The vertices of a ∆ ABC are A (4, 6), B (1, 5) and C (7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that AD/AB = AE/AC = 1/4. Calculate the area of the ∆ ADE and compare it with area of ∆ ABC. (Recall Theorem 6.2 and Theorem 6.6)

Given that: A (4, 6), B (1, 5) and C (7, 2) are the vertices of a ∆ABC.

AD/AB = AE/AC = 1/4

AD/(AD + BD) = AE/(AE + EC) = 1/4

Points D and E divide AB and AC in a 1:3 ratio, respectively.

D’s coordinates can be computed using the formula below:

x = (m₁x₂ + m₂x₁)/(m₁ + m₂) and y = (m₁y₂ + m₂y₁)/(m₁ + m₂)

Given here, m₁ = 1 and m₂ = 3

Consider the line segment AB, which is divided at the ratio 1:3 by the point D.

x = [3(4) + 1(1)]/4 = 13/4

y = [3(6) + 1(5)]/4 = 23/4

E’s coordinates can be calculated in the same way as given below:

x = [1(7) + 3(4)]/4 = 19/4

y = [1(2) + 3(6)]/4 = 20/4 = 5

Find Area of triangle:

Using the formula: The area of a triangle  = 1/2 × [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]

The area of triangle ABC can be calculated as:

= ½ [4(5 – 2) + 1( 2 – 6) + 7( 6 – 5)]

= ½ (12 – 4 + 7) = 15/2 sq unit

The area of triangle ADE can be calculated as:

= ½ [4(23/4 – 5) + 13/4 (5 – 6) + 19/4 (6 – 23/4)]

= ½ (3 – 13/4 + 19/16)

= ½ ( 15/16 ) = 15/32 sq unit

As a result, the ratio of the area of triangle ADE to the area of triangle ABC is 1:16.