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by [email protected] » Tue Oct 13, 2015 8:28 am
Hi gmat_thingie,

This prompt is incomplete - unless you include an actual LENGTH for ANY of the line segments, there's no way to determine the lengths of the other line segments.

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by gmat_thingie » Tue Oct 13, 2015 9:39 am
Yes the edited figure is attached
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by gmat_thingie » Tue Oct 13, 2015 9:41 am
Same question for "perimeter"
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by theCEO » Tue Oct 13, 2015 2:53 pm
gmat_thingie wrote:Same question for "perimeter"
Two equations worthwhile to know:
Sin 30 = 1/2
Sin 60 = sqrt(3)/2

lets label the point on CD where the two points meet as F
lets label the point on BC where the two points meet as E

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Area of reactangle = AD X CD
Area of reactangle = (sqrt 3) x (1 + sqrt 3) = sqrt 3 + 3

Area of trapezium = 1/2 * (AD + CE) * CD
Area of trapezium = 1/2 * (sqrt 3 + 1) * (1 + sqrt 3))
Area of trapezium = 1/2 * (3 + 2 * sqrt 3 + 1) = 1/2 * (4 + 2 * sqrt 3) = 2 + sqrt 3

Area of shaded region = (sqrt 3 + 3) - (2 + sqrt 3) = 1


Perimeter:
If we continue we have the following for the sides of the shaded region:
AB + BE + AE
1+sqrt(3) + (sqrt(3) - 1) + sqrt(2^2 + 2^2)
1+sqrt(3) + (sqrt(3) - 1) + 2sqrt2
2*sqrt(3) + 2sqrt2
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