# 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 the area of ∆ABC. (Recall Theorem 6.2 and Theorem 6.6)

**Solution:**

Theorem 6.2: (The converse of Basic Proportionality Theorem): If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.

Theorem 6.6: The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

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.

P (x, y) = [mx_{2} + nx_{1 }/ m + n , my_{2} + ny_{1 }/ m + n] .....Equation (1)

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

= [13/4, 23/4]

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 {x_{1} (y_{2} - y_{3}) + x_{2} (y_{3 }- y_{1}) + x_{3} (y_{1} - y_{2})} ...Equation (2)

By substituting the vertices A, D, E in (2),

Area of ΔADE = [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 1:16.

Alternatively, we know that if a line segment in a triangle divides its two sides in the same ratio, then the line segment is parallel to the third side of the triangle [Converse of Basic Proportionality Theorem]. These two triangles so formed (here ∆ADE and ∆ABC) will be similar to each other.

Hence, the ratio between the areas of these two triangles will be the square of the ratio between the sides of these two triangles. [Theorem 6.6]

Therefore, the ratio between the areas of ∆ADE and ΔABC = (1/4)^{2} = 1/16

**Video Solution:**

## 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 the area of ∆ABC

### Maths NCERT Solutions Class 10 - Chapter 7 Exercise 7.4 Question 6:

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 the area of ∆ABC

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. Therefore, the ratio between the areas of ∆ADE and ΔABC = (1/4)² = 1/16