Last visit was: 23 Jul 2024, 02:20 It is currently 23 Jul 2024, 02:20
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
SORT BY:
Date
Tags:
Difficulty: 655-705 Level,   Geometry,   Inequalities,                  
Show Tags
Hide Tags
Director
Director
Joined: 14 Jul 2010
Status:No dream is too large, no dreamer is too small
Posts: 959
Own Kudos [?]: 5019 [0]
Given Kudos: 690
Concentration: Accounting
Send PM
Manager
Manager
Joined: 10 Jan 2021
Posts: 154
Own Kudos [?]: 30 [0]
Given Kudos: 154
Send PM
Intern
Intern
Joined: 22 Jun 2020
Posts: 21
Own Kudos [?]: 8 [0]
Given Kudos: 17
Location: United States
GMAT 1: 730 Q48 V42
GMAT 2: 650 Q41 V38
Send PM
VP
VP
Joined: 10 Jul 2019
Posts: 1384
Own Kudos [?]: 579 [0]
Given Kudos: 1656
Send PM
A certain right triangle has sides of length x, y, and z, where x < y [#permalink]
tonebeeze wrote:
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. \(y < \frac {\sqrt {3}}{4}\)



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*

Posted from my mobile device
Intern
Intern
Joined: 11 Jan 2022
Posts: 38
Own Kudos [?]: 1 [0]
Given Kudos: 104
Send PM
Re: A certain right triangle has sides of length x, y, and z, where x < y [#permalink]
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
Director
Director
Joined: 04 Jun 2020
Posts: 542
Own Kudos [?]: 74 [0]
Given Kudos: 623
Send PM
Re: A certain right triangle has sides of length x, y, and z, where x < y [#permalink]
BrentGMATPrepNow wrote:
tonebeeze wrote:
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. \(y < \frac {\sqrt {3}}{4}\)


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


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)?
GMAT Club Legend
GMAT Club Legend
Joined: 12 Sep 2015
Posts: 6804
Own Kudos [?]: 30852 [1]
Given Kudos: 799
Location: Canada
Send PM
Re: A certain right triangle has sides of length x, y, and z, where x < y [#permalink]
1
Kudos
Expert Reply
Top Contributor
woohoo921 wrote:

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: \(\sqrt{2}<y\).
Director
Director
Joined: 04 Jun 2020
Posts: 542
Own Kudos [?]: 74 [0]
Given Kudos: 623
Send PM
Re: A certain right triangle has sides of length x, y, and z, where x < y [#permalink]
BrentGMATPrepNow wrote:
woohoo921 wrote:

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?
GMAT Club Legend
GMAT Club Legend
Joined: 12 Sep 2015
Posts: 6804
Own Kudos [?]: 30852 [1]
Given Kudos: 799
Location: Canada
Send PM
Re: A certain right triangle has sides of length x, y, and z, where x < y [#permalink]
1
Kudos
Expert Reply
Top Contributor
woohoo921 wrote:

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?


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.
Intern
Intern
Joined: 25 Feb 2015
Posts: 19
Own Kudos [?]: 2 [0]
Given Kudos: 148
Send PM
Re: A certain right triangle has sides of length x, y, and z, where x < y [#permalink]
BrentGMATPrepNow wrote:
tonebeeze wrote:
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. \(y < \frac {\sqrt {3}}{4}\)


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?
Manager
Manager
Joined: 30 Sep 2020
Posts: 93
Own Kudos [?]: 10 [0]
Given Kudos: 95
GMAT 1: 610 Q40 V35
Send PM
Re: A certain right triangle has sides of length x, y, and z, where x < y [#permalink]
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.
Intern
Intern
Joined: 07 Apr 2023
Posts: 9
Own Kudos [?]: [0]
Given Kudos: 1
Location: India
Send PM
Re: A certain right triangle has sides of length x, y, and z, where x < y [#permalink]
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.
GMAT Club Bot
Re: A certain right triangle has sides of length x, y, and z, where x < y [#permalink]
   1   2 
Moderator:
Math Expert
94544 posts