In the figure, the vertices of triangle ABC are A(4,6), B(1,5), and C(7,2). A line-segment DE is drawn to intersect the sides AB and AC at D and E respectively such that AD/AB = AE/AC = 1/3. Calculate the area of triangle ADE.

Area of a triangle having vertices (x1,y1), (x2,y2) and (x3,y3) is given by A = 1/2[x1(y2-y3) + x2(y3-y1) + x3(y1-y2)]
∴Area of a triangle ABC is given by A = 1/2[4(5-2) + 1(2-6) + 7(6-5)]
A = 1/2[12 - 4 + 7] = 15/2 sq.units
In ΔADE and ΔABC, AD/AB = AE/EC = 1/3 and ∠DAE = ∠BAC (Common)
Hence, ΔADE ~ ΔABC by AA or
Area of ΔADE/Area of ΔABC = (AD/AB)² = (1/3)² = 1/9
Area of ΔADE/15/2 = 1/9
Area of ΔADE = (15/2) × (1/9) = 5/6 sq.units
However, there seems to be a calculation error in the provided solution. Let's recalculate the area of triangle ABC:
Area(ABC) = 0.5 * |(4(5-2) + 1(2-6) + 7(6-5))| = 0.5 * |12 - 4 + 7| = 0.5 * 15 = 7.5 square units
Since the ratio of areas of similar triangles is the square of the ratio of corresponding sides:
Area(ADE) / Area(ABC) = (1/3)² = 1/9
Area(ADE) = (1/9) * 7.5 = 5/6 square units
The closest option is 56 sq. units, but there's a discrepancy. The calculation of the area of triangle ABC and the subsequent application of similar triangles is correct, but there might be an error in the original provided solution or the options presented.