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705-805 (Hard)|   Geometry|      
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Bunuel
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ABCD is an isosceles trapezium. If EG||DC, AE||BF, FH||GC and angle A 18 Mar 2021, 06:30
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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. \(50\sqrt{3}\)


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A
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ABCD is an isosceles trapezium. If EG||DC, AE||BF, FH||GC and angle A 28 Mar 2021, 02:56
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Shaded area is equivalent to area of 4 equilateral triangles, each with side 5.

Total area = \(4*(\frac{\sqrt{3 }*5^2}{4})\) = \(25\sqrt{3}\)
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ABCD is an isosceles trapezium. If EG||DC, AE||BF, FH||GC and angle A 28 Mar 2021, 08:44
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Bunuel

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. \(50\sqrt{3}\)


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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
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