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Scoretop 13 - Question 31

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by akay » Fri Sep 28, 2007 7:20 am
A slightly different one...

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 > &#8730;2
B. &#8730;3/2 < y < &#8730;2 ------- sq. root of 3 by 2
C. &#8730;2 / 3 < y < &#8730;3 / 2 ------- sq. root of 2 by 3 & sq. root of 3 by 2
D. &#8730;3 / 4 < y < &#8730;2 / 3 --------- sq. root of 3 by 4 and sq. root of 2 by 3
E. y < &#8730;3 / 4 ------sq. root of 3 by 4.
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by gmatguy16 » Fri Sep 28, 2007 7:50 am
IMO A ...
y has to be greater than sqrt(2) so thats its greater then x and area is 1 ...
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by samirpandeyit62 » Fri Sep 28, 2007 8:01 am
x < y < z

so z is hypotenuse

hence area is is xy/2=1

so xy=2 & y>x

A. y > &#8730;2
B. &#8730;3/2 < y < &#8730;2 ------- sq. root of 3 by 2
C. &#8730;2 / 3 < y < &#8730;3 / 2 ------- sq. root of 2 by 3 & sq. root of 3 by 2
D. &#8730;3 / 4 < y < &#8730;2 / 3 --------- sq. root of 3 by 4 and sq. root of 2 by 3
E. y < &#8730;3 / 4 ------sq. root of 3 by 4.

In choices C,D,E max val of y will be less than 1, hence xy<2

eliminate these

In ch B max val of y =1.41, there may be a slight possibilty that if the value of y & x differ say the 5th or 6th decimal place this may satisfy the condition xy=2

howver it has a lower value of sqrt(3)/2 which would be less than 1 hence will not satisfy xy=2

so ans should be A
Regards
Samir
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by akay » Fri Sep 28, 2007 8:02 am
cool explanation. tks
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