Bunuel wrote:
What is the area of the triangle in the figure above?
A. \(2+2\sqrt{3}\)
B. \(4\sqrt{3}\)
C. 8
D. \(8\sqrt{3}\)
E. 12
Attachment:
The attachment 2015-12-27_2141.png is no longer available
Hi,
the Q may be 700+, since you require to know to play around with Geometrical figures..
the best way is to look how we can draw a 90 degree angle , which will help us to break the figure down to simpler terms[img]
http://gmatclub.com/forum/download/file.php?mode=view&id=28776[/img]..
here we see the third angle C shouold be 45, so we draw a perpendicular from B to AC..
now we have two triangles ABD and BCD..
lets see the first triangle BCD..
angle BCD and CBD are 45 and CDB is 90...
\(Hyp = BC= 2\sqrt{2}\)...
so the sides BD and CD=2..
area of BCD=1/2 * 2*2=2...
lets see the second triangle ABD..
it is 30-60-90 triangle...
and therefore the sides BD:AD:AB are in ratio \(1:\sqrt{3}:2\)..
BD is 2, so AD = \(2\sqrt{3}\)..
so \(area = 1/2 * 2\sqrt{3}*2=2\sqrt{3}\)..
\(total Area=2+ 2\sqrt{3}\)
ans A
Attachments
ANGLES.png [ 24.18 KiB | Viewed 2050 times ]
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