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

krakow
11-15-2004, 12:20 PM
In the fig above, if the area of the smaller square region is ½ 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)√2–1 b)1/2 c) (√2)/2 d)[ (√2)+1]/2 e) √2

======================
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

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

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

swethav12
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
√2