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area of the triangle (weird options)

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area of the triangle (weird options)

by rahul.s » Sun Jan 31, 2010 4:44 am
A right triangle with area 6 has sides of length r, s, and t. If r > s > t, what is the range of possible values for t?

A) 0 < t < 2√3
B) 3√2 < t < 3√3
C) √2 < t < √3
D) √3 < t < 2
E) t < 4

OA: A
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by thephoenix » Sun Jan 31, 2010 5:04 am
rahul.s wrote:A right triangle with area 6 has sides of length r, s, and t. If r > s > t, what is the range of possible values for t?

A) 0 < t < 2√3
B) 3√2 < t < 3√3
C) √2 < t < √3
D) √3 < t < 2
E) t < 4

OA: A
t*s=12
now s=12/t and s>t
hence when t=3.5 s is aprrox 3.4 which is violating the cond
hence t<3.5
so b) where max value of t=3*1.7=5.1 is wrng
c) is ok but we can have t less than 1 also as long as t*s=12
d) same as c
e) when t=3.5 s will be 3.4 hence not possible as s>t
hence a is true

now if consider a perfact triplet
then it will be a case of 3,4,5(t,s,r)
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by sanju09 » Sun Jan 31, 2010 11:36 pm
rahul.s wrote:A right triangle with area 6 has sides of length r, s, and t. If r > s > t, what is the range of possible values for t?

A) 0 < t < 2√3
B) 3√2 < t < 3√3
C) √2 < t < √3
D) √3 < t < 2
E) t < 4

OA: A
When s t = 12, s = 12/t.

With s > t, we can have 12 > t^2, and I ended with

(t - 2√3) (t + 2√3) < 0

Now, TWO cases

1. (t - 2√3) > 0 and (t + 2√3) < 0, which is impossible (must think, why?).

2. (t - 2√3) < 0 and (t + 2√3) > 0, which yields logic and -2√3 < t < 2√3, but! Can t be negative? What's the lowest value that t could take? Just paid a thought, and made [spoiler]0 < t < 2√3[/spoiler] my answer. [spoiler]A[/spoiler]
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by Stuart@KaplanGMAT » Sun Jan 31, 2010 11:42 pm
rahul.s wrote:A right triangle with area 6 has sides of length r, s, and t. If r > s > t, what is the range of possible values for t?

A) 0 < t < 2√3
B) 3√2 < t < 3√3
C) √2 < t < √3
D) √3 < t < 2
E) t < 4

OA: A
As noted, st = 12.

If s=t, then t would be root12. Since t must be less than s, t must be less than root12.

root12 = (root4)(root3) = 2(root3).

We also know that t must be positive (you can't have non-positive distances), so we have:

0 < t < 2(root3)... choose A.
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