Find the area of a parallelogram ABCD if three of its vertices are A(2

1.	Find the area of a parallelogram ABCD if three of its vertices are A(2, 4), B (2 + \(\sqrt3\) , 5) and C(2, 6).
Answer» It is given that A(2, 4), B(2 + \(\sqrt3\), 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) = \(\frac{1}2\) \|x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)\| Area of □ABCD = 2 × Area of ∆ABC Area of ∆ABC = \(\frac{1}2\) \|2(5 - 6)+ (2 + \(\sqrt3\))(6 - 4)+2(4 - 5)\| = \(\frac{1}2\) \|-2 + 4 + 2\(\sqrt3\) - 2\| = \(\frac{1}2\) × 2\(\sqrt3\) = \(\sqrt3\) sq. units ∴ Area of □ABCD = 2 × \(\sqrt3\) = 2\(\sqrt3\) sq. units Hence, the area of given parallelogram is 2\(\sqrt3\) sq. units

Find the area of a parallelogram ABCD if three of its vertices are A(2, 4), B (2 + \(\sqrt3\) , 5) and C(2, 6).

Answer»

It is given that A(2, 4), B(2 + \(\sqrt3\), 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)

= \(\frac{1}2\) |x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)|

Area of □ABCD = 2 × Area of ∆ABC

Area of ∆ABC = \(\frac{1}2\) |2(5 - 6)+ (2 + \(\sqrt3\))(6 - 4)+2(4 - 5)|

= \(\frac{1}2\) |-2 + 4 + 2\(\sqrt3\) - 2|

= \(\frac{1}2\) × 2\(\sqrt3\) = \(\sqrt3\) sq. units

∴ Area of □ABCD = 2 × \(\sqrt3\) = 2\(\sqrt3\) sq. units

Hence,

the area of given parallelogram is 2\(\sqrt3\) sq. units

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