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

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A certain right triangle has sides of length x, y, and z, where x < y [#permalink]
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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

Cheers,
Brent

Originally posted by BrentGMATPrepNow on 11 Sep 2017, 13:54.
Last edited by BrentGMATPrepNow on 21 Jan 2021, 15:57, edited 3 times in total.
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Re: A certain right triangle has sides of length x, y, and z, where x < y [#permalink]
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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. $$y < \frac {\sqrt {3}}{4}$$

Since it is a right triangle, z is the greatest side and hence the hypotenuse.

So area of the triangle will be (1/2)*xy = 1
xy = 2

Note that x < y.
If x were equal to y, $$x = y = \sqrt{2}$$
Since x is less than y, $$y > \sqrt{2}$$ and $$x < \sqrt{2}$$

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Re: A certain right triangle has sides of length x, y, and z, where x < y [#permalink]
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OA is A
since this is a rt angled triangle so z is th hypotnuse
and given xy = 2
so as x decreased y increases. Now if x is 1 then y is 2, when x is 1/2 y is 4.
Only option A supports this result.
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Re: A certain right triangle has sides of length x, y, and z, where x < y [#permalink]
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2flY wrote:
Bunuel wrote:
$$x=\frac{2}{y}$$. As $$x<y$$, then $$\frac{2}{y}<y$$

Could you explain why that is?

Just substitute x with $$\frac{2}{y}$$ in $$x<y$$ to get $$\frac{2}{y}<y$$.
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A certain right triangle has sides of length x, y, and z, wh [#permalink]
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Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

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) $$y<\frac{\sqrt{3}}{4}$$

First step will be to breakdown the options into recognizable decimal representations (assuming $$\sqrt{2} \approx 1.4$$, $$\sqrt{3} \approx 1.7$$)

A) y>1.4
B) 0.8<y<1.4
C) 0.5<y<0.8
D) 0.4<y<0.5
E) y<0.4

We are given that x<y<z and that 0.5*x*y=1 --> x*y=2

Now from the relation xy=2 --> go back to the options and test for y=1. You get x=2 but we are given that x<y ---> y MUST be > $$\approx$$1.4 such that x < y

For any value of y < 1.4 , you will end up getting x>y (try with y=0.5 or 0.75 etc).

Only A satisfies this condition and is hence the correct answer.

Hope this helps.
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Re: A certain right triangle has sides of length x, y, and z, where x < y [#permalink]
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mridulparashar1 wrote:
rohansharma wrote:
Bunuel wrote:
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 $$\frac{2}{y}<y$$ --> $$2<y^2$$ --> $$\sqrt{2}<y$$.

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.

Dear Bunuel,

While solving the question ,I assumed it to be the special 90,60,30 triangle.
Am I wrong in following that approach ?

Hi Bunuel,

The Q stem says that sides are x<y<z and it is a right angle triangle. So we can assume that it will be 30-60-90 triangle. Had it been 45-45-90 triangle then the 2 sides ie base and perpendicular would have been equal and therefore x=y and x,y<z

I guess it should be okay to assume that it is 30-60 -90 triangle.

Yes, if it were 45-45-90, then we would have that x=y<z. BUT, knowing that it's not a 45-45-90 right triangle does NOT mean that it's necessarily 30-60-90 triangle: there are numerous other right triangles. For example, 10-80-90, 11-79-90, 25-65-90, ...

Hope it's clear.
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Re: A certain right triangle has sides of length x, y, and z, where x < y [#permalink]
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One other way that I noticed to solve this problem is to check the length of $$y$$ when $$x=y$$, i.e. 45,45,90. In that case $$x=y=\sqrt{2}$$, however as $$y>x$$, it'd always need to be $$>\sqrt{2}$$.

HTH
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A certain right triangle has sides of length x, y, and z, where x < y [#permalink]
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Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

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) $$y<\frac{\sqrt{3}}{4}$$

If x < y < z, then our RIGHT triangle looks something like this (where the hypotenuse is always the longest side)

When we scan the answer choices (ALWAYS scan the answer choices before performing any calculations!), I see that B, C, D and E all provide a MAXIMUM value of y.
This should be a bit of a surprise, because there's no limit to the length of each leg of the triangle.

