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

Diagonal of a parallelogram divides the parallelogram in two triangles of equal areas.
So, if we find the area of one triangle and then double it we'll get the area of the parallelogram.

Area of a triangle whose sides are $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is given as,

Area of triangle = $\frac{1}{2}[(x_1{y}_2-x_2{y}_1)+(x_2{y}_3-x_3{y}_2)+(x_3{y}_1-x_1{y}_3)]$

$=\frac{1}{2}[10-8-4\sqrt{3}+12+6\sqrt{3}-10+8-12]$

$\Rightarrow$ $\frac{1}{2}\left[2\sqrt{3}\right]=\sqrt{3}$
$\because$ Area of parallelogram = $2$ (Area of triangle)

$\therefore$ Area of parallelogram = $2\sqrt{3}$