View Full Version : ETS question - Geometry

11-15-2004, 10:21 AM
Pls see the attachment

11-15-2004, 12:20 PM
In the fig above, if the area of the smaller square region is the area of the larger square region, then the diagonal of the larger square is how many inches longer than the diagonal of the smaller square?

a)√21 b)1/2 c) (√2)/2 d)[ (√2)+1]/2 e) √2

Answer: A
let a and b be sides of big and small squares adequate,

we know that
1/2 a^2 = b^2

and we are looking for diagonals' lenght difference,
big square's diagonal lenght is (from P.T.) sqrt(a^2 + a^2), and small square's diagonal lenght is sqrt(2 b^2)

difference is
sqrt(2 a^2) - sqrt(2 b^2) = sqrt(2)a - sqrt(2*1/2*a^2) = a (sqrt(2) - 1)
so it is (sqrt(2) - 1) fraction of a, hence
diagonal is (sqrt(2) - 1) inch longer


11-16-2004, 07:45 PM
how do I send attachment...?

11-16-2004, 07:46 PM
Simple answer - with concept of formula in attached file

11-19-2004, 12:59 AM
bigger side=1 inch
bigger square diagonal= √2
area of smaller square= 1/2 sq inch(half the area of bigger square)
so smaller square side =1/ √2
smaller square diagonal length = 1
difference between two diagonals is √2 -1