A certain right triangle has sides of length x, y, and z, where x \sqrt {2} B. \frac {\sqrt {3}} {2} < y < \sqrt {2} C. \frac {\sqrt {2}} {3} < y < \frac {\sqrt {3}} {2} D. \frac {\sqrt {3}} {4} < y < \frac {\sqrt {2}} {3} E. y < \frac {\sqrt {3}}{4}

x, \ y, \, and \ z \ are \ the \ sides \ of \ a \ right \ triangle \ where \ x 2 y>\sqrt{2} The answer is A.

So I was stuck on this for a bit but then it hit me. We know x * y = 4 And x radical2 Posted from my mobile device

BrentGMATPrepNow For what Bunuel mentions above "2 2√<y2<y ." Don't you need to flip the sign when taking the square root (or consider alternative versions)?

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

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tonebeeze

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. \(y > \sqrt {2}\)

B. \(\frac {\sqrt {3}} {2} < y < \sqrt {2}\)

C. \(\frac {\sqrt {2}} {3} < y < \frac {\sqrt {3}} {2}\)

D. \(\frac {\sqrt {3}} {4} < y < \frac {\sqrt {2}} {3}\)

E. \(y < \frac {\sqrt {3}}{4}\)

\(x, \ y, \, and \ z \ are \ the \ sides \ of \ a \ right \ triangle \ where \ x<y<z \ so \ z \ must \ be \ the \ hypotenuse.\)

\(The \ area = \frac{1}{2}xy=1\)

\(xy=2\)

\(x=\frac{2}{y}\)

\(Given,\)

\(x<y, \)

so, \(\frac{2}{y}<y\)

\(y^2>2\)

\(y>\sqrt{2}\)

The answer is \(A.\)

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(1/2)xy=1
=> xy=2

Also, y>x
Therefore, y cannot be less than 1
Therefore (A)

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02 Dec 2021, 15:04

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So I was stuck on this for a bit but then it hit me.

We know x * y = 4

And x < y

Options B C D E all give y constraints that are less than 2 AND if Y is less then two we will never be able to find an X value LESS than y that will allow us to multiply to a product of 4.

Therefore,

The only possible constraint would be y > radical2

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tonebeeze

Concept:
For any two positive, real numbers, it will always be true that the Arithmetic Mean of those two numbers will be greater than or equal to the Geometric Mean of those two numbers.

the A.M. will equal the G.M. when the two positive values are the SAME. Since we are told that X < Y, it must be the case that the A.M. exceeds the G.M. for these two positive values

We are given that: x < y < z

A.M. of (x ; y) > G.M. of (x ; y)

and since the hypotenuse is the longest side for any right triangle, the Legs of the right triangle must be: x and y

The Area = (1/2) (x) (y) = 1
Or

xy = 2

A.M. > G.M.

(x + y) / 2 > sqrt(x * y)

x + y > 2 * sqrt(xy)

We know that xy = 2 from the given information about the area of the right triangle.

x + y > 2 * sqrt(2)

And we can use the area formula again to substitute for the variable — x

x*y = 2
So: x = (2/y)

(2/y) + y > 2 * sqrt(2)

(2 + y^2) / y > 2 * sqrt(2)

—can multiply both side by positive Y and we get the quadratic:

(y)^2 — 2 * sqrt(y) + 2 > 0

—-factoring the quadratic we get:

(y - sqrt(2))^2 > 0

or

y - sqrt(2) > 0

y > sqrt(2)

*A*

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It’s a right Angle triangle to the largest Side is the hypotenuse which is Z, base and height can be x and y
So 1/2 base* hieght = area
1/2 x*y =1
X*y = 2

Let assume x,y to root 2
Root 2 * root 2 = 2
But since y>x
Y needs to greater than root 2
And x needs to be some value less than root 2
Hence option A is right

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BrentGMATPrepNow

tonebeeze

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. \(y > \sqrt {2}\)

b. \(\frac {\sqrt {3}} {2} < y < \sqrt {2}\)

c. \(\frac {\sqrt {2}} {3} < y < \frac {\sqrt {3}} {2}\)

d. \(\frac {\sqrt {3}} {4} < y < \frac {\sqrt {2}} {3}\)

e. \(y < \frac {\sqrt {3}}{4}\)

Aside: There's no need to calculate the actual value of z (which is √5), since the question isn't asking us about the possible values of z.

