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A (6, 1), B (8, 2) and C (9, 4) are three vertices of a parallelogram ABCD. If E is the midpoint of DC, find the area of ∆ ADE

Solution:

Given, ABCD is a parallelogram.

The vertices of the parallelogram are A(6, 1) B(8, 2) and C(9, 4).

E is the midpoint of DC.

We have to find the area of the triangle ADE.

Let the fourth vertex of the parallelogram be (x, y).

We know that the diagonals of a parallelogram bisect each other.

i.e., midpoint of AC = midpoint of BD

The coordinates of the mid-point of the line segment joining the points P (x₁ , y₁) and Q (x₂ , y₂) are [(x₁ + x₂)/2, (y₁ + y₂)/2]

Midpoint of A(6, 1) and C(9, 4) = [(6 + 9)/2, (1 + 4)/2]

= [15/2, 5/2]

Midpoint of B(8, 2) and D(x, y) = [(8 + x)/2, (2 + y)/2]

Now, [(8 + x)/2, (2 + y)/2] = [15/2, 5/2]

(8 + x)/2 = 15/2

8 + x = 15

x = 15 - 8

x = 7

(2 + y)/2 = 5/2

2 + y = 5

y = 5 - 2

y = 3

Therefore, the coordinates of the fourth vertex D = (7, 3)

Midpoint of D(7, 3) and C(9, 4) = E

E = [(7 + 9)/2, (3 + 4)/2]

E = [16/2, 7/2]

E = [8, 7/2]

The area of a triangle with vertices A (x₁ , y₁) , B (x₂ , y₂) and C (x₃ , y₃) is

1/2[x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)]

Here, (x₁ , y₁) = (6, 1) (x₂ , y₂) = (7, 3) and (x₃ , y₃) = (8, 7/2) = (8, 3.5)

Area of triangle ADE = 1/2[6(3 - 3.5) + 7(3.5 - 1) + 8(1 - 3)]

= 1/2[6(-0.5) + 7(2.5) + 8(-2)]

= 1/2[-3 + 17.5 - 16]

= 1/2[-19 + 17.5]

= 1/2[-1.5]

= -0.75

Area cannot be negative

So, area = 0.75 square units.

Therefore, the area of the triangle ADE is 0.75 square units.

✦ Try This:The fourth vertex D of a parallelogram ABCD whose three vertices are A(-2, 3), B(6, 7) and C(8, 3) is

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 7

NCERT Exemplar Class 10 Maths Exercise 7.4 Problem 2

Summary:

A (6, 1), B (8, 2) and C (9, 4) are three vertices of a parallelogram ABCD. If E is the midpoint of DC, the area of ∆ ADE is 0.75 square units

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