kaps786 wrote:If vertices of a triangle have coordinates (-1,0), (4,0), and (0,A) , is the area of the triangle greater than 15?

1) A < 3
2) The triangle is right

Refer to the following figure,


We can see that the length of the base of the triangle = (4 + 1) = 5
And length of the height of the triangle = |A|, we are taking absolute value of A in case A < 0

Thus, the area of the triangle is 5*|A|/2 = (2.5)*|A|

Statement 1: A < 3
If A = 2, area of the triangle = 2.5*2 = 5 < 15
If A = -10, area of the triangle = 2.5*|-10| = 25 > 15

Not sufficient

Statement 2: As the triangle is a right angled triangle, A must have a particular value which we can calculate. Knowing that value we can easily determine the area of the triangle.

Sufficient

The correct answer is B.

Note : If someone need further justification on the claim I made in analyzing statement 2...
As the triangle is a right angled triangle, the angle at the coordinate (0, A) must be the right angle. Hence, the line joining (-1, 0) and (0, A) must be perpendicular with the line joining (4, 0) and (0, A). Thus product of their slopes must be -1.

Slope of the line joining (-1, 0) and (0, A) = (A - 0)/(0 - (-1)) = A
Slope of the line joining (4, 0) and (0, A) = (A - 0)/(0 - 4) = -A/4

Hence, A*(-A/4) = -1 ----> A² = 4 ----> A = ±2

Therefore, area of the triangle = (2.5)*|±2| = (2.5)*2 = 10 < 15