6.3 Areas of Polygons – user's Blog!

6.3 Areas of Polygons

(A) Area of Triangle

Area of Triangle

1. When the given points are taken in an anticlockwise direction the result is positive; taken in a clockwise direction the result is negative. The answer for the area must be given as a positive value.

Example 1:

Calculate the area of ∆ ABC with the vertices A (-5, 5), B (-2, -4),

C (4, -1).

Solution:

\begin{array}{l} Area of △ A B C \\ = \frac{1}{2} | \begin{array}{l} - 5 - 2 4 \\ 5 - 4 - 1 \end{array} \begin{array}{l} - 5 \\ 5 \end{array} | \\ = \frac{1}{2} | (- 5) (- 4) + (- 2) (- 1) + (4) (5) \\ - (5) (- 2) - (- 4) (4) - (- 1) (- 5) | \\ = \frac{1}{2} | 20 + 2 + 20 + 10 + 16 - 5 | \\ = \frac{1}{2} | 63 | \\ = 31 \frac{1}{2} {unit}^{2} \end{array}

(B) Area of Quadrilateral

Area of Quadrilateral

(C) If the points A, B and C are collinear, then the area of ∆ ABC is 0.

Example 2:

Find the values of k if the points P (2, 1), Q (6, k) and R (3k,

\frac{9}{2}

) are collinear.

Solution:

\begin{array}{l} Area of △ P Q R = 0 \\ \frac{1}{2} | \begin{array}{l} 2 63 k \\ 1 k \frac{9}{2} \end{array} \begin{array}{l} 2 \\ 1 \end{array} | = 0 \\ \frac{1}{2} | 2 k + 27 + 3 k - 6 - 3 k^{2} - 9 | = 0 \\ | - 3 k^{2} + 5 k + 12 | = 0 \\ 3 k^{2} - 5 k - 12 = 0 \\ (3 k + 4) (k - 3) = 0 \\ k = - \frac{4}{3} or 3 \end{array}