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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, 16:01

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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. \(\frac {\sqrt {3}} {4} < y < \frac {\sqrt {2}} {3}\)

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

Answer: A.

Also note that max value of \(y\) is not limited at all. For example \(y\) can be \(1,000,000\) and in this case \(\frac{xy}{2}=\frac{x*1,000,000}{2}=1\) --> \(x=\frac{2}{1,000,000}\).

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

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06 Aug 2011, 04:41

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

Answer: A.

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 ?

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

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09 Feb 2012, 06:54

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.

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

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01 Mar 2013, 02:21

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

Answer: A.

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.

Please confirm
_________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

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

Answer: A.

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.

Please confirm

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, ...

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

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16 Dec 2013, 14:17

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?

We are told that this is a right triangle which right off the bat tells me one of two things, either we need to solve with some variation of a^2 + b^2 = c^2 or that we can find the area with base*height.

Because this is a right triangle and x < y < z we know that z is the hypotenuse and that x is the shortest leg. The area = 1 so:

a=1/2 b*h 1=1/2 b*h 2=b*h.

Y is the second longest measurement in this right triangle which means it must be longer than x but shorter than z. If we run through a few possible combinations of a and b we see that there isn't a limit on the length of y so long as y*x = 2 and y<x. For example, x=1 and y = 4 and z can = 5. This means that there is no upward limit on the value of y so answer choice E is out. This also means that D, C and B are out as well because all contain upward limits on the value of y can be any number so long as y*x = 2 and y<x. Therefore, A is the only answer choice.

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

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12 Feb 2016, 19:34

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 \(>\sqrt{2}\).

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

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25 May 2017, 11:23

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. \(\frac {\sqrt {3}} {4} < y < \frac {\sqrt {2}} {3}\)

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

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

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28 Sep 2017, 08:39

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 y is the longest of all sides, we can conclude that it's hypotenuse. Hence, x & y are relevant for Area. Since Area= 1, we can 1/2* x*y = 1. Therefore, xy = 1. Notice there could be any combination of x & y. What needs to be taken care of is x<y. Hence it could be 1 & 2, 1/2 & 4 etc.

Option A : It says Y> Sq.root 2 If we substitute same in the equation we get x also as sq.root 2. However, we know that x cannot equal to y. Hence, y can be anything greater than sq.root 2. This option satisfies the criteria. This doesn't leave any scope for Y value to differ as Y cannot be equal to less than sq.root of 2.

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

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29 Sep 2017, 11:22

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

The answer is A

Area =1/2 x*y as it is right triangle and according to the condition x < y < z therefore the two legs of the triangle are x and y

so x*y=2

Now taking the condition x < y < z Let us take x<y Multiply both sides by y we have xy<y^2 or 0<y^2-x*y 0<y^2 -2 or 0<(y-√2)*(y+√2)

As length can not be negative therefore we have y>√2 _________________

We are more often frightened than hurt; and we suffer more from imagination than from reality