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18 Mar 2021, 06:30
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
48% (02:58) correct
52%
(03:20)
wrong
based on 44
sessions
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ABCD is an isosceles trapezium. If EG||DC, AE||BF, FH||GC, BG = GC = HC, and \(\angle AEG = 60°\), what is the area of the shaded region?
A. \(25\sqrt{3}\)
B. \(30\sqrt{3}\)
C. \(35\sqrt{3}\)
D. \(40\sqrt{3}\)
E. \(50\sqrt{3}\)
Attachment:
2021-03-08_14-25-01.png [ 44.14 KiB | Viewed 3053 times ]
28 Mar 2021, 02:56
Kudos
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Attachment:
Untitled.png [ 6.58 KiB | Viewed 2755 times ]
Shaded area is equivalent to area of 4 equilateral triangles, each with side 5.
Total area = \(4*(\frac{\sqrt{3 }*5^2}{4})\) = \(25\sqrt{3}\)
28 Mar 2021, 08:44
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Bunuel
Connect AF, triangle BFG is an equilateral triangle so BF = BG = 5. We know AB = 5 too so we can find triangle ABF is an equilateral triangle too. Thus the top portion consists of 3 equilateral triangles, the bottom portion would consist of 5 equilateral triangles. The bottom shaded region would have 3 equilateral triangles and the top shaded region is one equilateral triangle.
The shaded area would be \(4 * \frac{\sqrt{3}}{4}*5^2 = 25\sqrt{3}\).
Ans: A
Attachments
triangles.png [ 20.79 KiB | Viewed 2651 times ]
