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

Question

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 z?

(A) z > 2
(B) √2 < z < 2
(C) √2 < z < √3
(D) 1 < z < √2
(E) z < 1

Bunuel · Answer

This is a modified question of the the following OG question: https://gmatclub.com/forum/a-certain-ri ... 68727.html

GMATinsight · Answer

Right triangle has sides x, y, z and z being the highest must be Hypotenuse

i.e. (1/2)*x*y = 1
i.e. x*y = 1
  @ y_{min}    ,  x must be maximum  and   z must be Minimum  
and then first unacceptable solution will be 
@x = y, i.e.  y^2 = 2 
@x = y, i.e.  y = \sqrt{2}

@x=y.  z_{min} = 2

but since x<y, therefore z must be greater than 2

Answer: option A

GMATinsight · Answer

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

brs1cob · Answer

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.

Dkingdom · Answer

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!

zvazviri · Answer

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

shiva1325 · 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

mserramo · Answer

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.

Fdambro294 · Answer

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

Posted from my mobile device

Fdambro294 · Answer

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

Posted from my mobile device

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

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