Step-by-step explanation:

Given:

AD/AB = AE/AC = 1/4

Therefore, D and E are two points on side AB and AC respectively such that they divide side AB and AC in a ratio of 1:3.

By Section formula,

P (x, y) = [(mx₂ + nx₁) / (m + n) , (my₂ + ny₁) / (m + n)] .....(1)

Considering A(4, 6) and B(1, 5),

Coordinates of Point D = [(1 × 1 + 3 × 4) / (1 + 3), (1 × 5 + 3 × 6) / (1 + 3)]

= (13/4, 23/4)

Considering A(4, 6) and C(7, 2),

Coordinates of point E = [(1 × 7 + 3 × 4) / (1 + 3), (1 × 2 + 3 × 6) / (1 + 3)]

= (19/4, 20/4)

Area of a triangle = 1/2 [x₁ (y₂ - y₃) + x₂ (y₃ - y₁) + x₃ (y₁ - y₂)] ....(2)

By substituting the vertices A(4, 6), B(1, 5), and C(7, 2) in equation (2),

Area of ΔABC = 1/2 [4 (5 - 2) + 1 (2 - 6) + 7 (6 - 5)]

= 1/2 [4 (3) + 1 (-4) + 7 (1)]

= 1/2 [12 - 4 + 7]

= 15/2 square units

By substituting the vertices A(4, 6), D(13/4, 23/4), and E(19/4, 20/4) in equation (2),

Area of ΔADE = 1/2 [4 (23/4 - 20/4) + 13/4 (20/4 - 6) + 19/4(6 - 23/4)]

= 1/2 [3 - 13/4 + 19/6]

= 1/2 [(48 - 52 + 19)/16]

= 15/32 square units

Clearly, the ratio between the areas of ∆ADE and ∆ABC is = (15/32) : (15/2) = 1:16.