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# If each of the line segments in the figure above has a length equal to

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Manager
Joined: 23 Sep 2016
Posts: 220
If each of the line segments in the figure above has a length equal to  [#permalink]

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14 Mar 2018, 00:54
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Difficulty:

25% (medium)

Question Stats:

86% (02:35) correct 14% (02:02) wrong based on 21 sessions

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If each of the line segments in the figure above has a length equal to 4, and the angle a,b and c are each 60 degrees, what is the total area of the shaded figure?

A. 16
B. $$8\sqrt{3}$$
C. $$24\sqrt{2}$$
D. 36
E. $$24\sqrt{3}$$

Source :- globalexperts

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Screenshot (72).png [ 6.28 KiB | Viewed 524 times ]
Manager
Joined: 23 Sep 2016
Posts: 220
If each of the line segments in the figure above has a length equal to  [#permalink]

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14 Mar 2018, 01:10
rishabhmishra wrote:

If each of the line segments in the figure above has a length equal to 4, and the angle a,b and c are each 60 degrees, what is the total area of the shaded figure?

A. 16
B. $$8\sqrt{3}$$
C. $$24\sqrt{2}$$
D. 36
E. $$24\sqrt{3}$$

Source :- globalexperts

Attachment:
The attachment Screenshot (72).png is no longer available

If you look at the attached photo we divide the figure into triangles
SO if each side is 4 and a,b and c=60
then take one triangle lets take c triangle
2 sides are equal and one is 60 then 2 sides is also 60 as if we take both side as x then 2x+60=180
and that means 3rd side is also equal to 4
now we can say these things will true in all triangles and each triangle is equilateral triangle
so area of equilateral triangle is $$\sqrt{3}side^2/4$$ (root3*sidesquare/ 4)
so$$\sqrt{3}*4^2*6(total triangles in the figure)/4$$
$$24\sqrt{3}$$
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InkedScreenshot (72)_LI.jpg [ 806.94 KiB | Viewed 373 times ]

Math Expert
Joined: 02 Aug 2009
Posts: 6973
If each of the line segments in the figure above has a length equal to  [#permalink]

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14 Mar 2018, 02:13
1
rishabhmishra wrote:

If each of the line segments in the figure above has a length equal to 4, and the angle a,b and c are each 60 degrees, what is the total area of the shaded figure?

A. 16
B. $$8\sqrt{3}$$
C. $$24\sqrt{2}$$
D. 36
E. $$24\sqrt{3}$$

Source :- globalexperts

Attachment:
The attachment Screenshot (72).png is no longer available

Hi...
All three figures are parallelogram with side 4..
Just take one of them..
Draw a PERPENDICULAR from a point to get height of this llgm.
We get a 30-60-90 triangle where hypotenuse is 4..
So side opposite 60 will be $$\frac{√3*4}{2}=2√3$$
Area of that llgm =$$4*2√3=8√3$$

Area of all three llgms = $$3*8√3=24√3$$

E
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gc1.png [ 8.3 KiB | Viewed 317 times ]

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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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If each of the line segments in the figure above has a length equal to &nbs [#permalink] 14 Mar 2018, 02:13
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