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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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16 Jan 2011, 15:01
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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. $$y < \frac {\sqrt {3}}{4}$$
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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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16 Jan 2011, 15:12
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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. $$y < \frac {\sqrt {3}}{4}$$

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.
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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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06 Aug 2011, 03:41
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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09 Feb 2012, 03:42
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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09 Feb 2012, 05:54
4
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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28 Feb 2013, 08:55
Bunuel wrote:
$$x=\frac{2}{y}$$. As $$x<y$$, then $$\frac{2}{y}<y$$

Could you explain why that is?
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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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28 Feb 2013, 09:01
3
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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Re: A certain right triangle has sides of length x, y, and z, where x < y  [#permalink]

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01 Mar 2013, 01:21
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]

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01 Mar 2013, 01:50
1
1
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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12 Feb 2016, 18:34
1
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 $$>\sqrt{2}$$.

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

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27 Apr 2016, 19:00
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5
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) $$y<\frac{\sqrt{3}}{4}$$

You can think about it in terms of transition points too.

x < y< z so this means that x and y are the legs of the right triangle and z is the hypotenuse.
The area of the triangle will be (1/2)*xy = 1
xy = 2

Now, if x = y, then both x and y would be equal to $$\sqrt{2}$$.
But y is greater than x, so y would be at least a slight bit greater than $$\sqrt{2}$$ and x would be a slight bit less than $$\sqrt{2}$$. In all options other than (A), y takes values less than 1.414.
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A certain right triangle has sides of length x, y, and z, wh  [#permalink]

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27 Apr 2016, 19:12
1
1
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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A certain right triangle has sides of length x, y, and z, where x < y  [#permalink]

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Updated on: 16 Apr 2018, 11:36
1
Top Contributor
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. $$y < \frac {\sqrt {3}}{4}$$

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 see what conclusions we can draw.

Here's one such right triangle:

This meets the conditions that the area is 1 AND x < y < z
With this triangle, y = 4

When we check the answer choices, only one (answer choice A) allows for y to equal 4

Cheers,
Brent
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Originally posted by GMATPrepNow on 11 Sep 2017, 12:54.
Last edited by GMATPrepNow on 16 Apr 2018, 11:36, 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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14 Dec 2017, 04:17
4
1
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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A certain right triangle has sides of length x, y, and z, wh  [#permalink]

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12 Feb 2019, 11:09
1
Top Contributor
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

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

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26 Nov 2019, 03:22
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
A certain right triangle has sides of length x, y, and z, where x < y   [#permalink] 26 Nov 2019, 03:22
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