Area

Area of Triangle

The area of a triangle with vertices at \(A(x_1, y_1)\), \(B(x_2, y_2)\), and \(C(x_3, y_3)\) is given by:

\[ \text{Area}(\triangle ABC) = \frac{1}{2} \left| \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \right| \]

Proof:

Consider trapezoids \(ABML\), \(ACNL\), and \(BCNM\) with bases on the \(x\)-axis and vertices \(A\), \(B\), and \(C\) having coordinates \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\), respectively. The vertical lines \(AM\), \(BN\), and \(CL\) are drawn from vertices \(A\), \(B\), and \(C\) perpendicular to the \(x\)-axis, meeting at points \(M\), \(N\), and \(L\) on the \(x\)-axis. These perpendiculars define the heights of the respective trapezoids.

The area of trapezoid \(ABML\) is given by the average of the lengths of the parallel sides (\(y_1\) and \(y_2\)) multiplied by the distance between them (\(x_2 - x_1\)):

\[ \text{Area}_{ABML} = \frac{1}{2} (y_1 + y_2)(x_2 - x_1) \]

Similarly, the area of trapezoid \(ACNL\) is:

\[ \text{Area}_{ACNL} = \frac{1}{2} (y_1 + y_3)(x_3 - x_1) \]

And the area of trapezoid \(BCNM\) is:

\[ \text{Area}_{BCNM} = \frac{1}{2} (y_2 + y_3)(x_3 - x_2) \]

As can be seen in the figure, the area of triangle \(ABC\) is the sum of the areas of trapezoids \(ABML\) and \(ACNL\) minus the area of trapezoid \(BCNM\):

\[ \text{Area}_{ABC} = \text{Area}_{ABML} + \text{Area}_{ACNL} - \text{Area}_{BCNM} \]

Substituting the area expressions and simplifying, the resulting area of triangle \(ABC\) is:

\[ \text{Area}_{ABC} = \frac{1}{2} [y_1(x_2 - x_3) + y_2(x_3 - x_1) + y_3(x_1 - x_2)] \]

The absolute value is taken since the area should be positive:

\[ \text{Area}_{ABC} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| \]

This can be written in determinant form as:

\[ \text{Area} = \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \]

Area of polygon

Area Determination of Polygons via Triangulation

Let a polygon \( P \) with vertices \( V_1, V_2, \ldots, V_n \) be given. Then the area \( A(P) \) of polygon \( P \) is equal to the sum of the areas of the triangles formed by connecting a chosen vertex \( V_i \) to all non-adjacent vertices of \( P \).

Proof:

Begin with selecting vertex \( V_i \) from polygon \( P \). This vertex acts as a common vertex for the triangulation process. Construct segments from \( V_i \) to all non-adjacent vertices \( V_{i+2}, \ldots, V_{i-1} \) (indices considered modulo \( n \)), thereby dividing \( P \) into \( n-2 \) triangles.

Denote the area of a triangle with vertices \( V_i, V_j, V_k \) as \( A(V_iV_jV_k) \). The area of \( P \), which can be represented as the sum of the areas of these triangles, is given by:

\[ A(P) = A(V_iV_{i+1}V_{i+2}) + A(V_iV_{i+2}V_{i+3}) + \cdots + A(V_iV_{i-1}V_n) \]

By the principle of additivity of areas, the sum of the areas of the triangles equals the area of \( P \). Since these triangles are non-overlapping and cover \( P \) entirely, no area is counted twice, and no region within \( P \) is omitted. Hence, the area of polygon \( P \) is the sum of the areas of the triangulated sections.

This completes the proof.

Example

To find the area of a simple polygon such as the one depicted, which is a pentagon, one can employ the triangulation method. This process involves creating non-overlapping triangles that together compose the original polygon and calculating the sum of their areas.

For the given pentagon \( ABCDE \), select a vertex, say \( A \), and draw diagonals \( AC \) and \( AD \). These diagonals divide the pentagon into three triangles: \( \triangle ABC \), \( \triangle ACD \), and \( \triangle ADE \).

If the Cartesian coordinates of the vertices are known, apply the determinant formula to each triangle:

\[ \text{Area}(\triangle XYZ) = \frac{1}{2} \left| \begin{vmatrix} x_X & y_X & 1 \\ x_Y & y_Y & 1 \\ x_Z & y_Z & 1 \end{vmatrix} \right| \]

The total area of the pentagon is then the sum of the areas of triangles \( \triangle ABC \), \( \triangle ACD \), and \( \triangle ADE \):

\[ \text{Area}(ABCDE) = \text{Area}(\triangle ABC) + \text{Area}(\triangle ACD) + \text{Area}(\triangle ADE) \]

Shoelace Formula

To apply the Shoelace Formula to a polygon with vertices \(A_1, A_2, ..., A_n\), list the coordinates in a tabular form with two columns, arranging the abscissas (x-coordinates) in the first column and the ordinates (y-coordinates) in the second column. Include an additional row at the bottom with the coordinates of the first vertex \(A_1\) repeated to close the loop.

The Shoelace Formula is then executed by cross-multiplying the entries diagonally and summing the products. Specifically, for each row, multiply the x-coordinate by the y-coordinate of the next row and sum these values. Then, subtract the sum of the products obtained by multiplying the y-coordinate of each row by the x-coordinate of the next row.

Formally, for vertices \( (x_1, y_1), (x_2, y_2), ..., (x_n, y_n) \), the area \( A \) is given by the absolute value of:

\[ A = \frac{1}{2} \left|\sum_{i=1}^{n} (x_i y_{i+1} - x_{i+1} y_i)\right| \]

where \( x_{n+1} = x_1 \) and \( y_{n+1} = y_1 \) to ensure the loop is closed.

This expands to:

\[ A = \frac{1}{2} \left| (x_1 y_2 + x_2 y_3 + ... + x_n y_1) - (y_1 x_2 + y_2 x_3 + ... + y_n x_1) \right| \]