Find the area of a parallelogram ABCD if three of its vertices are A (2, 4), B (2 + √3, 5) and C (2, 6).

Given: Three vertices of a triangle are A (2, 4), B (2 + √3, 5) and C (2, 6).

To find: Area of a parallelogram.

Formula Used:

Area of the triangle having vertices (x₁, y₁), (x₂, y₂) and (x₃, y₃)
= 1/2 |x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂) |

Explanation:

It is given that A (2, 4), B (2 + √3, 5) and C (2, 6) are the vertices of the parallelogram ABCD.

Area of the triangle having vertices (x₁,y₁), (x₂,y₂) and (x₃,y₃)
= 1/2 |x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|

Area of □ABCD = 2 × Area of ∆ABC

Area of ∆ABC = 1/2 |2(5 - 6) + (2 + √3 )(6 - 4) + 2(4 - 5) |

= 1/2 | - 2 + 4 + 2√3 - 2|

= 1/2 × 2√3 = √3 sq. units

∴ Area of □ABCD = 2 × √3 = 2√3 sq. units

Hence, the area of given parallelogram is 2√3 sq. units