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05 May 2020, 11:55
Kudos
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
37% (02:04) correct
63%
(02:05)
wrong
based on 68
sessions
History
Date
Time
Result
Not Attempted Yet
If ABCDEFGH is a regular octagon, what fraction of the octagon is shaded?
A) 1/12
B) 1/8
C) 1/6
D) 1/4
E) 3/8
06 May 2020, 05:14
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BrentGMATPrepNow
If we focus on rectangle BCFG, we might recognize that the area of the shaded portion is half the area of the rectangle BCFG
How do we know this?
Well, we know that the area of the rectangle = (base)(height)
And we know that the area of the triangle = (base)(height)/2
Since both shapes have the same base and height, it must be the case that the area of the shaded portion is half the area of the rectangle BCFG
With this in mind, let's divide rectangle BCFG into TWO EQUAL rectangles.
From here, we can add two smaller triangles.
Since the area of each small triangle is HALF the area of each small rectangle, we can conclude that the combined areas of the two small triangles is HALF the combined areas of the two small rectangles.
In other words, the combined area of the two shaded triangles is the same as the area of the original triangle (triangle YGF)
At this point, when we add a couple of extra diagonals to the diagram...
.... we can see that the shaded portions account for 2/8 of the entire octagon.
Since 2/8 = 1/4, the correct answer is D.
Cheers,
Brent
General Discussion
05 May 2020, 18:24
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As base and height of triangle YGF is equal to those of triangle BGY, Area of triangle YGF = Area of triangle BGY = 1/2 (area of rectangle BCGF)
\(BI = AI = \frac{a}{\sqrt{2}}\)
Hence area of rectangle BCGF = \(2*a*\frac{a}{\sqrt{2}} + a^2= (\sqrt{2}+1)a^2\)
Area of octagon = \(2*(\sqrt{2}+1)a^2 \)= 2*area of rectangle BCGF =2*2*(Area of triangle YGF)
(Area of triangle YGF)/Area of octagon = 1/4
\(BI = AI = \frac{a}{\sqrt{2}}\)
Hence area of rectangle BCGF = \(2*a*\frac{a}{\sqrt{2}} + a^2= (\sqrt{2}+1)a^2\)
Area of octagon = \(2*(\sqrt{2}+1)a^2 \)= 2*area of rectangle BCGF =2*2*(Area of triangle YGF)
(Area of triangle YGF)/Area of octagon = 1/4
BrentGMATPrepNow
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