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14 Mar 2018, 00:54
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E
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Difficulty:
25%
(medium)
Question Stats:
78% (02:05) correct
22%
(01:49)
wrong
based on 76
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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
Attachment:
Screenshot (72).png [ 6.28 KiB | Viewed 5280 times ]
14 Mar 2018, 01:10
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rishabhmishra
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\(\sqrt{3}*4^2*6(total triangles in the figure)/4\)
answer is
\(24\sqrt{3}\)
Attachments
InkedScreenshot (72)_LI.jpg [ 806.94 KiB | Viewed 4790 times ]
14 Mar 2018, 02:13
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rishabhmishra
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
Attachments
gc1.png [ 8.3 KiB | Viewed 4722 times ]
