Question

Hexagon ABCDEF has vertices A(-2, 4), B(0,4), C(2,1) D(5,1) E(5,2) and F(-2,2). How do you sketch the figure on a coordinate plane. What is the area of the hexagon?

Answer 1

`\text{Area }=18\ \unit^2`

Explanation:

The hexagon ABCDEF with the vertices `A(-2, 4)`, `B(0,4)`, `C(2, 1)`, `D(5, 1)`, `E(5, 2)` & `F(-2, 2)` can be divided into four triangles `\Delta ABC. \Delta ACD, \Delta ADE` & `\Delta AEF`

The area `\Delta_1` of `\Delta ABC` with vertices `A(-2, 4)\equiv(x_1, y_1), B(0, 4)\equiv(x_2, y_2)` & `C(2, 1)\equiv(x_3, y_3)` is given by following formula

`\Delta_1=1/2|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|`

`=1/2|-2(4-1)+0(1-4)+2(4-4)|`

`=3`

Similarly, the area `\Delta_2` of `\Delta ACD` with vertices `A(-2, 4), C(2, 1)` & `D(5, 1)` is given as follows

`\Delta_2=1/2|-2(1-1)+2(1-4)+5(4-1)|`

`=4.5`

Similarly, the area `\Delta_3` of `\Delta ADE` with vertices `A(-2, 4), D(5, 1)` & `E(5, 2)` is given as follows

`\Delta_3=1/2|-2(1-2)+5(2-4)+5(4-1)|`

`=3.5`

Similarly, the area `\Delta_4` of `\Delta AEF` with vertices `A(-2, 4), E(5, 2)` & `F(-2, 2)` is given as follows

`\Delta_4=1/2|-2(2-2)+5(2-4)-2(4-2)|`

`=7`

hence, the area of hexagon ABCDEF is equal to the sum of areas of above triangles, given as

`\Delta_1+\Delta_2+\Delta_3+\Delta_4`

`=3+4.5+3.5+7`

`=18`