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

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

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13 Mar 2014, 01:22
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Question Stats:

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

Problem Solving
Question: 157
Category: Geometry; Algebra Triangles; Area; Inequalities
Page: 83
Difficulty: 600

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

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15 Mar 2014, 09:46
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SOLUTION

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 $$\frac{xy}{2}=\frac{x*1,000,000}{2}=1$$ --> $$x=\frac{2}{1,000,000}$$.
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A certain right triangle has sides of length x, y, and z, wh  [#permalink]

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20 Nov 2015, 13:11
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) $$y<\frac{\sqrt{3}}{4}$$

This was a tough problem for me, to be honest I couldn't find a solution like Bunuel did, but I have tested aswer choices to derive at the correct answer.

We know that $$x*y=2$$ actually let's square this expression $$x^2*y^2=4$$ so let's test the values given in the answer choices with Min/Max approach (we ca also square values in the answer choices, as sides of a triangle can not be -ve)

(B) so $$y^2$$ can be max 2 and $$x^2$$= max $$\frac{3}{4}$$--> multiply $$x^2*y^2$$=1,5. So B is out because we need a 4 when $$x^2 and y^2$$ are multiplied
(C) $$y^2$$ max = $$\frac{3}{4}$$, $$x^2$$ max=2/9 --> $$\frac{2}{9}*\frac{3}{4} < 4$$ , C is out
(D) $$\frac{2}{9}*\frac{3}{16} < 4$$
(E) $$y^2$$ < $$\frac{3}{16}$$, so we know that y > x, thus when y < $$\frac{3}{16}$$ $$x^2$$ is also < $$\frac{3}{16}$$ and their product is < 4, E is also out and we are left with Answer Choice A
(A)
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Re: A certain right triangle has sides of length x, y, and z, wh  [#permalink]

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27 Apr 2016, 16:59
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) $$y<\frac{\sqrt{3}}{4}$$

Thank you![/textarea]

Hi,
from the area of triangle we know x*y = 2.
So there is one possibility that x could be 1 and y could be 2, but none of the answer choices ,except A confirms to y =2 possibility.

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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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3
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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Joined: 03 Aug 2017
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A certain right triangle has sides of length x, y, and z, wh  [#permalink]

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29 Apr 2018, 00:25
VeritasPrepKarishma wrote:
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.

Hi can you Please help me understand why have you assumed if x= y then the value for both is equal to root 2 ?
I understood that x < y< z and if z is the Hypotenuese and therefore in case its a 30 60 90 triangle then hypotenuse coud hold the value of root 2
But how Y could hold the value higher than root 2 and x an even smaller value than root 2 ? also in the equation above we dervided xy=2 don't we use this finding ?
A certain right triangle has sides of length x, y, and z, wh &nbs [#permalink] 29 Apr 2018, 00:25
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