shasadou 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?

A. 81/16

B. 9/4

C. 9/5

D. 5/4

E. 4/5

HI,

Two ways to find the answer

lets take each side of ABCD as 3

1)what you have as a parallelogram inside is also a square EFGH..

what is each side of the square EFGH ?

Take TRiangle EHA..

its sides are \(AE=\frac{1}{3} *3=1\) and \(AH = 3-\frac{1}{3}*3= 2\)

so HYP or side EH = \(\sqrt{1^2+2^2}\)=\(\sqrt{5}\)

so area of EFGH= \(\sqrt{5}\)*\(\sqrt{5}\)=5

and area of ABCD=3*3=9

ans = 9/5..

2) when we know

Take TRiangle EHA..

its sides are AE= \(\frac{1}{3} *3=1\)and \(AH = 3-\frac{1}{3}*3= 2\)

the area of this triangle = 1/2 * 1*2 =1..

there are 4 such triangles at each vertices A,B,C,D..

so Area =1*4..

area of ABCD =9..

area of EFGH= area of ABCD - Area of 4 triangles= 9-4=5

ans=\(\frac{9}{5}\)