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Triangle T has vertices of (-2,6), (6,2) and (0,k) where k [#permalink]
 12 Jul 2012, 17:29
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25% (01:43) wrong based on 0 sessions
Triangle T has vertices of (-2,6), (6,2) and (0,k) where k is negative. Is k an integer? (1) The area of T in square units is a multiple of 3 (2) The area of T in square units is a multiple of 5 Can this formula be used to find the area of any triangle with three vertices? Or must the triangle be a right angle for this equation to be applicable. Does anyone know? I found it used by another user on Gmatclub, not sure if its correct. Area=1/2 * [x1(y3-y2) + x2(y1-y3) +x3(y2-y1] The other method I could think of was posted by Bunuel, Hero's formula. But I tried using it, the equation was too sloppy for this answer. Once you have 3 sides (a,b,c), then the area is calculated using A = sqrt(s(s-a)(s-b)(s-c)) where s = (a+b+c)/2 Help!! Thank you!
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Re: The Area of a Triangle with only 3 Vertices Known [#permalink]
12 Jul 2012, 18:35
Use the formula:
1/2|(x1-x3)(y2-y1)- (x1-x2) (y3-y1)|
= |4k-20|
Since K is negative
1) Area is a multiple of 3 , so k=-1,-2,-3
yields area= 24,28,32 ....
and 24,36 are multiples of 3
Not sufficient as K can be integer as well as real
2)Area is a multiple of 5 , so k=-1,-2,-3
yields area= 25,30,35 ....
and 25,30,35 are multiples of 5
Not sufficient as K can be integer as well as real
1&2)Area is a multiple of 3,5
for k = -10
Area= 60
Not sufficient as K can be integer as well as real
Hence , the answer is E.
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Re: The Area of a Triangle with only 3 Vertices Known [#permalink]
13 Jul 2012, 19:11
The formula seems correct but GMAT doesn't rely on such criptic formulas.
I would calculate the area by first considering the triangle with vertices (-2, 6), (6,2) and the origin (0, 0). As usual in GMAT, it turns out a special triangle. It is an isosceles right triangle (in other words 45-45-90 triangle) with legs = root(6^2+2^2) = root(40). It's area is leg*leg/2 = 20.
Then I would consider the triangle with vertices (-2, 6), (k, 0) and the origin (0 , 0). It's base=|k| and height to that base = 2. So it's area is |k|*2/2 = |k|.
Last, I would consider the triangle with vertices (2, 6), (k, 0) and the origin (0 , 0). It's base=|k| and height to that base = 6. So it's area is |k|*6/2 = 3|k|.
Triangle T is the sum of the 3 considered triangles and has area T = 20 + 4|k| = 20 - 4k, which coinsides with the formula the other poster used.
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Re: The Area of a Triangle with only 3 Vertices Known [#permalink]
15 Jul 2012, 23:59
alphabeta1234 wrote: Triangle T has vertices of (-2,6), (6,2) and (0,k) where k is negative. Is k an integer?
(1) The area of T in square units is a multiple of 3
(2) The area of T in square units is a multiple of 5
Can this formula be used to find the area of any triangle with three vertices? Or must the triangle be a right angle for this equation to be applicable. Does anyone know? I found it used by another user on Gmatclub, not sure if its correct.
Area=1/2 * [x1(y3-y2) + x2(y1-y3) +x3(y2-y1]
The other method I could think of was posted by Bunuel, Hero's formula. But I tried using it, the equation was too sloppy for this answer.
Once you have 3 sides (a,b,c), then the area is calculated using
A = sqrt(s(s-a)(s-b)(s-c)) where s = (a+b+c)/2
Help!! Thank you! This question is painful without the formula but GMAT doesn't require you to know anything other than the basic formulas. So I wouldn't expect anything like this in actual GMAT. Using the formula, you can do it quite easily. Area = |1/2 ( x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))| When you put in the values of the coordinates, you get Area = |4k - 20| Say, if area = 15 (a multiple of 3 and 5), k = 35/4 (not an integer) If area = 60 (a multiple of 3 and 5), k = 20 (an integer) So k can be an integer or non integer. Answer (E)
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Re: The Area of a Triangle with only 3 Vertices Known
[#permalink]
15 Jul 2012, 23:59
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