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

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17 Jan 2010, 02:39

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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. \(y<\frac{\sqrt{3}}{4}\)

Re: Possible Values of side of Right Angle Triangle [#permalink]

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17 Jan 2010, 02:55

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.

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..!!

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

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 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

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

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.
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