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

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,

\[\text{ ar} \left( ∆ ABC \right) = \frac{1}{2}\left| x_1 \left( y_2 - y_3 \right) + x_2 \left( y_3 - y_1 \right) + x_3 \left( y_1 - y_2 \right) \right|\]
\[ = \frac{1}{2}\left| 2\left( 5 - 6 \right) + \left( 2 + \sqrt{3} \right)\left( 6 - 4 \right) + 2\left( 4 - 5 \right) \right|\]
\[ = \frac{1}{2}\left| - 2 + 4 + 2\sqrt{3} - 2 \right|\]
\[ = \frac{1}{2} \times 2\sqrt{3}\]
\[ = \sqrt{3}\text{ square units } \]

∴ 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 .