Guys ,
Pls help me out with the below problem..
If a sandbox in the shape of a right triangle has a hypotenuse of h feet, an area of a square feet, and one leg of length x feet, which of the following must be true?
1) x^2-4a^2/x^2-h^2=0
2) x^2+4a^2/x^2+h^2=0
3) x^2+4a^2/x^2-h^2=0
4) x^2+2a^2/x^2-h^2=0
5) x^2+2a^2/x^2+h^2=0
OA3
Thanks
Shreyans
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Answer: Option C i.e. x^2+(4a^2/x^2)-h^2=0j_shreyans wrote:Guys ,
Pls help me out with the below problem..
If a sandbox in the shape of a right triangle has a hypotenuse of h feet, an area of a square feet, and one leg of length x feet, which of the following must be true?
1) x^2-4a^2/x^2-h^2=0
2) x^2+4a^2/x^2+h^2=0
3) x^2+4a^2/x^2-h^2=0
4) x^2+2a^2/x^2-h^2=0
5) x^2+2a^2/x^2+h^2=0
OA3
Thanks
Shreyans
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Hi j_shreyans,
This question is perfect for TESTing Values:
We're dealing with a RIGHT TRIANGLE and we're given the variables...
H = hypoteneuse
X = one of the legs
A = area of the triangle
Let's choose a 3/4/5 right triangle...
H = 5
X = 3
A = (1/2)(3)(4) = 6
Now we just have to test those values in the answers and find the one that's true...
A) 9 - 144/9 - 25 = something negative (not 0)
B) 9 + 144/9 + 25 = something positive (not 0)
C) 9 + 144/9 - 25 = 0
D) 9 + 72/9 - 25 = -8
E) 9 + 72/9 + 25 = something positive (not 0)
Final Answer: C
GMAT assassins aren't born, they're made,
Rich
This question is perfect for TESTing Values:
We're dealing with a RIGHT TRIANGLE and we're given the variables...
H = hypoteneuse
X = one of the legs
A = area of the triangle
Let's choose a 3/4/5 right triangle...
H = 5
X = 3
A = (1/2)(3)(4) = 6
Now we just have to test those values in the answers and find the one that's true...
A) 9 - 144/9 - 25 = something negative (not 0)
B) 9 + 144/9 + 25 = something positive (not 0)
C) 9 + 144/9 - 25 = 0
D) 9 + 72/9 - 25 = -8
E) 9 + 72/9 + 25 = something positive (not 0)
Final Answer: C
GMAT assassins aren't born, they're made,
Rich
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Algebraically ...
Let's call the two legs b and x.
We know that
b² + x² = h²,
(b*x)/2 = a
Since all of the answers are equal to 0 and do NOT include the variable b, we'll try to combine these so b goes away and the whole thing equals 0.
From the first equation,
b² + x² - h² = 0
From the second,
b = (2a)/x
Substituting (2a)/x for b in the first equation ...
((2a)/x)² + x² - h² = 0
4a²/x² + x² - h² = 0
So C.
Let's call the two legs b and x.
We know that
b² + x² = h²,
(b*x)/2 = a
Since all of the answers are equal to 0 and do NOT include the variable b, we'll try to combine these so b goes away and the whole thing equals 0.
From the first equation,
b² + x² - h² = 0
From the second,
b = (2a)/x
Substituting (2a)/x for b in the first equation ...
((2a)/x)² + x² - h² = 0
4a²/x² + x² - h² = 0
So C.