This triangle meets the condition that its area is 1, and that x < y < z
With this particular triangle, y = 2

When we check the answer choices, only one option (answer choice A) allows for y to equal 2

Answer: A

Cheers,
Brent

BrentGMATPrepNow
For what Bunuel mentions above "2<y^2--> 2√<y2<y ."
Don't you need to flip the sign when taking the square root (or consider alternative versions)?

BrentGMATPrepNow

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23 Dec 2022, 15:33

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woohoo921

BrentGMATPrepNow
For what Bunuel mentions above "2<y^2--> 2√<y2<y ."
Don't you need to flip the sign when taking the square root (or consider alternative versions)?

Here are the steps Bunuel took:

Start: \(\frac{2}{y}<y\)

Multiply both sides by y to get \(2<y^2\)

Square root both sides: \(\sqrt{2}<y\).

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BrentGMATPrepNow

woohoo921

BrentGMATPrepNow
For what Bunuel mentions above "2<y^2--> 2√<y2<y ."
Don't you need to flip the sign when taking the square root (or consider alternative versions)?

Here are the steps Bunuel took:

Start: \(\frac{2}{y}<y\)

Multiply both sides by y to get \(2<y^2\)

Square root both sides: \(\sqrt{2}<y\).

BrentGMATPrepNow
Thank you for your reply! To follow-up, I was confused as to what happens with the number of cases/sign changes when you have a variable that you take a square root of within in inequality.

Don't we have to consider the positive and negative versions of root(2) because we took the square root?

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woohoo921

If this were a algebra question where x, y and z were ANY numbers, then you would have to consider positive and negative values.
However, in this geometry question, x, y and z represent the LENGTHS of the sides of a triangle, in which case we can be certain that x, y and z are all POSITIVE numbers, in which case it would be incorrect to consider y as having a negative value.

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BrentGMATPrepNow

tonebeeze

Area of triangle \(= \frac{base \times height}{2}\)

There are infinitely many right triangles that have an area of 1.
So, one approach is to find a triangle that meets the given conditions, and then see what conclusions we can draw.

Here's one such right triangle:

Aside: There's no need to calculate the actual value of z (which is √5), since the question isn't asking us about the possible values of z.

This triangle meets the condition that its area is 1, and that x < y < z
With this particular triangle, y = 2

When we check the answer choices, only one option (answer choice A) allows for y to equal 2

Answer: A

Cheers,
Brent

Hi Brent,

How did you come to the conclusion that Y=2?Can you please explain it for me?

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As all the experts have already pointed we can get to xy=2 and then substitute x for y in the initial inequality.

One more way of looking *at this is a number properties logic. When we multiply 2 numbers we can either have the x*x or x*y. For ex - 16 = 4*4 or 8*2. Notice that in the case of the latter either x>y or y>x.
Similarly, In this question, we cannot have x=y (which is root2*root2)as it is already stated in the question that x<y<z.
Therefore it must be true that y>root 2 because if both numbers are not equal then one number must be greater than the other such that neither of them are root 2. (please refer the example I have mentioned above). Hence, the range is y>root.

I hope this helps.

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We know XY=2 (From Area of triangle).
Now one possible value for Y can be 2 when X=1 . (also X<Y)
If we see only option which allows Y to have value above 2 is a.
None of the option allow value of 2 hence they are incorrect.

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is this question part of focus? Bunuel

Bunuel

The area of the triangle is \(\frac{xy}{2}=1\) (\(x<y<z\) means that hypotenuse is \(z\)) --> \(x=\frac{2}{y}\). As \(x<y\), then \(\frac{2}{y}<y\) --> \(2<y^2\) --> \(\sqrt{2}<y\).

Answer: A.

Also note that max value of \(y\) is not limited at all. For example \(y\) can be \(1,000,000\) and in this case \(\frac{xy}{2}=\frac{x*1,000,000}{2}=1\) --> \(x=\frac{2}{1,000,000}\).

Hope it helps.