Question

Find the area of a parallelogram $ABCD$ if three of its vertices are $A(2,4),B(2+\sqrt{3},5)$ and $C(2,6)$

Answer 1

We know that the diagonal of a parallelogram divides the parallelogram into $2$ equal triangles.

So, Area of parallelogram,

$ABCD=2\times$ area of $\Delta ABC$ ......... (1)

We know that,

Area of $\Delta ABC$ with vertices $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is

$A=\frac{1}{2}\times [x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

So, Area of $\Delta ABC$ with vertices $[A(2,4),B(2+\sqrt{3},5),C(2,6)]$ is

$A=\frac{1}{2}\times [2(5-6)+(2+\sqrt{3})(6-4)+2(4-5)]$

$A=\frac{1}{2}\times [-2+2(2+\sqrt{3})-2]$

$A=\frac{1}{2}\times [-2+4+2\sqrt{3}-2]$

$A=\sqrt{3}$

Put this value in equation (1), we get,

Area of parallelogram,

$ABCD=2\times$ area of $\Delta ABC$

$ABCD=2\sqrt{3}$ sq units

Hence, area of the parallelogram is $2\sqrt{3}$ sq units.