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). [CBSE 2013]

Answer 1

It is given that A(2, 4), B(2 + $\sqrt{3}$, 5) and C(2, 6) are the vertices of the parallelogram ABCD.



We know that the diagonal of a parallelogram divides it into two triangles having equal area.

∴ Area of the parallogram ABCD = 2 × Area of the ∆ABC

Now,
$$\begin{gathered} \mathrm{ar}\left(∆\mathrm{ABC}\right)=\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right| \\ =\frac{1}{2}\left|2\left(5-6\right)+\left(2+\sqrt{3}\right)\left(6-4\right)+2\left(4-5\right)\right| \\ =\frac{1}{2}\left|-2+4+2\sqrt{3}-2\right| \\ =\frac{1}{2}\times 2\sqrt{3} \\ =\sqrt{3}\mathrm{square}\mathrm{units} \end{gathered}$$

∴ Area of the parallogram ABCD = 2 × Area of the ∆ABC = 2 × $\sqrt{3}$ = 2$\sqrt{3}$ square units

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