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# In the given figure ABCD is a rectangle and EFGHIJ is a regular hexago

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Intern
Joined: 15 Nov 2018
Posts: 22
Location: United States
Concentration: Finance, Entrepreneurship
GPA: 3.76
In the given figure ABCD is a rectangle and EFGHIJ is a regular hexago  [#permalink]

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14 Jan 2019, 11:18
6
00:00

Difficulty:

65% (hard)

Question Stats:

58% (03:01) correct 42% (02:55) wrong based on 44 sessions

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In the given figure ABCD is a rectangle and EFGHIJ is a regular hexagon. If the length of AD = 10cm then, find the area of the hexagon.

A) $$\frac{50}{√3}$$

B) $$50√3$$

C) $$100√3$$

D) $$100$$

E) $$100√3$$
Intern
Joined: 11 Dec 2018
Posts: 9
Re: In the given figure ABCD is a rectangle and EFGHIJ is a regular hexago  [#permalink]

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14 Jan 2019, 13:58
1

Posted from my mobile device
Intern
Joined: 27 Nov 2018
Posts: 30
Re: In the given figure ABCD is a rectangle and EFGHIJ is a regular hexago  [#permalink]

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14 Jan 2019, 15:51
2
1
Given EFGHIJ is a regular hexagon, we can conclude that
1) all sides are equal, which makes four triangles similar triangles, which means AJ=JD=BG=GC=10/2=5
2) each angle is 120 degrees. Which makes <AEJ, BFG, GHC, JID = 60 degrees.
Hence the four triangles are all 30-60-90 triangle, which follows the ratio of 1:√3:2
Let a=side of hexagon, we then get AJ/a=5/a=(√3)/2, a=10/(√3)

Area of a regular hexagon: (3√3)/2 * a^2
Sub in 10/(√3), A=50√3

B is correct.
Re: In the given figure ABCD is a rectangle and EFGHIJ is a regular hexago   [#permalink] 14 Jan 2019, 15:51
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