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Difficult Math Problem #116 - Triangles

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Difficult Math Problem #116 - Triangles

by 800guy » Mon Apr 09, 2007 7:42 am
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 > ROOT2

B. ROOT3/2 < y < ROOT2

C. ROOT2/3 < y < ROOT3/2

D. ROOT3/4 < y < ROOT2/3

E. y < ROOT3/4

from diff math doc, oa coming when people respond with explanations
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by rajesh_ctm » Mon Apr 09, 2007 5:55 pm
The problem itself might be difficult, but that does not mean getting it right on the exam is difficult.

Area = xy/2=1 hence xy = 2
Now
x < y < z

x can be very very small while y might be very large while keeping xy = 2.
Think of a very thin and long triangle.

That means x very near to 0 and y has no upper limit. Eliminate choices which put an upper limit to y.
Surprise Surprise! B,C,D and E are gone, and you are left with A, which must be the correct answer! That was easy! :-)

For the sake of ego, let us arrive at the correct answer ourselves rather than looking at the choices. :-)

Think of the upper limit for x. Since x < y, the maximum value x can reach is very near to y, almost y, but less than y. So in this case, xy will be almost y squared. y squared is almost 2, so y is almost ROOT2, but slightly higher. So y>ROOT2.
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by jayhawk2001 » Mon Apr 09, 2007 5:55 pm
Since this is a right angled triangle, the hypotenuse should be the largest
of the 3 sides. We are given x < y < z which implies z is the hyp.

Area is 1/2 * x * y (we don't care whether x is base or y is base)

1/2 * x * y = 1
xy = 2

If x < y, y has to be > root-2 (else the product cannot equal 2). So,
we have y > root-2

Scanning the answer choices -- B, C, D and E all indicate that y < root-2.

Is A the answer ?

I can't seem to figure out the upper-bound on y with the info given. We
know that y > z-x and y < z+x (property of triangles) but this doesn't
give me the upper-bound for y. Not sure if it is needed for the question.
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by Cybermusings » Tue Apr 10, 2007 1:04 am
The triangle is a right angle triangle and x < y < z
Thus z for sure is the hypotenuese for z^2 = x^2 + y^2
Area of a triangle = 1/2 * Base * Height
1/2 * Base * Height = 1
Thus Base * height = 2
Hence we know that the base (one side of the triangle) and the height (the other side of the triangle) = 2
In case of an isoceles right angle triangle base = height = sqr rt. 2
Since it's not an isoceles right angle triangle (with x<y) the greater side has to be more than sqr rt 2
Hence Choice A
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OA:

by 800guy » Wed Apr 11, 2007 12:41 pm
OA:

right triangle with sides x<y<z and area of 1 => z = hypotenuse and xy/2 = 1
i.e xy = 2

If x were equal to y, we would have had xy = y^2 = 2. And y = root2

But, x<y and so y>root2.
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