Difficult Math Problem #116 - Triangles

The problem itself might be difficult, but that does not mean getting it right on the exam is difficult.

Area = xy/2=1 hence xy = 2
Now
x < y < z

x can be very very small while y might be very large while keeping xy = 2.
Think of a very thin and long triangle.

That means x very near to 0 and y has no upper limit. Eliminate choices which put an upper limit to y.
Surprise Surprise! B,C,D and E are gone, and you are left with A, which must be the correct answer! That was easy!

For the sake of ego, let us arrive at the correct answer ourselves rather than looking at the choices.

Think of the upper limit for x. Since x < y, the maximum value x can reach is very near to y, almost y, but less than y. So in this case, xy will be almost y squared. y squared is almost 2, so y is almost ROOT2, but slightly higher. So y>ROOT2.

Since this is a right angled triangle, the hypotenuse should be the largest
of the 3 sides. We are given x < y < z which implies z is the hyp.

Area is 1/2 * x * y (we don't care whether x is base or y is base)

1/2 * x * y = 1
xy = 2

If x < y, y has to be > root-2 (else the product cannot equal 2). So,
we have y > root-2

Scanning the answer choices -- B, C, D and E all indicate that y < root-2.

Is A the answer ?

I can't seem to figure out the upper-bound on y with the info given. We
know that y > z-x and y < z+x (property of triangles) but this doesn't
give me the upper-bound for y. Not sure if it is needed for the question.

The triangle is a right angle triangle and x < y < z
Thus z for sure is the hypotenuese for z^2 = x^2 + y^2
Area of a triangle = 1/2 * Base * Height
1/2 * Base * Height = 1
Thus Base * height = 2
Hence we know that the base (one side of the triangle) and the height (the other side of the triangle) = 2
In case of an isoceles right angle triangle base = height = sqr rt. 2
Since it's not an isoceles right angle triangle (with x<y) the greater side has to be more than sqr rt 2
Hence Choice A

OA:

right triangle with sides x<y<z and area of 1 => z = hypotenuse and xy/2 = 1
i.e xy = 2

If x were equal to y, we would have had xy = y^2 = 2. And y = root2

But, x<y and so y>root2.