himang wrote:
hoping_for_stern wrote:
Contradicting statement 2 :
We can keep the fact in mind that triangles between two parallel lines have same area.(draw a line ll to AC / BC's value will vary and still satisfy the area being constant at 30 ,as base and height in this case are constant (AC) and distance between ll lines remain the same , which in this case would be the height of the triangle)
Heron's formula makes one jump to a quick conclusion about statement 2 being sufficient but an important point is overlooked by the fact using it to find the value of BC would give some polynomial ( x3 or x4 or whatever ....) and that will lead to multiple values of x
(at least 2 in this case ).
Hence insufficient.
Statement 1 is sufficient
Hence A
Answer Should Be "D" since area is equals= 0.5*A*B*SinQ; so from here we can get the included angle between the given two sides and then use Cosine law to get the third side easily.....
+1 KUDDUS if I'm Right
That's not correct. The answer to this question is A, not D.
The formula you posted is correct: \(Area=\frac{1}{2}BA*CA*sinA\) --> \(sinA=\frac{12}{13}\), but from this we cannot calculate angle \(A\), as \(sin A=Sin (180-A)\). Meaning that this won't give us ONLY one value for angle A, thus we won't have ONLY one value for CB --> A can be acute angle, making B right angle AND making CB equal to 12 OR A can be obtuse angle, making B acute angle AND making CB greater then 12.
Check here:
http://gmatclub.com/forum/what-is-the-l ... l#p1105437Hope it helps.
, why can't we find third side of triangle using Heron's formula i.e.
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