Question

Question 4
Find the area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, -5), (3, -2) and (2, 3).

Answer 1

Let the vertices of the quadrilateral be A (- 4, - 2), B ( - 3, - 5), C (3, - 2), and D (2, 3).
Join AC to form two triangles$\Delta ABCand\Delta ACD.$
Area of a triangle
=$\frac{1}{2}\left\{x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right\}$
Area of $\Delta ABC$
$=\frac{1}{2}\left[\right(-4)\left\{\right(-5)-(-2\left)\right\}+(-3)\left\{\right(-2)-(-2\left)\right\}$
$+3\left\{\right(-2)-(-5\left)\right\}]$
$=\frac{1}{2}(12+0+9)$
$=\frac{21}{2}squareunits$
Area of $\Delta ACD$
$=\frac{1}{2}\left[\right(-4)\left\{\right(-2)-(3\left)\right\}+3\left\{\right(3)-(-2\left)\right\}$
$+2\left\{\right(-2)-(-2\left)\right\}]$
$=\frac{1}{2}(20+15+0)$
$=\frac{35}{2}squareunits$
Area of $\Delta ABCD$
$=Areaof\Delta ABC+Areaof\Delta ACD$
$=(\frac{21}{2}+\frac{35}{2})$ square units = 28 square units