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29 Nov 2008, 06:09
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If a triangle has sides 8, 10, and z, which of the following cannot be the area of the triangle?
(A) 3
(B) 6
(C) 12
(D) 24
(E) 46
would you please explain a fast method to solve this problem? this can be time consuming if you want to start experimenting with different combination.
thanks
(A) 3
(B) 6
(C) 12
(D) 24
(E) 46
would you please explain a fast method to solve this problem? this can be time consuming if you want to start experimenting with different combination.
thanks
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29 Nov 2008, 06:49
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Answer is A
easy method tht striked my mind is the triangle rule,
a+b > c and a-b <c
So by the above rule, 2<z<18 --- 1)
So the area cannot be 3. since z has to be multiplied with 8/2 -- 4 or 10/2 -- 5
easy method tht striked my mind is the triangle rule,
a+b > c and a-b <c
So by the above rule, 2<z<18 --- 1)
So the area cannot be 3. since z has to be multiplied with 8/2 -- 4 or 10/2 -- 5
29 Nov 2008, 06:58
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E
sum of two sides of the triangle must be more than a value of third side, so z<18. If z=16, perimetr (p) = 34. By Heron's formula, where s=p/2, Area=sqrt ( s(s-a)(s-b)(s-c) ) = sqrt (17 (9 x 7 x 1)) = sqrt 1008 = approx.32, almost the max. possible area of this triangle, and we can conclude that answer E (46) is impossible to be the one. This is not a perfect explanation, but I can't find the shortest way to solve this problem.
sum of two sides of the triangle must be more than a value of third side, so z<18. If z=16, perimetr (p) = 34. By Heron's formula, where s=p/2, Area=sqrt ( s(s-a)(s-b)(s-c) ) = sqrt (17 (9 x 7 x 1)) = sqrt 1008 = approx.32, almost the max. possible area of this triangle, and we can conclude that answer E (46) is impossible to be the one. This is not a perfect explanation, but I can't find the shortest way to solve this problem.
