D IMO

(1) tells you diff between 2 sides is 10. The third side should be > than 10. So we can tell perimeter will be >20
(2) for a triangle with given area, perimeter is minimum for an eq triangle. Even for a eq. triangle the side is approx 6.8, so 3(6.8) is again > 20

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.

shankar.ashwin wrote: Even for a eq. triangle the side is approx 6.8,

how did u get 6.8?

I don't think (2) is sufficient. I would go with A.

The second statement says that the area is 20, which means that 20=.5*base*height, so 40=base*height. If you consider a case where one side is the base and the other is the height, we could have a base and height of 10 and 4 or of 8 and 5. If 10 and 4, the final side could be no smaller than 7 (since no one side can be greater than the sum of the other two), and thus the perimeter would be greater than 20. But if the sides were 8 and 5, then the third side could be 4, which means that the perimeter would be less than 20. Thus statement 2 is insufficient.

rohit_gmat wrote:

shankar.ashwin wrote: Even for a eq. triangle the side is approx 6.8,

how did u get 6.8?

for a equilateral triangle,
area = sqrt(3)*side*side/4
=> area = .43301 * side ^2
=>side = sqrt(2.3094 * area )
=> side = 6.79 in this case


Btw I rarely see so much calculation involved in a GMAT question.
so IMO
D

Small break from the forum.

I think Amiable Scholar has answered your question. Given the area you could find the side of an equilateral triangle,

Area = Sqrt (3)/4 * (a^2)

rohit_gmat wrote:

shankar.ashwin wrote: Even for a eq. triangle the side is approx 6.8,

how did u get 6.8?