A certain right triangle has sides of length x, y and z, where x < y <

ALTERNATIVELY

Take any random case

i.e. x = 1/8, y = 16 and calculate z i.e. z > 16

All options eliminated except the first option

Answer: Option A

the answer is A.

since we know that area is 1 and it is given that x < y < z for the right angled triangle, we can say that z is the longest triangle. x*y = 2. let's say y(2) > x(1), then z>2.

Hi brs1cob,

I do understand your approach, though I had a doubt.
For any given triangle if two sides are given then the third side has to be less than the sum and greater than the difference of the two sides.
So in the current case since xy=2, if we take x =1, y=2, then the value of z can only be greater than 1 and less than 3.

Where am I going wrong?

Thanks!

none of the explanations are satisfying. I like the original question.I understand that and can figure out the answer.

as per question z is hypotenuse
=> 1/2*x*y=1
x*y=2

now we know that AM>=GM
(x^2+y^2)/2 >= sqrt(x^2 * y^2)
z^2/2 >= x*y (right angled triangle)
z^2 >= 4
(z+2)(z-2)>=0
z<=-2 or z>=2
z>=2 (since z cannot be negative)

Hence choice A

Unless option A is z>= 2, my guess is that none of the options is correct.

Let's assume x = 2/(sqrt 2) and y = sqrt 2. In this case, (x*y/2) = 2, which is consistent with an area of 1. In addition, using these values and the Pythagorean theorem, we get that z = 2. So none of the options is correct.

We are given a product: the area of the right triangle

(1/2) (x) (y) = 1

Or

XY = 2

By the Pythagorean theorem:

(X)^2 + (Y)^2 = (Z)^2

Given the constant product, if we were to make both Legs EQUAL, this would minimize the Hypotenuse Z

Or

Following the concept underlying the fact that the (Arithmetic Mean) is always greater than or equal to the (Geometric Mean) ——-> the AM will be minimized when the 2 factors are equal

(The basic concept: given a constant product, when you make the variables equal ——> the addition of those variables will be MINIMIZED)

Let X = Y

XY = 2 ——> becomes: (X)^2 = 2

And

X = sqrt(2)

Substituting this value into the Pythagorean theorem, also setting both Legs Equal:

(X)^2 + (Y)^2 = (Z)^2

(X)^2 + (X)^2 = (Z)^2

2(X)^2 = (Z)^2 ——-> where X = sqrt(2)

2 * [ sqrt(2) ]^2 = (Z)^2

2 * 2 = (Z)^2

Z = 2

The minimum value that the hypotenuse Z could take is 2

Or
An Isosceles Right Triangle (given a set, Constant AREA) will have the minimum hypotenuse possible

However, we are told that the right triangle can NOT be an Isosceles Right Triangle since:
X < Y < Z

This means the values of X and Y can vary from being equal ——-> and the value of Z will have to be GREATER than 2

*A*
Z > 2


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2 / sqrt(2) = sqrt(2)

It is just two ways of writing the same value (conjugate the Denominator of the first term and you will see what I mean)

The key is the Given information: X < Y < Z

The right triangle can NOT be Isosceles, in which X = Y

Thus, Z can not take the value of 2.

However as you adjust the values of X and Y away from being Equal, Z will increase accordingly

Z > 2

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