# 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.

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.

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 (∆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

**☛ Check: **Class 10 Maths NCERT Solutions Chapter 7

**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

**Summary:**

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. The area of ΔADE is 15/2 square units and that of ΔABC is 15/32 square units. Therefore, the ratio between the areas of ∆ADE and ΔABC is 1:16.

**☛ Related Questions:**

- Determine the ratio in which the line 2x + y - 4 = 0 divides the line segment joining the points A (2, - 2) and B (3, 7).
- Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.
- Find the centre of a circle passing through the points (6, - 6), (3, - 7) and (3, 3).
- The two opposite vertices of a square are (- 1, 2) and (3, 2). Find the coordinates of the other two vertices.

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