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605-655 (Medium)|   Geometry|      
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Bunuel
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In the figure above, triangles ADE and BFG are equilateral triangles 22 Jan 2021, 06:00
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In the figure above, triangles ADE and BFG are equilateral triangles each with side 2 cm and EF = 2 cm. What is the area of the pentagon CDEFG?


A. \(15 \sqrt{3}\) cm^2

B. \(12 \sqrt{3}\) cm^2

C. \(10 \sqrt{3}\) cm^2

D. \(8 \sqrt{3}\) cm^2

E. \(7 \sqrt{3}\) cm^2


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E
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In the figure above, triangles ADE and BFG are equilateral triangles 23 Jan 2021, 07:25
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Area of Equilateral triangle =\(\frac{\sqrt{3}}{4} a^2\)

Area of ADE & BFG = \(\frac{\sqrt{3}}{4}\)*4*2 = 2\(\sqrt{3}\)

Area of ACB = \(\frac{\sqrt{3}}{4}\)*36 = 9\(\sqrt{3}\)

Area of CDEFG = 9\(\sqrt{3}\) - 2\(\sqrt{3}\) = 7\(\sqrt{3}\)

Option E
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In the figure above, triangles ADE and BFG are equilateral triangles 23 Jan 2021, 08:33
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The answer to this question is D \(8\sqrt{3}\)
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In the figure above, triangles ADE and BFG are equilateral triangles 23 Jan 2021, 08:44
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Area of Equilateral triangle =\(\frac{\sqrt{3}}{4} a^2\)

Area of ADE & BFG = \(\frac{\sqrt{3}}{4}\)*4*2 = 2\(\sqrt{3}\)

Area of ACB = \(\frac{\sqrt{3}}{4}\)*36 = 9\(\sqrt{3}\)

Area of CDEFG = 9\(\sqrt{3}\) - 2\(\sqrt{3}\) = 7\(\sqrt{3}\)

Option E

Bunuel Is ABC also an Equilateral triangle.
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Since ADE & BFG are equilateral \(\triangle\)s, \(\angle\) A and \(\angle\) B are 60 each, so in \(\triangle ABC\) \(\angle\) A = \(\angle\) B =60, and the third angle too becomes 60. Thus ABC too is an equilateral \(\triangle\).
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In the figure above, triangles ADE and BFG are equilateral triangles 23 Jan 2021, 08:48
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I missed it in the beginning, but figured it out later.
Thanks.
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In the figure above, triangles ADE and BFG are equilateral triangles 23 Jan 2021, 08:58
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Bunuel

In the figure above, triangles ADE and BFG are equilateral triangles each with side 2 cm and EF = 2 cm. What is the area of the pentagon CDEFG?


A. \(15 \sqrt{3}\) cm^2

B. \(12 \sqrt{3}\) cm^2

C. \(10 \sqrt{3}\) cm^2

D. \(8 \sqrt{3}\) cm^2

E. \(7 \sqrt{3}\) cm^2


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Look at the attached figure.
Since ADE & BFG are equilateral \(\triangle\)s, \(\angle A\) and \(\angle B\) are 60 each, so in \(\triangle ABC\), \(\angle A\) \(=\angle B = 60\) , and the third angle, \(\angle C\), too becomes 60. Thus ABC too is an equilateral \(\triangle\) with each side = 2+2+2=6

Area of \(\triangle ABC\) =\(\frac{\sqrt{3}}{4}*a^2\) =\(\frac{\sqrt{3}}{4}*36=9\)\(\sqrt{3}\)

Area of \(\triangle ADE=\) \(\triangle BFG\) = \(\frac{\sqrt{3}}{4}*4=\) \(\sqrt{3}\)


Area of CDEFG = (Area of \(\triangle ABC\))-( Area of \(\triangle ADE+\) \(\triangle BFG\))=\(9\sqrt{3}\) - 2\(\sqrt{3}\) = 7\(\sqrt{3}\)

E
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In the figure above, triangles ADE and BFG are equilateral triangles 15 Feb 2021, 10:49
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Similar triangle problem. Took me a brutal 10 minutes.


Step 1: Determine the height of the smaller triangle

A(ADE) = A(FGB) = √3 / 4 x s2 = √3/4 x 22 = √3

√3 = bh/2 --> √3 = 2h/2 --> h = √3

Step 2: Determine the ratio between the height of the larger triangle ABC and set this equal to the ratio between the base AB and AE

x + √3 / √3 = 6/2 = 3/1 --> x = 3√3 - √3

Step 3: Determine the area of the full triangle
3 √3 x 6 / 2 = 9 √3

Step 4: Subtract the area of the two smaller triangles from the larger triangle

9 √3 - 2 √3 = 7 √3 = A(pentagon)
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