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Find the area of the pentagon of the five vertices $(1,2), (4,1), (5,3), (3,7), (2,6)$ . Please, use the way of using determinant.

My idea is to cut the pentagon into some triangles, then calculate each triangle, then sum them. I wonder if there is some other way to directly calculate it using a bigger matrix calculating its determinant?

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Your idea to cut into triangles should work. –  Joel Cohen Dec 24 '12 at 21:10
    
the question if my idea is the right way to do it. If there is any other ways of solving this ? –  user53800 Dec 24 '12 at 21:12
    
I have another question is that, if any operator that satisfies the universal properties of the axioms of determinant, then we can use determinant to calculate this operator, like the volume of vectors. –  user53800 Dec 24 '12 at 21:17
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I don't know an answer. However, if your polygon is a simple convex polygon and $v_1=v_{n+1},v_2,\ldots,v_n\in\mathbb{R}^2$ are its vertices arranged in anticlockwise order, then the area of the polygon can be simplified as $\frac12\sum_{i=1}^n \det(v_i, v_{i+1})$. –  user1551 Dec 25 '12 at 6:04

1 Answer 1

If you label your points such that they form a clockwise rotation about a central point you can use the following shoelace formula:
$$A =\frac12\bigg| \sum_{i=1}^{n} x_iy_{i+1} - x_{i+1}y_i \bigg| = \frac12\bigg|\sum_{i=1}^n\det\bigg(\begin{matrix} x_i & x_{i+1} \\ y_i & y_{i+1} \end{matrix}\bigg)\bigg|$$

Where $x_0 = x_n,\ x_{n+1} = x_1$ and similarly, $y_0 = y_n,\ y_{n+1} = y_1$

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