revised
the question can be rephrased Is P=AB+BC+AC>20? where P is perimeter of triangle ABC.
st(1) BC-AC=10 means BC=10+AC. The perimeter must be now P=(AB+10+2AC). Per the triangle sides' rule two smaller sides must be greater than the third side. Now possible solutions for AB,BC and AC are
AB=12, AC=1, BC=11 ==> P=12+14+24>20 OR AB=11, AC=3, BC=13 ==> P=11+3+13>20. We can test further,as long as one side is bigger of another by 10 which makes half of 20 this allows for us to conclude that two small sides added by the biggest side will be greater than 20. Sufficient
st(2) S(ABC)=20 means that either AB, or BC, or AC taken as a base and drawn the height to its line (base line) will result in height with S equal to (1/2 *h*base line).
2S=h*base line OR 40=h*base line. When the base is minimized h (the height is maximized) for S to be equal to 40, hence if any side is lower than its base or vice verse we have situation of 40 factored into two numbers and the third side must be greater than greater of these two numbers factored out of 40. Considering that the height will be shortest distance between two lines P(ABC) will be always greater than 20 for S(ABC)=20
I think Shankar is right with D and st(2) is also sufficient.
GmatKiss wrote:Is the perimeter of triangle ABC greater than 20?
(1) BC-AC=10.
(2) The area of the triangle is 20.
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