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  4. Find the area of the pentagon whose vertices taken in order are (0, 4), (3, 0), (6, 1), (7, 5) and (4, 9).

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Q. Find the area of the pentagon whose vertices taken in order are (0, 4), (3, 0), (6, 1), (7, 5) and (4, 9).

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Solution:

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$A _1=\frac{1}{2}\begin{vmatrix}0 & 4 & 1 \\ 3 & 0 & 1 \\ 4 & 9 & 1\end{vmatrix}=\frac{1}{2}|-4(-1)+1(27)|=\frac{31}{2}$
$A _2=\frac{1}{2}\begin{vmatrix}3 & 0 & 1 \\ 4 & 9 & 1 \\ 6 & 1 & 1\end{vmatrix}=\frac{1}{2}|3 \cdot(9-1)+1 \cdot(4-54)|=\frac{1}{2}|24-50|=13$
$ A _3=\frac{1}{2}\begin{vmatrix}4 & 9 & 1 \\ 7 & 5 & 1 \\ 6 & 1 & 1\end{vmatrix}=\frac{1}{2}|4 \times 4-9(1)+1(7-30)| $
$ =\frac{1}{2}|16-9-23|=\frac{16}{2}=8 $
$ \therefore \quad \text { Area of pentagon }=\frac{31}{2}+13+8=\frac{73}{2}=36.5 \text { sq. units }$