6.3 Areas of Polygons

Posted on April 21, 2020 by user

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:

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