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Find the area of a parallelogram $ABCD$ if three of its vertices are $A (2, 4), B (2 + \sqrt3, 5)$ and $C (2, 6)$.
Given:
Three vertices of a parallelogram are $A (2, 4), B (2 + \sqrt3, 5)$ and $C (2, 6)$.
To do:
We have to find the area of the parallelogram.
Solution:
Join $A$ and $C$ to get two triangles $ABC$ and $ADC$.
We know that,
A diagonal bisects a triangle into two triangles of equal area.
This implies,
Area of parallelogram $ABCD=$ Area of triangle $ABC+$ Area of triangle $ADC$.
$=2\times$ Area of triangle $ABC$
We know that,
Area of a triangle with vertices $(x_1,y_1), (x_2,y_2), (x_3,y_3)$ is given by,
Area of $\Delta=\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$
Therefore,
Area of triangle \( ABC=\frac{1}{2}[2(5-6)+(2+\sqrt3)(6-4)+2(4-5)] \)
\( =\frac{1}{2}[2(-1)+(2+\sqrt3)(2)+2(-1)] \)
\( =\frac{1}{2}[-2+4+2\sqrt3-2] \)
\( =\frac{1}{2} \times (2\sqrt3) \)
\( =\sqrt3 \) sq. units.
Therefore,
The area of the parallelogram $ABCD=2\times \sqrt3=2\sqrt3$ sq. units.
The area of the parallelogram $ABCD$ is $2\sqrt3$ sq. units.
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