Question

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

Answer 1

The vertices of a ΙΙgm (parallelogram) are A(2,4) ,B(2+-√3,5) ,and C(2,6)
Area of ΙΙgm ABCD = ?
Solution:
We know that the diagonal of a ΙΙgm divides the ΙΙgm into two equal triangles.
So, Area of ΙΙgm ABCD = 2 (Area of ΔABC)   ......... (1)
We know that, 
Area of ΔABC with vertices (x₁,y₁) , (x₂,y₂) and (x₃,y₃) is
A = 1/2 [x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂) ]
So, Area of ΔABC with vertices A(2,4) ,B(2+-√3,5) ,and C(2,6) is
A = 1/2 [2 (5-6) + (2+-√3) (6-4) + 2 (4-5) ]
A = 1/2 [-2 + 2(2+-√3) -2]
A = 1/2 (-4 + 4 +-2√3)
A = 1/2(+-2√3)
A = +-√3
Put this value in equation (1), we get:
Area of ΙΙgm ABCD = 2 (Area of ΔABC)
Area of ΙΙgm ABCD = 2 (+-√3)
Area of ΙΙgm ABCD = +-2√3
Hence, are of a parallelogram = +-2√3
This is the required answer.
Thanks.

Answer 2

Thank you for asking this question. Here is your answer:

If there is a diagonal in a parallelogram it bisects into two equal triangles:

which will be:

ar(llgm ABCD) = 2ar (ΔABC) -- (this will be an equation 1)

The area of triagnle with the vertices (x₁,y₁),(x₂,y₂) and (x₃,y₃)

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

Area of ΔABC = 1/2 [2(5-6) + (2+√3) (6-4) + 2(4-5)]

= 1/2 [-2 + 4 + 2√3-2]

= 1/2 (2√3)

= √3

so ar(llgm ABCD) = 2ar (ΔABC)

= ar(llgm ABCD) = 2 x √3  

= 2√3 is the final answer for this question.