joconnor wrote:A certain right triangle has sides of length x, y, z where x<y<z. Area of tri is 1. Which indicates all the possible values of y?

y > sq rt 2

sq rt 3 / 2 < y < sq rt 2

sq rt 2 / 3 < y < sq rt 3 / 2

sqrt3 / 4 < y < sqrt2 / 3

y < sqrt3 / 4

thanks!

[spoiler]Is it A?[/spoiler]

in a right angled triangle, the area with above conditions is 1/2 * x * y
=> xy/2 = 1 (from given)
xy = 2
since x<y
y should be atleast greater than sqrt(2) to follow that condition.

user123321

A certain right triangle has sides of length x, y, z where x<y<z. Area of tri is 1. Which indicates all the possible values of y?.

Since x<y<z and length of Hypotenuse > length of the side(any side), we know that z is the length of the hypotenuse.
Area of triangle = 1/2 * base * height = 1/2 (Side X) * (Side Y) = (1/2)*x*y = 1 Implies xy = 2 and x = 2/y

From the question x<y.
Implies 2/y<y
Implies y-(2/y) > 0
Implies (y^2-2)/y > 0
Implies (y-√2)(y+√2)/y > 0
Checkpoints here are at y = √2, -√2 and 0
Since y is length of a side of a triangle, y > 0, so we need not check for y < 0

Condition 1

0<y<√2
(y-√2)(y+√2)/y < 0, doesn't satisfy the inequality.

Condition 2

y>√2
(y-√2)(y+√2)/y > 0, satisfies the inequality.

IMO option A

answer is a

thanks!