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Geometry Square and Parellelogram

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by kop » Wed Nov 20, 2013 12:06 am
A square ABCD is drawn and point E is marked on AB such that AE=AB/3. Similarly points F, G and H are marked on the sides of the square such that BF=BC/3, CG=CD/3 and DH=DA/3. If the points E, F, G, and H are connected to make a parallelogram, what is the ratio of the area of square ABCD to the area of parallelogram EFGH?

81/16
9/4
9/5
5/4
4/5
Last edited by kop on Wed Nov 20, 2013 3:08 am, edited 1 time in total.
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by theCodeToGMAT » Wed Nov 20, 2013 12:40 am
kop wrote:A square ABCD is drawn and point E is marked on AB such that AE=AB3. Similarly points F, G and H are marked on the sides of the square such that BF=BC3, CG=CD3 and DH=DA3. If the points E, F, G, and H are connected to make a parallelogram, what is the ratio of the area of square ABCD to the area of parallelogram EFGH?

81/16
9/4
9/5
5/4
4/5
Is the question transcribed correctly?
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by kop » Wed Nov 20, 2013 3:09 am
sorry !! edited now
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by GMATGuruNY » Wed Nov 20, 2013 4:06 am
kop wrote:A square ABCD is drawn and point E is marked on AB such that AE=AB/3. Similarly points F, G and H are marked on the sides of the square such that BF=BC/3, CG=CD/3 and DH=DA/3. If the points E, F, G, and H are connected to make a parallelogram, what is the ratio of the area of square ABCD to the area of parallelogram EFGH?

81/16
9/4
9/5
5/4
4/5
PLUG IN an easy figure that satisfies all of the given constraints:
Image

Square ABCD:
Area = s² = 3² = 9.

∆AEH, ∆BEF, ∆CFG, and ∆DGH:
Area of each triangle = (1/2)bh = (1/2)*2*1 = 1.
Combined area of the 4 triangles = 4*1 = 4.

EFGH:
Area = (square ABCD) - (combined area of the 4 triangles) = 9-4 = 5.

Resulting ratio:
ABCD/EFGH = 9/5.

The correct answer is C.
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by Mathsbuddy » Thu Nov 21, 2013 10:01 am
ABCD area = 3^2 = 9

Each triangle has area (1 * 2)/2 = 1
4 triangles area = 1 * 4 = 4

9-4 = 5

Answer = 9/5
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