The triangle is right, hence its biggest side z ought to be a hypotenuse and x and y ought to be its legs. The area of a right triangle is determined to be the half of the product of its legs: 1/2 x y = S = 1, therefore x y = 2.
We know that x < y, so we start to check the answers to find a pair of values of x and y with a product equal to 2. We're not going to check all the possible values, but we're going to check whether 2 is in the range of minimal and maximal values of x and y product. If the maximal value of such product is less than 2, then we simply cut out the option.

A. y > √2, then x ≤ √2: MAX(x y) = √2√2 = 2. O.K.!

B. √3 / 2 < y < √2, then x ≤ √3 / 2: the maximum value that can be obtained is x y = √2 √3 / 2 = √6 / 2. The root of 6 is sure to be less than 4, divided by 2 it gives a number certainly less than 2.

C. √2 / 3 < y < √3 / 2, then x ≤ √2 / 3. We act in the same way as above: x y = (√2 / 3) (√3 / 2) = √6 / 6 = 1 / √6 < 2.

D. √3 / 4 < y < √2 / 3, then x ≤ √3 / 4. We act in the same way as in B: x y = (√3 / 4) (√2 / 3) = √6 / 12 = 1 / 2√6 < 2.

E. y < √3 / 4, assuming y to be the closest number to √3 / 4, we get that maximal x has also to be the closest number to √3 / 4, hence their maximal product is the closest number to (√3 / 4)(√3 / 4) = 3/16, which is no doubt less than 2.

Thanks for ur reply ...........good method..ton of thanks

pardeep_10sharma wrote:A certain right triangle has sides of length x, y, and z, where x < y < z. If the area of this triangular region is 1, which of the following indicates all of the possible values of y ?
A. y > Suare root (2)
B. {Square root( 3) }/ 2 < y < Square root (2)
C. {Square root( 2) }/ 3 < y < {Square root( 3) }/ 2
D. {Square root( 3) }/ 4 < y < {Square root( 2) }/ 3
E. y < {Square root( 3) }/ 4

Ans is :- A ............Plz help

Also, the two legs measure x and y, so the area is xy/2

Since x < y, xy/2 < y^2/2

Thus y^2/2 > 1 and therefore y^2 > 2 i.e. y > sqrt(2)

z is hypotaneous and x and y are legs
so 1/2xy=1 so xy=2
since x<y, so xy/2<y^2/2 so y>root2
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