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Re: A certain right triangle has sides of length x, y, and z, wh [#permalink]
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Carcass 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. \(y < \frac {\sqrt {3}}{4}\)


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:
Image

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

Answer: A

Cheers,
Brent
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Re: A certain right triangle has sides of length x, y, and z, wh [#permalink]
Carcass 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. \(y < \frac {\sqrt {3}}{4}\)


Can we approach the solution using a different manner besides the one shared by grenico?
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Re: A certain right triangle has sides of length x, y, and z, wh [#permalink]
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The explanation by brent I believe is the fastest approach sir
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Re: A certain right triangle has sides of length x, y, and z, wh [#permalink]
1
Area = 1/2 * x * y

we know x < y thus Area = 1/2 * (<y) * y

Area = 1 , thus 1 = (< y^2/2)

2 < y^2

as both are positives we can take square root without changing sign of inequality.

y > sqrt 2
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