{3}\)
Area of shaded portion = 2*(A of \(\triangle aed\)-A of \(\triangle ged\)=\(2(18\sqrt{3}-6\sqrt{3})=24\sqrt{3}\)
E
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2021-03-08_14-18-12.png [ 20.55 KiB | Viewed 3808 times ]
20 Mar 2021, 17:47
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Bunuel
The area of the hexagon is 6 times that of an equilateral triangle of length 6. so the area is \(6*\frac{\sqrt{3}}{4}*6^2 = 54\sqrt{3}\). We can subtract twice the area of triangle AFE and twice the area of the right triangle with a base of ED.
Triangle AFE: Split AFE into two right triangles. Each triangle is a 90-60-90 triangle. AF = 6 so the corresponding base a right triangle is \(6*\frac{\sqrt{3}}{2}= 3\sqrt{3}\) and the height is \(3\). The area of the two right triangles added is \(3*3\sqrt{3} = 9\sqrt{3}\). Two of AFE is \(18\sqrt{3}\).
Right triangle with a base of ED: Base is ED = 6, height is \(\frac{6}{\sqrt{3}} = 2\sqrt{3}\). Two of these right triangles has an area of \(6*2\sqrt{3} = 12\sqrt{3}\).
Thus finally the area left is \((54 - 30)\sqrt{3}\).
Ans: E
