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03 Jun 2019, 21:41
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Difficulty:
75%
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Question Stats:
45% (03:13) correct
55%
(03:39)
wrong
based on 20
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Points E, F, G, H lie on the sides AB, BC, CD, and DA, respectively, of a square ABCD. If EFGH is also a square whose area is 62.5% of that of ABCD and CG is longer than EB, then the ratio of length of EB to that of CG is
A. 1:4
B. 1:3
C. 2:5
D. 4:9
E. 2:3
A. 1:4
B. 1:3
C. 2:5
D. 4:9
E. 2:3
04 Jun 2019, 23:23
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nick1816
look at the attached sketch.
Since both are squares, each side will be cut in the same ratio, so \(AE=BF=CG=DH....and.... EB=FC=GD=HA\)
Area of smaller square EFGH = \(EF^2=BE^2+BF^2=BE^2+GC^2\)
Area of larger square ABCD = \(AB^2=(AE+BE)^2=BE^2+AE^2+2*AE*BE=BE^2+CG^2+2*BE*CG\)
It is given that \(EF^2=\frac{625}{1000}AB^2\)
\(BE^2+GC^2=\frac{5}{8}(BE^2+CG^2+2*BE*CG)\)
\(8BE^2+8GC^2=5(BE^2+CG^2+2*BE*CG)\)........
\(3BE^2+3CG^2-10*BE*CG=0\).......
\((3BE-CG)(3CG-BE)=0\)
SO, either 3BE=CG or 3CG=BE.
Since CG>BE, CG=3BE.....\(\frac{BE}{CG}=\frac{1}{3}\)
B
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