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Someone pl explain why A cannot be the answer? If we map the triangle on a coordinate plane we have the base on X-axis and apex on Y-axis. Going by formula: Area = 1/2 (base x height) we have the base as 5 units and in order to have the area bigger than 15 we need the height to be >6 units. S1 says A<3, so height of triangle is <3 too hence area is less than 15. S1 SUFFICIENT - consider \(A=-10\), what is the area then?

From S2 we know the triangle is right hence if the side on X-axes is 5 units the other 2 sides will be 3 & 4 units. Again, area <15. S2 SUFFICIENT If above are correct then answer is D. Comments welcome.

Hope that helps!

This certainly is very helpful, thank you & Bunuel!!

I suppose here D would be logical option My apporach Statement 1 A<3 now according to distance formula -1,0 and 4,0 distance is 5 therefore area 5A/2>15 , means A<6 hence SUFFICENT sTATEMENT 2RIGHT ANGLE if we take 3,4,5 triplet the area will be 3*4/2=6 and any subsequent relation gives area <15 Hence Sufficient Therefore D (Correct me if I am wrong )

I suppose here D would be logical option My apporach Statement 1 A<3 now according to distance formula -1,0 and 4,0 distance is 5 therefore area 5A/2>15 , means A<6 hence SUFFICENT sTATEMENT 2RIGHT ANGLE if we take 3,4,5 triplet the area will be 3*4/2=6 and any subsequent relation gives area <15 Hence Sufficient Therefore D (Correct me if I am wrong )

From the coordinates of 3 vertices, the triangle's area = [4-(-1)]a/2 = 5a/2 (a: length of OA) 1) A<3: if A = -10 -> area = 25>15 but if A = 2 -> area = 5<15. Hence insufficient 2) The triangle is right: distances from (0,A) to (-1,0) and to (4,0) are sqrt(a^2+1) and sqrt (a^2+16) respectively In a right triangle, we have: area = [sqrt(a^2+1) x sqrt (a^2+16)]/2 = 5a/2 => (a^2+1)(a^2+16) = 25a^2 => a^2 = 4 => a = 2 and area = 5<15 => sufficient