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A certain right triangle has sides of length x, y, and z [#permalink]
17 Jan 2010, 01:39

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Difficulty:

35% (medium)

Question Stats:

71% (02:10) correct
29% (01:22) wrong based on 67 sessions

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}

Re: Possible Values of side of Right Angle Triangle [#permalink]
17 Jan 2010, 01:55

vibhaj 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 > sqr root (2) B. sqr root (3)/2 < y < sqr root (2) C. sqr root (2)/3 < y < sqr root (3)/2 D. sqr root (3)/4 < y < sqr root (2)/3 E. y < sqr root (3)/4

OA after discussion.

--Edited to denote the square roots more clearly

IMO it should be A,

area of the triangle=1 = 1/2 X*Y

or X*Y = 2

Now , also X<Y , now we by using the POE, we can see that apart from 1 option rest all doesn't satisfy this equation.

Re: Possible Values of side of Right Angle Triangle [#permalink]
17 Jan 2010, 11:30

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Expert's post

vibhaj wrote:

Thanks Nitish.is there a better way to solve this , without POE ..?

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}.

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. (sqrt 3)/2 < y <sqrt 2 C. (sqrt 2)/3 < y < (sqrt 3)/2 D. (sqrt 3)/4 < y < (sqrt 2)/3 E. y < (sqrt 3)/4

I'M LOOKING FOR THE ALGEBRAIC SOLUTION, WITHOUT SUPPOSING (PLUGGING) THE VALUES OF X,Y AND Z..!!

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. (sqrt 3)/2 < y <sqrt 2 C. (sqrt 2)/3 < y < (sqrt 3)/2 D. (sqrt 3)/4 < y < (sqrt 2)/3 E. y < (sqrt 3)/4

I'M LOOKING FOR THE ALGEBRAIC SOLUTION, WITHOUT SUPPOSING (PLUGGING) THE VALUES OF X,Y AND Z..!!

Some questions .... please help [#permalink]
07 May 2010, 21:05

1.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. sqrt.3/2 < y < sqrt. 2 C. sqrt.2/3 < y < sqrt.3/2 D. sqrt.3/4 < y < sqrt.2/3 E. y < sqrt.3/4

Re: Some questions .... please help [#permalink]
07 May 2010, 22:42

ok question 1 doesnt seem GMATesque but i have come up with the solution ...let me know what the source is and what the OA is

1) we know that area of triangle is 2 so xy/2 = 1 so xy = 2 moreover x < y .....so think about the two graphs ...y =2/x and x=y ....there will be a point on x-axis beyond which y<x ...to find this point lets equate the two eqns

2/x = x --> x = sqrt(2)

beyond this point y<x ...so y > sqrt(2)

so answer should be A ...in all other choices the value of y is less than sqrt(2)

A certain right triangle [#permalink]
21 May 2010, 02:17

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. sqrt.3/2 < y < sqrt. 2 C. sqrt.2/3 < y < sqrt.3/2 D. sqrt.3/4 < y < sqrt.2/3 E. y < sqrt.3/4

Re: A certain right triangle [#permalink]
21 May 2010, 02:24

dimitri92 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. sqrt.3/2 < y < sqrt. 2 C. sqrt.2/3 < y < sqrt.3/2 D. sqrt.3/4 < y < sqrt.2/3 E. y < sqrt.3/4

This has already been discussed, please follow the link

Re: Possible Values of side of Right Angle Triangle [#permalink]
10 Mar 2011, 20:03

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Expert's post

gmat1220 wrote:

Solved this intuitively - pls verify this reasoning. The above stem means that one angle > 45 and another angle is < 45

At x = y (hypothetically) i.e. Both angles equal. x = z sin (45) = z/ sqrt(2) y = z cos (45) = z/ sqrt(2)

1/2 x y = 1 Hence z = 2

x = sqrt(2) y = sqrt(2)

If one angle > 45 and another angle is < 45, And knowing x < y < z So 0 < x < sqrt(2) and sqrt(2) < y < Infinity

Yes, it can be solved intuitively. It is a very interesting way of thinking. There is one small thing: z may not be 2. e.g. When x = 1 and y = 2, z = sqrt(5)

Think this way: We know x < y and (1/2)xy = 1 (z is the hypotenuse) So xy = 2 If x = y, then x = y = sqrt(2) But since x < y, x is less than sqrt(2) and y is greater than sqrt(2). Makes sense. Good thinking. _________________