GIVEN: The triangle has area 1
Area = (base)(height)/2
So, here's one possible triangle with area 1:

ASIDE: Area = (10)(0.2)/2 = 2/2 = 1 (voila!)

In this triangle, x = 0.2, y = 10 and z = some number greater than 10
Since it's possible for y to equal 10, we can eliminate answer choices B, C, D and E

RELATED VIDEO

Originally posted by BrentGMATPrepNow on 12 Feb 2019, 12:09.
Last edited by BrentGMATPrepNow on 17 May 2021, 07:25, edited 1 time in total.
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Re: A certain right triangle has sides of length x, y, and z, where x < y [#permalink]
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Bunuel 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. $$y < \frac {\sqrt {3}}{4}$$

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 $$\frac{2}{y}<y$$ --> $$2<y^2$$ --> $$\sqrt{2}<y$$.

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.

Dear Bunuel,

While solving the question ,I assumed it to be the special 90,60,30 triangle.
Am I wrong in following that approach ?
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Re: A certain right triangle has sides of length x, y, and z, where x < y [#permalink]
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rohansharma wrote:
While solving the question ,I assumed it to be the special 90,60,30 triangle.
Am I wrong in following that approach ?

We are just told that the triangle is right, not that it's a special kind like 30-60-90 or 45-45-90.
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Re: A certain right triangle has sides of length x, y, and z, where x < y [#permalink]
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Bunuel wrote:
$$x=\frac{2}{y}$$. As $$x<y$$, then $$\frac{2}{y}<y$$

Could you explain why that is?
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A certain right triangle has sides of length x, y, and z, where x < y [#permalink]
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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. $$y < \frac {\sqrt {3}}{4}$$

Since it is a right triangle, z is the greatest side and hence the hypotenuse.

So area of the triangle will be (1/2)*xy = 1
xy = 2

Note that x < y.
so for XY to be 2 either x=2 and Y=1 or Y =2 and X =1

we are already told x < y. so Y has to be 2.... and Root2 = 1.4 and since we now know y =2 option A
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Re: A certain right triangle has sides of length x, y, and z, where x < y [#permalink]
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Given, the area of this triangular region is 1 and x < y < z.
So, area of triangle= 1/2*x*y ( as it's a right angled triangle, so z must be the hypotenuse, and x and y can be either base or height)
=> 1/2*x*y=1
=> x*y=2
now, lets find the minimum value for y (or maximum value of x).
As, its a right angled triangle, the maximum value of x will be when its a isosceles right triangle i.e. 45:45:90 triangle.
=> x=y<z
so, x*y=2
=>x*x=2 (i.e minimum value of y)
=> x^2=2
=> x=√2
But, as we know that y>x,
so, y>√2. Option A
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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: $$\sqrt{2}<y$$.
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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.
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Re: A certain right triangle has sides of length x, y, and z, where x < y [#permalink]
rohansharma wrote:
Bunuel 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. $$y < \frac {\sqrt {3}}{4}$$

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 $$\frac{2}{y}<y$$ --> $$2<y^2$$ --> $$\sqrt{2}<y$$.

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.

Dear Bunuel,

While solving the question ,I assumed it to be the special 90,60,30 triangle.
Am I wrong in following that approach ?

Hi Bunuel,

The Q stem says that sides are x<y<z and it is a right angle triangle. So we can assume that it will be 30-60-90 triangle. Had it been 45-45-90 triangle then the 2 sides ie base and perpendicular would have been equal and therefore x=y and x,y<z

I guess it should be okay to assume that it is 30-60 -90 triangle.

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Re: 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. $$y < \frac {\sqrt {3}}{4}$$

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

Asked: If the area of this triangular region is 1, which of the following indicates all of the possible values of y?

Area of right triangle = xy/2 = 1
xy = 2
x = 2/y < y
$$y^2 >2$$
$$y > \sqrt{2}$$

IMO A
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