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SHADED REGION (GEOMETRY_Uploaded image)

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Source: — Problem Solving |
Junior | Next Rank: 30 Posts
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by gmat_thingie » Tue Oct 13, 2015 4:23 am
Clearer image
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Area of shaded triangle_ABF.PNG
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by theCEO » Tue Oct 13, 2015 2:32 pm
gmat_thingie wrote:Clearer image
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

Image

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

Ans = b
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by Matt@VeritasPrep » Wed Oct 14, 2015 11:08 pm
theCEO wrote: Two equations worthwhile to know:
Sin 30 = 1/2
Sin 60 = sqrt(3)/2
These equations are not at all worthwhile on the GMAT, as trigonometry is not tested.

The proportions of a 30-60-90 are essential, but you can find them by splitting an equilateral in half and using the Pythagorean Theorem should you forget them; do NOT dust off your trig books and start learning about sines and cosines!
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by gmat_thingie » Thu Oct 15, 2015 7:42 am
Hi

I think I went through this question again and found it fairly simple.

AD = ROOT 3 AND AE = 2 (30/60/90 Triangle calculation)
EF = 2, AF = 2 ROOT 2 (45,45,90 Triangle traits), AE IS known from above
FOR ECF, EC = ROOT 3 AND FC = 1 on account of EF = 2
BF = ROOT 3 - 1 (ROOT 3 - FC)
AB = 1 + ROOT 3

PERIMETER = 2 ROOT 2 + 2 ROOT 3 (OPTION 5)

SQUARE OF the shaded triangle ABF = 1
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