Question

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

Answer 1

Answer: 36.5

$A_1=\frac{1}{2}\left|\begin{array}{ccc}0& 4& 1\\ 3& 0& 1\\ 4& 9& 1\end{array}\right|=\frac{1}{2}|-4(-1)+1(27)|=\frac{31}{2}$
$A_2=\frac{1}{2}\left|\begin{array}{ccc}3& 0& 1\\ 4& 9& 1\\ 6& 1& 1\end{array}\right|=\frac{1}{2}|3\cdot (9-1)+1\cdot (4-54)|=\frac{1}{2}|24-50|=13$
$A_3=\frac{1}{2}\left|\begin{array}{ccc}4& 9& 1\\ 7& 5& 1\\ 6& 1& 1\end{array}\right|=\frac{1}{2}|4\times 4-9(1)+1(7-30)|$
$=\frac{1}{2}|16-9-23|=\frac{16}{2}=8$
$\therefore \text{ Area of pentagon }=\frac{31}{2}+13+8=\frac{73}{2}=36.5\text{ sq. units }$