Question

find the area of hexagon whose vertices are taken (5,0)(4,2)(1,3)(-2,2)(-3,-1)(0,-4) is

Answer 1

The area of hexagon whose vertices are taken (5,0), (4,2), (1,3), (-2,2), (-3,-1), (0,-4) is "zero(0)".

Step-by-step explanation:

Here,

$x_{1} = 5, y_{1} = 0, x_{2}= 4, y_{2} = 2, x_{3} = 1, y_{3} = 3, x_{4} = - 2, y_{4} = 2, x_{5} = - 3, [y_{5} = - 1$ and  $x_{6} = 0, y_{6} = 4$

We know that,

Area of hexagon = $\dfrac{1}{2}$ × {5(2 - 3) + 4(3 - 2) + 1(2 + 1) - 2(-1 - 4) - 3(4 - 0) + 0(0 - 2)}

= $\dfrac{1}{2}$ × {- 5 + 4 + 3 + 10 -12 + 0}

= $\dfrac{1}{2}$ × {- 5 + 4 + 3 + 10 - 12 + 0}

= $\dfrac{1}{2}$ × {0}

= 0

Hence,  the area of hexagon whose vertices are taken (5,0), (4,2), (1,3), (-2,2), (-3,-1) and (0,-4) is "zero(0)".