Question

The area of a quadrilateral whose vertices taken in order are (–4, –2), (–3, –5), (3, –2) and (2, 3) is _______.

• A. 28

Answer 1

The correct option is A 28

Consider the points
$A(-4,-2),B(-3,-5),C(3,-2)$ and $D(2,3).A\left(△\right)=\frac{1}{2}\times |x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$

$\therefore$ Area of $△ABC$

$=\frac{1}{2}|(-4)(-5+2)-3(-2+2)+3(-2+5)|=\frac{1}{2}|20-8-6+15|=\frac{21}{2}A\left(△ABC\right)=10.5\text{sq. units}$

Similarly, area of $△ACD$

$=\frac{1}{2}|(-4)(-2-3)+3(3+2)+2(-2+2)|=\frac{1}{2}|20+15|=\frac{35}{2}A\left(△ACD\right)=17.5\text{sq. units}$

$\text{Area of quadrilateral}ABCD=A\left(△ABC\right)+A\left(△ACD\right)=(10.5+17.5)=28\text{sq. units}$